Recently I am writing a products catalog for my company. I use the dropdown list to provide a filter to a grid views product list that has pagination. The filter is work fine if products not more than 1 page. When more then one page of search result, the browser will show all products same as no filter applied. Below is my code of the index.php. Please help because I already stuck in here for a week. Thank in advance.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<?php session_start();
if (!isset($_SESSION['user_name']))
{
echo "<script>window.open('../login.php','_self')</script>";
}
$user_name=isset($_SESSION['user_name'])?$_SESSION['user_name']:'';
?>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.7.0/css/font-awesome.min.css">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title></title>
<link href="css/gallery.css" rel="stylesheet" type="text/css" />
<script src="js/jquery-3.2.1.min.js"></script>
<meta name="theme-color" content="#444" />
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<?php
include_once('../database/db_conection.php');//Include the database connection
?>
<div class="container">
<div class="col-md-2"> <br/>
<i class="fa fa-home" style="font-size:28px"></i>Home <br/>
<br/>
<br/>
<form action="buckle.php'" method="post" >
<div class="cf_header5">MATERIAL<br/>
(e.g.Brass, Alloy...)</div>
<select name="material" style="width:150px; height:25px;font-size:16px">
<option value="">Please select ....</option>
';
<?php
$sql_ml = "SELECT distinct material FROM products where material !='' order by material";
$records_ml=mysqli_query($dbcon,$sql_ml);
while ($row_ml=mysqli_fetch_array($records_ml)){
echo '<option value="'.$row_ml['material'].'">'.$row_ml['material'].'</option>';
}
?>
</select>
<br/>
<br/>
<div class="cf_header5">DESCRIPTION</div>
<select name="description" style="width:150px; height:25px;font-size:16px">
<option value="">Please select ....</option>
';
<?php
$sql_dn = "SELECT distinct catagory FROM products where catagory !='' order by catagory";
$records_dn=mysqli_query($dbcon,$sql_dn);
while ($row_dn=mysqli_fetch_array($records_dn)){
echo '<option value="'.$row_dn['catagory'].'">'.$row_dn['catagory'].'</option>';
}
?>
</select>
<br/>
<br/>
<div class="cf_header5">SIZE</div>
<select name="size" style="width:150px; height:25px;font-size:16px">
<option value="">Please select ....</option>
<?php
$sql_se = "SELECT distinct size FROM products where size !='' order by size";
$records_se=mysqli_query($dbcon,$sql_se);
while ($row_se=mysqli_fetch_array($records_se)){
echo '<option value="'.$row_se['size'].'">'.$row_se['size'].'</option>';
}
?>
</select>
<br/>
<br/>
<input type="submit" name="submit" value="Search" />
</form>
<?php //get the value from dropdown list and stored it to a variable
$material=isset($_POST['material']) ? $_POST['material'] : "";
$size=isset($_POST['size']) ? $_POST['size'] : "";
$description=isset($_POST['description']) ? $_POST['description'] : "";
?>
</div>
<div class="col-md-10">
<div id="main">
<header>
<div class="logo-outer">
<div class="logo"> <img src="../images/logo4.png" /> </div>
<?php
//////SET UP THE TOTAL images PER PAGE & CALCULATIONS:
$per_page = 12;// Number of images per page, change for a different number of images per page
// Get the page and offset value:
if (isset($_GET['page'])) {
$page = $_GET['page'] - 1;
$offset = $page * $per_page;
}
else {
$page = 0;
$offset = 0;
}
//Display no of items in cart
echo '<a href="viewcart.php?page=';echo $page+1;echo '&type=buckle';echo'">';
echo "You have";
$t_sql="select count(product_id) as nocart from cart where user_name='$user_name'";
$t_run=mysqli_query($dbcon,$t_sql);
$t_row=mysqli_fetch_array($t_run);
$cart_count=$t_row['nocart'];
echo ' ';
echo '<span class="badge" >';echo $cart_count; echo '</span>';
echo ' items on list</a>';
//end of display number of items in cart
?>
</div>
</div>
</header>
<div class="margin">
<div id="pagination"><!-- #pagination start -->
<?php
$images_sql="select count(id) from products where product_type='buckle'";
if(isset($_POST['material'])and $material!='')
{$images_sql=$images_sql." and material='$material'";}
if(isset($_POST['description'])and $description!='')
{$images_sql=$images_sql." and catagory='$description'";}
if(isset($_POST['size'])and $size!='')
{$images_sql=$images_sql." and size='$size'";}
$images_sql=$images_sql." ORDER by id ASC";
$result_images = mysqli_query($dbcon, $images_sql);
$row_images = mysqli_fetch_array($result_images);
$total_images = $row_images[0];
// Calculate the number of pages:
if ($total_images > $per_page) {//If there is more than one page
$pages_total = ceil($total_images / $per_page);
$page_up = $page + 2;
$page_down = $page;
$display ='';//leave the display variable empty so it doesn't hide anything
}
else {//Else if there is only one page
$pages = 1;
$pages_total = 1;
$display = ' class="display-none"';//class to hide page count and buttons if only one page
}
////// DISPLAY THE PAGE COUNT AND BUTTONS:
echo '<h2'.$display.'>Page '; echo $page + 1 .' of '.$pages_total;//Page out of total pages
$i = 1;//Set the $i counting variable to 1
echo '<div id="pageNav"'.$display.'>';// $display variable will do nothing if more than one page
// Show the page buttons:
if ($page) {
echo '<button><<</button>';//Button for first page [<<]
echo '<button><</button>';//Button for previous page [<]
}
for ($i=1;$i<=$pages_total;$i++) {
if(($i==$page+1)) {
echo '<button class="active">'.$i.'</button>';//Button for active page, underlined using 'active' class
}
//In this next if statement, calculate how many buttonso show.
