Hello im new to php but i cant make this dynamiclist to work here is the code..
<?php
if (isset($_GET['id'])) {
include "DocumentSystem/scripts/connect_to_mysql.php";
$id = preg_replace('#[^0-9]#i', '', $_GET['id']);
$dynamicList = "";
$sql = mysql_query("SELECT * FROM documents WHERE id='$id' LIMIT 1");
$documentCount = mysql_num_rows($sql); // count the output amount
if ($documentCount > 0) {
while($row = mysql_fetch_array($sql)){
$id = $row["id"];
$document_name = $row["document_name"];
$full_document = $row["full_document"];
$dynamicList .= '<div id="slidepic">
<img src="images/document_images/'.$id.'.jpg" width="550" height="350" />
<div id="slideshow">
<h1> '.$document_name.'</h1>
<br />
<p1>'.$full_document.'</p1>
</div>
</div>';
}
} else {
$dynamicList = "We have no documents listed in the database";
}
mysql_close();
?>
It is getting the correct id from the url but the query isnt working for me
thx for any answers!
Try
to usr mysql_fetch_array($sql,MYSQL_ASSOC)or mysql_fetch_assoc($sql) instead of mysql_fetch_array($sql)
By default mysql_fetch_array return non associative array.
Pay attention:
Do not SELECT * FROM table_name. It's better for performance to use SELECT column1, colum2 FROM table_name
Mysql module is deprecated. It's better to use PDO mysql module page MySQL API comparison
Please use you have get the query result
<?php
$con = mysql_connect("localhost","username","password");
mysql_select_db("your_database_name", $con);
$result = mysql_query("SELECT * FROM documents WHERE id = '$id' LIMIT 1");
while($row = mysql_fetch_array($result))
{
echo = $row["id"];
echo = $row["document_name"];
echo = $row["full_document"];
}
mysql_close($con);
?>
Related
I wrote simple code to fetch data from MySQL using PHP.
This is the code:
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
{
echo "<p>";
echo "$row[id_kendaraan]";
echo "<p>";
echo "$row[mangkal]";
}
?>
The script is working, if the data is returned by the db call, I can see the results displayed. But, if the query result has no data, the script just displays blank. I want to show a message that says - 'Data not found'. How can I do that?
I have more than one record for a query but the script displays just one data. Please help me to show all records.
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
I figured out the way to link to the page and set what ID i would like to call:
CLICK TEST **(IS THIS RIGHT?)**
But then I need page.php to pull the id, this is what I am using at the moment to pull the id manually. How would I make the following code pull it form the link?
<?php
$query = "select * from Drinklist where id = 10";
$result = mysqli_query($conn,$query);
while($Drinklist = mysqli_fetch_array($result)){
echo "<head>";
echo "<title>".$List['name']." - Site Name</title>";
}
?>
I tried the following (didn't work):
$query = "select * from List where id = . $id";
Seems like I can only find the way to do it with MYSQL and not MYSQLI... Any help would be appreciated.
UPDATED CODE:
<?php
$query = "select * from Drinklist where id = ?";
$result = mysqli_prepare($conn,$query);
mysqli_stmt_bind_param($result, 'i', $_GET['id']);
mysqli_stmt_execute($result);
while($Drinklist = mysqli_fetch_array($result)){
echo "<head>";
echo "<title>".$Drinklist['name']." - Mixed Drinks Station</title>";
}
?>
Getting error:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,
object given in public_html/page.com/test/inc/drink-page.php on line 6
Ok so I ended up figuring this one out with a ton of trial and error...
Thank you chris85 for the early support and hopefully this can help you out a little. This is fun ;)
<?php
$server = "localhost";
$user = "user";
$pass = "password";
$dbname = "database";
//Creating connection for mysqli
$conn = new mysqli($server, $user, $pass, $dbname);
//Checking connection
if($conn->connect_error){
die("Connection failed:" . $conn->connect_error);
}
$article_id = $_GET['id'];
if( ! is_numeric($article_id) )
die("Looks like you are lost! <a href='#'>Back to Home</a> ");
$query = "SELECT * FROM `Whatever` WHERE `ID` =$article_id LIMIT 0 , 30";
$info = mysqli_query($conn,$query);
while($row = mysqli_fetch_array($info, MYSQL_ASSOC))
{
$name = $Whatever['name'];
$description = $Whatever['description'];
$keywords = $Whatever['keywords'];
$lrgpic = $Whatever['lrgpic'];
$vid = $Whatever['vid'];
$name = htmlspecialchars($row['name'],ENT_QUOTES);
$description = htmlspecialchars($row['description'],ENT_QUOTES);
$keywords = htmlspecialchars($row['keywords'],ENT_QUOTES);
$lrgpic = htmlspecialchars($row['lrgpic'],ENT_QUOTES);
$vid = $row['vid']; //Use <-- to be able to have HTML in your database otherwise the above would remove <, >, /, ', ", ect.
echo "<head>";
echo "<title>$name - Site title</title>";
echo "<meta name='description' content='$description'>";
echo "<meta name='keywords' content='$keywords'>";
include 'inc/head.php'; //includes already are in a php file ;)
echo "</head>";
echo "<body>";
include 'inc/header.php';
include 'inc/nav.php';
echo "<div class='wrapper'>";
echo "<div id='drink-name'>";
echo "<h2>$name</h2>";
echo "</div>";
// AND SO ON
}
?>
I am building a site, and essentially what this PHP algorithm will do is look at a product (row in MySQL database) one at a time, and do a process accordingly.
I put a lot of research into this but couldn't find anything, any help would be greatly appreciated!
