I have problem with my (made with tutorial) PHP comment box.
The problem is this number 1. I think that problem is in last code, but as I am beginner I don't know where is problem.
I'm posting all 3 php files without header and footer file.
PrtSc of problem on page
Source code in chrome of page
This is full code in php....
First php file
<?php
$title = "FilmoviRecenzije";
$content = '
<img src="Slike_filmovi/novo_Mockingjay2.jpg" class="imgLeft"/>
<h2>The Hunger Games: Mockingjay - Part 2 (2015)</h2>
<p><i>"Žanr: Avantura, Znanstvena fantastika"</i></p>
<p>Režija: Francis Lawrence<p>
<h4>
U grand finalu spektakularne kino-hit franšize Igre gladi Katniss Everdeen (Jennifer Lawrence),
nakon što je postala vođa masovne pobune potlačenih, zajedno sa Distriktom 13 pokreće
sveopću revoluciju protiv autokratskog Kapitola. Katniss nastavlja svoju ulogu heroja
koji se u nevjerojatnoj bitci za preživljavanje bori za svoj narod!
</h4>
<dev class="videoCenter">
<iframe width="720" height="405" src="https://www.youtube.com/embed/n-7K_OjsDCQ" frameborder="0" allowfullscreen></iframe>
</dev>
';
include ('Template_Mockingjay2.php');
?>
Next file which is connected through "include" is:
<?php include('Header.php'); ?>
<div id="wrapper">
<div id="banner">
</div>
<nav id="navigation">
<ul id="nav">
<li>Početna </li>
<li>Novi Filmovi</li>
<li>Kontakt</li>
</ul>
</nav>
<div id="content_area">
<?php echo $content; ?>
</div>
<div id="sidebar">
<?php echo include('novo_Mockingjay2_komentar.php'); ?>
</div>
<?php include ("Footer.php"); ?>
And last file for Mysql, same included is:
<?php
mysql_connect("localhost","root","");
mysql_select_db("db_komentari");
if (isset($_POST['Ime']) && isset($_POST['Komentar'])){
// $Ime = isset($_POST['Ime']) ? $_POST['Ime'] :
$Ime = $_POST['Ime'];
$Komentar = $_POST['Komentar'];
$submit = $_POST['submit'];
$dbLink = mysql_connect("localhost","root","");
if($submit)
{
if($Ime&&$Komentar)
{
$insert=mysql_query("INSERT INTO novo_mockingjay2 (Ime,Komentar) VALUES ('$Ime','$Komentar') ");
echo "<meta HTTP-EQUIV='REFRESH' content='0; url=novo_Mockingjay2.php'>";
}
else
{
echo "Molim Vas ispunite oba polja";
}
}
}
?>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Komentari</title>
<center>
<form action="novo_Mockingjay2.php" method="POST">
<table>
<tr><td>Ime: <br><input type="text" name="Ime"/></td></tr>
<tr><td colspan="2">Komentar: </td></tr>
<tr><td colspan="5"><textarea name="Komentar" rows="1" cols="18"></textarea></td></tr>
<tr><td colspan="2"><input type="submit" name="submit" value="Objavi"></td></tr>
</table>
</form>
<?php
$dbLink = mysql_connect("localhost","root","");
$getquery=mysql_query("SELECT * FROM novo_mockingjay2 ORDER BY ID DESC");
while($rows=mysql_fetch_assoc($getquery))
{
$ID=$rows['ID'];
$Ime=$rows['Ime'];
$Komentar=$rows['Komentar'];
echo $Ime . ':<br/>' . $Komentar . '<br/>' . '<br/>' . '<hr size="4"/>';}
?>
Thank You in Advance!
As include language construct returns 1 on successful file including - this value is passed to your echo, so line:
echo include('novo_Mockingjay2_komentar.php');
means - include file, execute it and echo the result of inclusion. As result is 1 - it's echo'ed.