if(($i!=$page+1)&&($i<=$page+3)&&($i>=$page-1)) {//This is set for two below and two above the current page
echo '<button>'.$i.'</button>'; }
}
if (($page + 1) != $pages_total) {
echo '<button>></button>';//Button for next page [>]
echo '<button>>></button>';//Button for last page [>>]
}
echo '</div>';// #pageNav end
?>
</div>
<div id="gallery"><!-- Gallery start-->
<?php
// DISPLAY THE images:
//Define the SQL statement based on the dropdown list
if ($total_images>0)
{
$img_sql="select * from products where product_type='buckle'";
if(isset($_POST['material'])and $material!='')
{$img_sql=$img_sql." and material='$material'";}
if(isset($_POST['description'])and $description!='')
{$img_sql=$img_sql." and catagory='$description'";}
if(isset($_POST['size'])and $size!='')
{$img_sql=$img_sql." and size='$size'";}
$img_sql=$img_sql." ORDER by id ASC LIMIT $offset, $per_page";
$result = mysqli_query($dbcon, $img_sql);
while($row = mysqli_fetch_array($result)) {//Open the while array loop
//Define the image variable:
$image=$row['images'];
$product_id=$row['product_id'];
$product_type=$row['product_type'];
$size=$row['size'];
$material=$row['material'];
$description=$row['catagory'];
echo '<div class="img-container">';
echo '<div class="img">';
echo '<img src="images/'.$image.'">';
echo '</a>';
echo '</div>';
echo $product_id;echo '    ';
echo 'Add to List';
echo '<br/>Size:';echo $size;
echo '    Material: '; echo $material ;
echo '<br/>Description: ';echo $description;
echo '</div>';// .img-container end
}//Close the while array loop
}
else
{
echo'<br/><br>';
echo '<h1>Sorry! No result is matched, please search again.</h1>';
echo'<br/><br>';
}
?>
</div>
<!-- Gallery end-->
<div class="clearfix"></div>
</div>
</div>
<!-- #main end -->
</div>
</body>
</html>
Your logic seems OK but your problem is when you navigate trough pages you make a new call of the document, which doesn't include variables like filters but the current page ($_GET['page']).
Either you send filters as variables trough URL as well, which isn't the cleanest way to do table pagination.
Or you gather results in a javascript array, and then on click on a pagination button display the previous/next rows. By doing this you also avoid refreshing the whole document every time, which isn't necessary.
I wonder why the 2nd output $devNO gives me bool(false) which is empty. i have no idea about it. Am I wrong at the mysql_query() part?