My Code (currently returning nothing for echo variables):
<?php
include_once 'dbconnect.php';
$query = "SELECT * FROM track";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$pro_code = mysql_result(mysql_fetch_array(mysql_query('SELECT product_code FROM track')));
$currency = mysql_fetch_array(mysql_query('SELECT currency FROM track'));
$cc = mysql_fetch_array(mysql_query('SELECT cctld FROM track'));
$initial_price = mysql_fetch_array(mysql_query('SELECT initial_price FROM track'));
$url = 'test';
}
echo $pro_code;
echo $currency;
echo $initial_price;
?>
First of all, try the advice about PDO and stuff from Jay Blanchard some day.
Secondly I've tried to answer your question anyway and I've tried to interpret your complete intention. I put comments in the code:
<?php
include_once 'dbconnect.php';
$query = "SELECT * FROM track";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
//You need to read the row variable as an array
$pro_code = $row['product_code'];
$currency = $row['currency'];
$cc = $row['cctld'];
$initial_price = $row['initial_price'];
//$url is not used.. I asume a test to get the external source ;-)
$url = 'test';
if ($url == $cc) {
//if you want to print every row, you must echo inside the while loop
echo $pro_code;
echo $currency;
echo $initial_price;
} elseif ($url == 'test') {
//do something else here
} else {
//do something for all the other cases here
}//end if
}//end while
?>
Why do you query the same table multiple times, your code should be written like this:
include_once 'dbconnect.php';
$query = "SELECT product_code, currency, cctld, initial_price FROM track";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo $row['product_code'];
echo $row['currency'];
echo $row['cctld'];
echo $row['initial_price'];
}
and please upgrade to mysqli or PDO
I have a problem to do the code work...I don't know why its not working...I want to display from the database title and many other by id...The name of the file is view.php...If I want to show id 1 then I type on url view.php?id=1 or id 2 view.php?id=2...here is my code please help me out...
<?php
$getid = $_GET["id"];
$dbuser = 'user';
$userpass = 'pass';
$database = 'db';
error_reporting(0);
$con = mysql_connect("localhost",$dbuser,$userpass);
if(!$con){die('Could not connect:'.mysql_error());}
mysql_select_db($database, $con);
mysql_query("SET NAMES utf8");
date_default_timezone_set('Europe/Athens');
$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
if(mysql_num_rows($query)){
$result = mysql_fetch_assoc($query);
$id = $result['id'];
$title = $result['title'];
$description = $result['description'];
echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
}else{
echo 'Error';
exit;
}
?>
Change this:
if(mysql_num_rows($query)){
to this:
if(mysql_num_rows($query) >= 1){
keep in mind that you've got to have things in your database with the ID that you give,
or it will just go to else again.
(mysql_num_rows returns FALSE or an integer with the amount of rows, and since the if statement didnt check for an integer but for true, it always went down the ELSE path)
Source: PHP.net
<?php
$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
$count = mysql_num_rows($query);
if($count != 0){
while($result = mysql_fetch_array($query)){
$id = $result['id'];
$title = $result['title'];
$description = $result['description'];
echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
}
}else{
echo 'Error';
}
?>
i'm a newbie to php still.
I'm using phpmyadmin as my database. I have a table called 'lessonno' and a column named 'lesson' in it. I tried using this code to retrieve out the number inside 'lesson'. But it's not printing out anything. Can someone help?
<?php
$server = 'localhost';
$username = '';
$password = '';
$database = 'project';
mysql_connect($server,$username,$password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
$sql = "SELECT 'lesson' FROM 'lessonno'";
$lesson = $_POST['lesson'];
$result = mysql_query($sql);
?>
<?php
for($i = 1; $i <= $lesson; $i++) {
echo "<div>
<span>Lesson ".$i."</span>
</div>
<br>";
}
?>
You can use something like this:
$sql = "SELECT lesson FROM lessonno";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
echo $row['lesson'];
}
If you would like to only print out a specific lesson with an certan ID, you can use something along the lines:
$id = $_GET['lessonid']; // If you would have something like index.php?lessonid=36 and you'd like it to only fetch the data for the lesson with the id of 36.
$sql = "SELECT lesson FROM lessonno WHERE id='$id'";
(by looking at the $_POST['lesson'] part, I suppose that's something you might be trying to do as it's in the for loop as well)
Also, I suggest you use mysqli.
And, this:
echo "<div>
<span>Lesson ".$i."</span>
</div>
<br>";
Will just echo the $i as both lesson= and the span with Lesson, which won't grab any information from the actual database but just go with the current number it's at, from the for loop you have.
i have made some changes in your code try this
<?php
$server = 'localhost';
$username = 'root';
$password = '';
$database = 'project';
$conn = mysql_connect($server,$username,$password) or die(mysql_error());
mysql_select_db($database, $conn) or die(mysql_error());
$sql = "SELECT `lesson` FROM `lessonno`";
$lesson = $_POST['lesson'];
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
$lesson_no = $row['lesson'];
echo "<div>
<span>Lesson ".$lesson_no."</span>
</div>
<br>";
}
?>
Note : mysql_* is deprecated. use mysqli_* OR PDO
For getting Values from DB you need to use something like this
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
}
For further reference please visit http://in2.php.net/manual/en/function.mysql-fetch-assoc.php
For counting the number of data in your database, just insert this code
$sql = "SELECT 'lesson' FROM 'lessonno'";
$lesson = $_POST['lesson'];
$result = mysql_query($sql);
$count=mysql_num_rows($result);//this will count the number of rows in your table.
echo "<div>
<span>Lesson ".$count."</span>
</div>
<br>";