Change it to:
include('novo_Mockingjay2_komentar.php'); // without echo
I have a database which contains 2 tables. In the second table there are numerous records and I'm trying to echo out the latest id (MAX id) to the page using PHP. There is something wrong with my code and I don't know what that is:
$sqlCount = "SELECT MAX(id) FROM records";
$sql = "SELECT id,preview,description FROM records";
$pn = (isset($_GET['pn'])) ? $_GET['pn'] : 1;
$res = upagination($con, $sql, $sqlCount, $pn, 6);
$list .="<table border=0>";
foreach ($res['rows'] as $row) {
$list .="<tr>";
$list .="<td>" . $row['id'] . "</td><td>" . $row['preview'] . "</td><td>" . $row['description'] . "</td>";
$list .="</tr>";
}
$list .="</table>";
$paginationCtrls = "";
if (isset($res['backLink'])) {
$paginationCtrls .= "Previous";
}
if (isset($res['numbers'])) {
foreach ($res['numbers'] as $number) {
if ($number === $res['current']) {
$paginationCtrls .= "$number";
}
(some deleted code to shorten this a little)
}
}
?>
<!DOCTYPE html>
<html>
<head>
<style type="text/css"></style>
<title>Humorweb</title>
<meta name="keywords" content="videos">
<meta name="description" content="Humorwebsite.org -collection of the funniest videos and photos from the internet!-">
<link rel="stylesheet" type="text/css" href="https://www.humorwebsite.org/style_main.css">
</head>
<body>
<div id="container">
<div id="logo"> </div>
<div id="sidebar">
<div id="links">
HOME
VIDEOS
PHOTOS
ABOUT
</div>
</div>
<div id="context">
<div id="context-text"></div>
<div id="context-kuva">
<br>
<div id="nakki">
<?php echo $list; ?>
</div>
</div>
</div>
</div>
The last row I'm trying to echo out is the $list variable, but I get nothing. I have made a CSS on this file and define width, height, position, and float properties, but end up getting nothing from database.
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>";
This question already has an answer here:
upload image to mysql database php
(1 answer)
Closed 9 years ago.
this is code of the page that I will get the image from(work perfectly)
<?php
ob_start();
session_start();
include('connect.php');
$id = $_GET['id'];
$query = mysql_query("SELECT * FROM news WHERE id=$id");
$row = mysql_fetch_assoc($query);
header("Content-type: image/jpeg");
echo $row['image'];
?>
and this is my page that i get the image to in
<?php
ob_start();
session_start();
include('includes/connect.php');
include('includes/phpCodes.php');
$id = $_GET['id'];
function showNews() {
$data = array( 'id' => $id );
$base = "includes/getImage.php";
$url = $base. "?" . "id=36";
echo $url;
echo '<img src=includes/getImage.php class="newsImage">';
echo '<h1><p class="subjecTitle">هنا العنوان</p></h1>
<div class="newsContent">
hihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihi
</div>
';
}
?>
<!DOCTYPE html>
<html>
<head>
<title>عينٌ على الحقيقة</title>
<meta charset="utf-8">
<link rel="stylesheet" type="text/css" href="css/mainstyle.css">
<link rel="stylesheet" type="text/css" href="css/showstyle.css">
<script lang="javascript">
function logout( myFrame ) {
myFram.submit();
}
</script>
</head>
<body>
<div class="wrapper">
<?php headerCode(); ?>
<div class="content" dir="rtl">
<?php showNews(); ?>
</div>
</div>
</body>
</html>
i think my wrong is in , can someone tell me how can I solve it?, sorry for my bad english
Cleared it up for you:
echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';
Should 100% work (if the $id param has a value of course).
Update to fix the missing $id var:
<?php
ob_start();
session_start();
include('includes/connect.php');
include('includes/phpCodes.php');
$id = $_GET['id'];
function showNews(){
$id = $_GET['id'];
$base = "includes/getImage.php";
$url = $base. "?" . "id=36";
echo $url;
echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';
echo '
<h1><p class="subjecTitle">??? ???????</p></h1>
<div class="newsContent"></div>
';
}
?>
why are you using two different pages?
put both code in single page and simply do this
<img src=includes/<?php echo $row['image']; ?> class="newsImage">
Change :
echo ' <img src=includes/getImage.php class="newsImage">';
To :
echo ' <img src="includes/getImage.php?id='.$id.'" class="newsImage">';
Please note src has " and also make sure that includes/getImage.php return a image path
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