$developer = $_POST['dev'];
$platform = $_POST['plat'];
$genre = $_POST['gen'];
var_dump($developer);
echo "<br>";
$devNO = mysql_query("SELECT No FROM developer WHERE Developer = $developer");
$platNO = mysql_query("SELECT No FROM platform WHERE Platform = $platform");
$genNO = mysql_query("SELECT No FROM genre WHERE Genre = $genre");
var_dump($devNO);
Here's my output:
Here I will show my full code for "games.php":
<!DOCTYPE html>
<html lang="en" xmlns="http://www.w3.org/1999/html">
<html>
<head lang="en">
<meta charset="utf-8" />
<title>Game List</title>
<link rel="stylesheet" href="css.css">
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.1/css/bootstrap.min.css">
<!-- Optional theme -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.1/css/bootstrap-theme.min.css">
<style>
legend {font-family: "Palatino Linotype", "Book Antiqua", Palatino, serif}
</style>
</head>
<body>
<div id="big_wrapper">
<header id="top_header">
<h1>Dandy's Game Library</h1>
</header>
<nav id="top_menu">
<ul>
<li><strong></stron>Home</strong></li>
<li><strong>Game List</strong></li>
</ul>
</nav>
<div id="game_wrapper">
<section id="filter">
<form action="games.php" method="post" name="search_form">
<fieldset>
<legend><h3><strong>Search</strong></h3></legend>
<strong>Developer</strong><br>
<select name="dev">
<option value="">--Select--</option>
<?php
include("dbConnection.php");
mysql_connect("localhost","root","");
mysql_select_db("games");
$sql = mysql_query("SELECT Publisher FROM publisher");
while($row = mysql_fetch_array($sql)){
echo "<option value='".$row['Publisher']."'>" .$row[Publisher]. "</option>";
}
?>
</select>
<br/><br/><strong>Game Platform</strong><br>
<select name="plat">
<option value="">--Select--</option>
<?php
$sql = mysql_query("SELECT Platform FROM platform");
while($row = mysql_fetch_array($sql)){
echo "<option value='".$row['Platform']."'>" .$row[Platform]. "</option>";
}
?>
</select>
<br/><br/><strong>Genre</strong><br>
<select name="gen">
<option value="">--Select--</option>
<?php
$sql = mysql_query("SELECT Genre FROM genre");
while($row = mysql_fetch_array($sql)){
echo "<option value='".$row['Genre']."'>" .$row[Genre]. "</option>";
}
$developer = $_POST['dev'];
$platform = $_POST['plat'];
$genre = $_POST['gen'];
?>
</select>
<br><br><input type="submit" name="search" value="Search"></input>
</fieldset>
</form>
</section>
<aside id="items">
<fieldset>
<legend><h3><strong>Game List</strong></h3></legend>
<?php
var_dump($developer);
$devNO = mysql_query("SELECT No FROM developer WHERE Developer = $developer");
$platNO = mysql_query("SELECT No FROM platform WHERE Platform = $platform");
$genNO = mysql_query("SELECT No FROM genre WHERE Genre = $genre");
var_dump($devNO);
$sql = sprintf("SELECT Title, Release_Year, Language, Price FROM games WHERE Developer_NO = $devNO");
$result = mysql_query($sql);
$game_title = 'Title';
$game_year = 'Release_Year';
$game_lan = 'Language';
$game_price = 'Price';
?>
<div id="gamelist">
<?php
if(!$result) {
die(mysql_error());
}
while($row = mysql_fetch_array($result)) {
?>
<div class="row">
<div class="cell"><?php echo $row[$game_title]?></div>
<div class="cell"><?php echo "Year : ".$row[$game_year]?></div>
<div class="cell"><?php echo "Language : ".$row[$game_lan]?></div>
<div class="cell"><?php echo "Price : RM".$row[$game_price]?></div>
</div>
<?php
}
?>
</div>
</fieldset>
</aside>
</div>
</div>
<div id="btm_wrapper">
<footer id="foot">
<strong></strong>
</footer>
</div>
</body>
</html>
The full output will be like this after i search from dropdown:
The method mysql_query() , if successful, returns a resultset, which has to be made an array even if only a single value is being returned from the query.
Ideally there should be a for-each iteration on this array to get result, however if you're sure you'll get at least one value, the first row with desired index('No' in your case).
$dataset= mysql_query("SELECT No FROM developer WHERE Developer = '$developer'");
$row = mysql_fetch_assoc($dataset);
$devNO = $row['No'];
Also note the quote I've used for varchar. It was correctly pointed in the other answer. Try Now.
If $developer is a string use quotes. You need to first fetch results from result set and than use it in code:
<?php
$devNOResult = mysql_query("SELECT No FROM developer WHERE Developer = '$developer'");
$platNOResult = mysql_query("SELECT No FROM platform WHERE Platform = '$platform'");
$genNOResult = mysql_query("SELECT No FROM genre WHERE Genre = '$genre'");
$devNO = mysql_fetch_row($devNOResult);
$sql = sprintf("SELECT Title, Release_Year, Language, Price FROM games WHERE Developer_NO = $devNO");
$result = mysql_query($sql);
$game_title = 'Title';
$game_year = 'Release_Year';
$game_lan = 'Language';
$game_price = 'Price';
?>
<div id="gamelist">
<?php
if(!$result) {
die(mysql_error());
}
while($row = mysql_fetch_array($result)) {
?>
I am working on a way to pull data from my mySQL database and display the text as an array however I have it so that only one string of data is available and each instance is separated by a comma example. 1,2,3,4 is there a way to have it break or start a new line after each comma instead of having a conglomerated mess that looks like this:
1 package KRAFT Deluxe Macaroni & Cheese Dinner Sharp Cheddar Cheese Sauce, 1 lb Ground Beef, 1 Jar Spaghetti Sauce, 1/2 Cup KRAFT Shredded Mozzarella Cheese
Code:
<?php
require_once '../scripts/app_config.php';
require_once '../../scripts/authorize.php';
require_once '../scripts/database_connection.php';
// Get the user ID of the user to show
$User_id = $_REQUEST['User_id'];
// Gets User Id
$select_query = "SELECT * FROM users WHERE User_id = " . $User_id;
// Run the query to get User Name
$result1 = mysql_query($select_query);
if ($result1) {
$row = mysql_fetch_array($result1);
$first_name = $row['First_name'];
$last_name = $row['Last_name'];
$email = $row['Email'];
$User_id = $row['User_id'];
} else {
handle_error("There was a problem finding your " .
"information in our system.",
"Error locating user with ID {$User_id}");
}
$Recipe_Id = $_REQUEST['Id'];
$result = mysql_query( "SELECT * FROM recipes WHERE Id = " . $Recipe_Id )
or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$Id = $row['Id'];
$Name = $row['Name'];
$Ingredients = $row['Ingredients'];
$Category1 = $row['Category1'];
$Category2 = $row['Category2'];
$Directions = $row['Directions'];
$Creator = $row['User_Name'];
?>
<html>
<head>
<title>Recipes</title>
<meta name="viewport" content="width=device-width">
<meta http-equiv="Content-Language" content="English">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<link rel="stylesheet" type="text/css" href="../style/index.css" media="screen">
<script type="text/javascript" src="jquery-1.9.1.min.js">
</script>
</head>
<body>
<div id="header">
</div>
<center>
<span class="nav">Home</span> <span class="nav">New Recipe </span> <span class="nav">Sign Out </span>
</center>
<br>
<div id="content">
<div class="user_profile">
<h1><?php echo "{$first_name} {$last_name}";?></h1>
</div>
<div id="feature">
<?php
echo "<H1>". $row['Name'] ."</H1>";
echo "<p>" . $row['Category1'] . "</p>";
echo "<p>" . $row['Category2'] . "</p>";
echo "<p>" . $row['Creator'] . "</p>";
?> </div>
<div class="column1">
<h1>My Recipes</h1>
<?php
echo "<p>" . $row['Ingredients'] . "</p>";
?>
</div>
<div class="column2">
<h1>Directions</h1>
<?php
echo "<p>" . $row['Directions'] . "</p>";
}
?>
</div>
<div class="column3">
<h1>Recipes I Like</h1>
</div>
</div>
<div id="footer"></div>
</body>
</html>
If I got it correctly you just need something like this:
echo "<p>" . str_replace(', ',', <br />',$row['Ingredients']) . "</p>";
I created a page in PHP
The database is connected online, so when I run the site local / but the sql database connected online
everything works good. But as soon as I run the page Online, and tries to do anything with the database it says
The website encountered an error while retrieving http://www..... It
may be down for maintenance or configured incorrectly.
So I understands that something is wrong, but I cant seem to understand why I can upload things to my SQL database when I drive the page localy.
Any tips on what I should be thinking on ?
I Will add my upload.php file
the thing is that even if i delete my include('/core/inc/init.inc.php'); file.
Its still get the same error. so its must be something with my query's ?
This is my code.
<?php
// Connects to your Database
include('/core/inc/init.inc.php');
if($_FILES['file_name'])) {
$file_name = mysql_real_escape_string($_FILES['file_name']['name']);
//Writes the information to the database
mysql_query("INSERT INTO `files` (`file_name`) VALUES ('$file_name')") ;
move_uploaded_file($_FILES['file_name']['tmp_name'], "core/files/{$_FILES['file_name']['name']}");
}
// sid antal
$page = ($_GET['page'])) ? (int)$_GET['page'] : 1;
//Retrieves data from MySQL
$data = mysql_query("SELECT * FROM files") or die(mysql_error());
//Puts it into an array
$files = mysql_query("SELECT * FROM files") or die(mysql_error());
/*pagination */
$per_page = 5;
$pages_query = mysql_query("SELECT COUNT('user_id') FROM files");
$pages = ceil(mysql_result($pages_query, 0) /$per_page);
$page = ($_GET['page'])) ? (int)$_GET['page'] : 1;
$start = ($page - 1) * $per_page;
/*pagination */
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Pixeltouch</title>
<link rel="stylesheet" type="text/css" href="ext/style.css" />
</head>
<body>
<div id="page-wrap">
<div id="main-content">
<br/>
<?php include_once('template/head.inc.php');
?>
<div id="menu">
<?php include_once('template/nav.inc.php');
?>
</div>
<!-- SKRIVBOX-->
<div>
<form enctype="multipart/form-data" action="" method="post">
<p>
<input type="file" name="file_name" /><br/>
</p>
<p>
<input type="submit" value="Add" />
</p>
</form>
<!-- Listan -->
</div>
<!--Pagination-->
<?php
$data = mysql_query("SELECT * FROM files LIMIT $start, $per_page")
or die(mysql_error());
echo "<table border cellpadding=1>";
while($info = mysql_fetch_array( $data ))
{ ?><?php
echo "<tr>";
// echo "<td>".$info['user_id'] . "</td> ";
echo "<td>".$info['user_name'] . "</td> ";
echo "<td>".$info['file_name'] . "</td> ";
?>
<td> <?php echo $info['file_name']; ?></td>
<?php
echo "<td>".date('d/m/Y') . "</td>" ;
}
echo "</tr>";
echo "</table>";
if($pages >= 1 && $page <=$pages){
for ($x = 1; $x<=$pages; $x++){
echo ($x == $page) ? '<strong>' .$x. ' </strong>' : '' .$x. ' ';
}
}
?>
<!--Pagination-->
<!--PAGINATION-->
<!-- SKRIVBOX END-->
</div>
<?php include_once('template/foot.inc.php');
?>
</div>
</body>
</html>
Init.inc.php file
<?php
session_start();
mysql_connect("server", "username", "password") or die(mysql_error()) ;
mysql_select_db("pixeltouch2") or die(mysql_error()) ;
$path = dirname(__FILE__);
include("{$path}/user.inc.php");
?>
<!--Registration/Login (START)-->
<?php
$exceptions = array('register', 'login', 'user_list', 'profile', 'edit_profile');
$page = substr(end(explode('/', $_SERVER['SCRIPT_NAME'])), 0, -4);
if(in_array($page, $exceptions) == false){
if (isset($_SESSION['username']) == false){
header('Location: login.php');
die();
}
}
?>
<!--Registration/Login (END)-->
<!--User Profile (START)-->
<?php
$_SESSION['uid'] = 1;
?>
<!--User Profile (END)-->
Make sure you have something like this:
mysql_connect($server, $username, $password)
in your /core/inc/init.inc.php.
Well this is the error message i get Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (2) – Felipe Otarola 1 hour ago
Felipe try start/restart your mysql server:
sudo service mysql start // ubuntu
sudo service mysqld start // centos
EDIT 1
Try this:
if ($_SERVER['HTTP_HOST'] != 'localhost') // running in remote server
$server = 'localhost';
else // running locally
$server = 'your-online-database-server-address';
mysql_connect($server, "username", "password") or die(mysql_error());
I wish to set the title of my webpage to Ultan.me - Whatever the post title. I want it to display the post title. The posts are submitted to a MySQL database and the title row is called "title". Any help is appreciated with this small question.
Update:
Here is the page itself now but it doesn't display the title. Should I open the php document and connect to my database somewhere different to it's current locations?
The Code (The only necessary piece is the beginning):
<html>
<head>
<meta name="keywords" content="Mac user Ultan Casey TheCompuGeeks UltanKC">
<title>Ultan.me - <?echo $title;?></title>
<link rel="stylesheet" href="css/styles.css" type="text/css" />
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script>
<script type="text/javascript"
src="js/jquery.labelify.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$(":text").labelify();
});
</script>
<style>
a {text-decoration:none}
</style>
</head>
<body>
<div id="main">
<!-- Menu Start -->
<div id="menu">
<ul>
<li>home</li>
<li>about me</li>
<li>archives</li>
<li>contact</li>
<li>gallery</li>
</ul>
</div>
<!-- Menu End -->
<img src="images/banner.png" />
<div id="content">
<div id="posts">
<?php
mysql_connect ('localhost', 'root', 'root') ;
mysql_select_db ('ultankc');
if (!isset($_GET['id']) || !is_numeric($_GET['id'])) {
die("Invalid ID specified.");
}
$id = (int)$_GET['id'];
$sql = "SELECT * FROM php_blog WHERE id='$id' LIMIT 1";
$result = mysql_query($sql) or print ("Can't select entry from table php_blog.<br />" . $sql . "<br />" . mysql_error());
while($row = mysql_fetch_array($result)) {
$date = date("l F d Y", $row['timestamp']);
$title = stripslashes($row['title']);
$entry = stripslashes($row['entry']);
$get_categories = mysql_query("SELECT * FROM php_blog_categories WHERE `category_id` = $row[category]");
$category = mysql_fetch_array($get_categories);
?>
<p><?php echo "<p id='post-title'><strong>" . $title . "</strong></p>"; ?><br /><br />
<div id="entry"><?php echo $entry; ?>
</div><br /><br />
<p id="date">Posted in <?php echo $category['category_name']; ?> on <?php echo $date; ?></p>
</p>
<h2 id="share-title">Share This Post</h2>
<div id="social-share">
<li id="link-right"><a href="http://twitter.com/home?status=
I just read <?php echo $title; ?> at http://ultan.me/post.php?id=<?php echo $id; ?>"><center>Twitter</center></a></li>
<li id="link-left"><center>Digg</center></li>
<br>
<li id="link-right"><center>Facebook</center></li>
<li id="link-left"><a href="http://www.google.com/buzz/post?url=http://ultan.me/post.php?id=<?php echo $id; ?>
"><center>Google Buzz</center></a></li>
<div class="clr"></div>
</div>
<h2 id="comments-title">Comments</h2>
<div id="comment-list">
<?php
}
$commenttimestamp = strtotime("now");
$sql = "SELECT * FROM php_blog_comments WHERE entry='$id' ORDER BY timestamp";
$result = mysql_query ($sql) or print ("Can't select comments from table php_blog_comments.<br />" . $sql . "<br />" . mysql_error());
while($row = mysql_fetch_array($result)) {
$timestamp = date("l F d Y", $row['timestamp']);
printf("<div class='comment-ind'><p id='comments'><a id='username' href=\"%s\">%s</a> %s</p>", stripslashes($row['url']), stripslashes($row['name']), $timestamp);
print("<p class='comments'>" . stripslashes($row['comment']) . "</p><div class='clr'><br></div></div>");
}
?>
<div class="clr"></div>
<form id="commentform" method="post" action="process.php">
<p><input type="hidden" name="entry" id="entry" value="<?php echo $id; ?>" />
<input type="hidden" name="timestamp" id="timestamp" value="<?php echo $commenttimestamp; ?>">
<input type="text" name="name" id="name" title="Name (required)" /><br />
<input type="text" name="email" id="email" title="Mail (will not be published) (required)" /><br />
<input type="text" name="url" id="url" title="Website" value="http://" /><br />
<br />
<textarea title="Your Comment Goes Here" name="comment" id="comment"></textarea></p>
<p><input type="submit" name="submit_comment" id="submit_comment" value="Add Comment" /></p>
</form>
</div>
<div id="pages">
<?php
$total_results = mysql_fetch_array(mysql_query("SELECT COUNT(*) AS num FROM php_blog"));
$total_pages = ceil($total_results['num'] / $blog_postnumber);
if ($page > 1) {
$prev = ($page - 1);
echo "<< Newer ";
}
for($i = 1; $i <= $total_pages; $i++) {
if ($page == $i) {
echo "$i ";
}
else {
echo "$i ";
}
}
if ($page < $total_pages) {
$next = ($page + 1);
echo "Older >>";
}
?>
</div>
</div>
</div>
<!-- Sidebar Start -->
<div class="sidebar">
<!-- Item 1 -->
<div id="side-item">
<h2>
<a href="http://www.dailybooth.com/UltanCasey">
<img src="images/db-icon.jpg">Dailybooth
</a></h2>
<div id="side-item-content">
<center>
<img src="http://dailybooth.com/UltanCasey/latest/medium.jpg" />
</center>
</div>
</div>
<!-- Item 2 -->
<div id="side-item">
<h2><img src="images/connect.jpg" />Connect</h2>
</div>
<div id="side-item-content">
<div class="tweet-title"><p>Latest Tweet:</p></div>
<div id="tweet">
<?php
function getTwitterStatus($userid){
$url = "http://twitter.com/statuses/user_timeline/$userid.xml?count=1";
function auto_link_twitter ($text)
{
// properly formatted URLs
$urls = "/(((http[s]?:\/\/)|(www\.))?(([a-z][-a-z0-9]+\.)?[a-z][-a-z0-9]+\.[a-z]+(\.[a-z]{2,2})?)\/?[a-z0-9._\/~#&=;%+?-]+[a-z0-9\/#=?]{1,1})/is";
$text = preg_replace($urls, " <a href='$1'>$1</a>", $text);
// URLs without protocols
$text = preg_replace("/href=\"www/", "href=\"http://www", $text);
// Twitter usernames
$twitter = "/#([A-Za-z0-9_]+)/is";
$text = preg_replace ($twitter, " <a href='http://twitter.com/$1'>#$1</a>", $text);
// Twitter hashtags
$hashtag = "/#([A-Aa-z0-9_-]+)/is";
$text = preg_replace ($hashtag, " <a href='http://hashtags.org/$1'>#$1</a>", $text);
return $text;
}
$xml = simplexml_load_file($url) or die("could not connect");
foreach($xml->status as $status){
$text = $status->text;
}
echo auto_link_twitter ($text);
}
getTwitterStatus("UltanKC");
?>
</div>
<br>
<ul>
<li id="social">YouTube</li>
<li id="social">Twitter</li>
<li id="social">LastFM</li>
<li id="social">Email</li>
</ul>
</div>
<!-- Item 2 End-->
<div id="side-item">
<h2><img src="images/archive.jpg" />Archives</h2>
</div>
<div id="archive-side">
<?php
mysql_connect ('localhost', 'root', 'root') ;
mysql_select_db ('ultankc');
$result = mysql_query("SELECT FROM_UNIXTIME(timestamp, '%Y') AS get_year, COUNT(*) AS entries FROM php_blog GROUP BY get_year");
while ($row = mysql_fetch_array($result)) {
$get_year = $row['get_year'];
$entries = $row['entries'];
echo "<li id='tag'>Entries from " . $get_year . " (" . $entries . ")<br /></li>";
}
$result1 = mysql_query("SELECT * FROM php_blog_categories ORDER BY category_name ASC");
while($row = mysql_fetch_array($result1)) {
$result2 = mysql_query("SELECT COUNT(`id`) AS entries FROM php_blog WHERE category = $row[category_id]");
$num_entries = mysql_fetch_array($result2);
echo '<li id="tag">' . $row['category_name'] . ' (' . $num_entries['entries'] . ')</li>';
}
?>
</div>
</div>
<div class="clr" />
</div>
<!-- Sidebar End -->
<div id="footer">
<p> © Ultan Casey 2010</p>
<p style="margin-top: -18px; float:right">Home | About Me | Email Me</p>
</div>
</div>
</div>
</body>
</html>
?>
Here's the method I use (for similar things, not just title):
<?
ob_start (); // Buffer output
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<title><!--TITLE--></title>
</head>
<body>
<?
$pageTitle = 'Title of Page'; // Call this in your pages' files to define the page title
?>
</body>
</html>
<?
$pageContents = ob_get_contents (); // Get all the page's HTML into a string
ob_end_clean (); // Wipe the buffer
// Replace <!--TITLE--> with $pageTitle variable contents, and print the HTML
echo str_replace ('<!--TITLE-->', $pageTitle, $pageContents);
?>
PHP usually works be executing any bits of code and printing all output directly to the browser. If you say "echo 'Some text here.';", that string will get sent the browser and is emptied from memory.
What output buffering does is say "Print all output to a buffer. Hold onto it. Don't send ANYTHING to the browser until I tell you to."
So what this does is it buffers all your pages' HTML into the buffer, then at the very end, after the tag, it uses ob_get_contents () to get the contents of the buffer (which is usually all your page's HTML source code which would have been sent the browser already) and puts that into a string.
ob_end_clean () empties the buffer and frees some memory. We don't need the source code anymore because we just stored it in $pageContents.
Then, lastly, I do a simple find & replace on your page's source code ($pageContents) for any instances of '' and replace them to whatever the $pageTitle variable was set to. Of course, it will then replace <title><!--TITLE--></title> with Your Page's Title. After that, I echo the $pageContents, just like the browser would have.
It effectively holds onto output so you can manipulate it before sending it to the browser.
Hopefully my comments are clear enough.
Look up ob_start () in the php manual ( http://php.net/ob_start ) if you want to know exactly how that works (and you should) :)
You parse the field from the database as usual.
Then let's say you put it in a variable called $title, you just
<html>
<head>
<title>Ultan.me - <?php echo htmlspecialchars($title);?></title>
</head>
EDIT:
I see your problem. You have to set $title BEFORE using it. That is, you should query the database before <title>...
header.php has the title tag set to <title>%TITLE%</title>; the "%" are important since hardly anyone types %TITLE% so u can use that for str_replace() later. then, you use output buffer like so
<?php
ob_start();
include("header.php");
$buffer=ob_get_contents();
ob_end_clean();
$buffer=str_replace("%TITLE%","NEW TITLE",$buffer);
echo $buffer;
?>
For more reference, click PHP - how to change title of the page AFTER including header.php?
What about using something like:
<?php
$page_title = "Your page tile";
include("navigation.php"); // if required
echo("<title>$page_title</title>");
?>
Move the data retrieval at the top of the script, and after that use:
<title>Ultan.me - <?php echo htmlspecialchars($title, ENT_QUOTES, 'UTF-8'); ?></title>
You need to set the value of $title before echoing it.
Also, you should really sanitize any data before using it in queries as this is a security risk
create a new page php and add this code:
<?php
function ch_title($title){
$output = ob_get_contents();
if ( ob_get_length() > 0) { ob_end_clean(); }
$patterns = array("/<title>(.*?)<\/title>/");
$replacements = array("<title>$title</title>");
$output = preg_replace($patterns, $replacements,$output);
echo $output;
}
?>
in <head> add code: <?php require 'page.php' ?> and on each page you call the function ch_title('my title');
The problem is that $title is being referenced on line 5 before it's being assigned on line 58. Rearranging your code isn't easy, because the data is both retrieved and output at the same time. Just to test, how does something like this work?
Because you're only retrieving one row, you don't need to use a while loop, but I left it with hopes that it'll make it easier for you to relate to your current code. All I've done is removed the actual output from your data retrieval, and added variables for category and category name which are then referred to as usual later on. Also, I haven't tested this. :)
It'll be tricky to rearrange your code to make this work, but I'll try :)
So, put this at the top of your code:
<?php require_once('mysql.php'); ?>
The top of the file should look like:
<?php require_once('mysql.php'); ?>
<html>
<head>
<meta name="keywords" content="Mac user Ultan Casey TheCompuGeeks UltanKC">
<title>Ultan.me - <?php echo htmlspecialchars($title); ?> </title>
Then, create a file called mysql.php in the same directory that the file which contains the code you quoted is in.
Put this is mysql.php:
<?php
mysql_connect ('localhost', 'root', 'root');
mysql_select_db ('ultankc');
if (!isset($_GET['id']) || !is_numeric($_GET['id'])) {
die("Invalid ID specified.");
}
$id = (int)$_GET['id'];
$sql = "SELECT * FROM php_blog WHERE id='$id' LIMIT 1";
$result = mysql_query($sql) or print ("Can't select entry from table php_blog.<br />" . $sql . "<br />" . mysql_error());
$res = mysql_fetch_assoc($result);
$date = date("l F d Y", $res['timestamp']);
$title = $res['title'];
$entry = $res['entry'];
$get_categories = mysql_query("SELECT * FROM php_blog_categories WHERE `category_id` = $res['category']");
$category = mysql_fetch_array($get_categories);
?>
Well, hope that helped :)
I know this is an old post but having read this I think this solution is much simpler (though technically it solves the problem with Javascript not PHP).
<html>
<head>
<title>Ultan.me - Unset</title>
<script type="text/javascript">
function setTitle( text ) {
document.title = text;
}
</script>
<!-- other head info -->
</head>
<?php
// Make the call to the DB to get the title text. See OP post for example
$title_text = "Ultan.me - DB Title";
// Use body onload to set the title of the page
print "<body onload=\"setTitle( '$title_text' )\" >";
// Rest of your code here
print "<p>Either use php to print stuff</p>";
?>
<p>or just drop in and out of php</p>
<?php
// close the html page
print "</body></html>";
?>
Simply add $title variable before require function
<?php
$title = "Your title goes here";
require("header.php");
?>
header.php
<title><?php echo $title; ?></title>
<?php echo APP_TITLE?> - <?php echo $page_title;?>
this should work fine for you
if you want to current script filename as your title tag
include the function in your project
function setTitle($requestUri)
{
$explodeRequestUri = explode("/", $requestUri);
$currentFileName = end($explodeRequestUri);
$withoutExt = preg_replace('/\\.[^.\\s]{3,4}$/', '', $currentFileName);
$explodeCurrentFileName = explode("-", $withoutExt);
foreach ($explodeCurrentFileName as $curFileValue)
{
$fileArrayName[] = ucfirst($curFileValue);
}
echo implode(" ", $fileArrayName);
}
and in your html include the function script
and replace your title tag with this
<title>Your Project Name -
<?php setTitle($_SERVER['REQUEST_URI']); ?>
</title>
it works on php7 and above but i dont have any idea about php 5.*
Hope it helps