get one variable out of database with PHP - php

Question

Ok obviously i'm a beginner.
What i'm trying to do here is probably simple.
I'm trying to get all the field of my database to echo on my website. So i use this to show only the basic details:
<?php
// Make a MySQL Connection
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("crnew") or die(mysql_error());
$result = mysql_query("SELECT * FROM releases")
or die(mysql_error());
while($row = mysql_fetch_array( $result )
) {
// Print out the contents of each row into a table
echo '<ul class="releaselist">';
echo '<li>';
echo $row['products_name'];
echo '</li>';
echo '<li>';
echo $row['products_title'];
echo '</li>';
echo '<li>';
echo '<img class="releaseimg" src="'.$row['products_image'].'">';
echo '</li>';
echo '</ul>';
}
?>
So far eveything works fine.
What i want to do with this is get my URL to look like this when i click on a single image.
www.mywebsite.com/detailed.php?id=1
and show everything available in the table RELEASES
The way i've done it which is not working is:
<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("crnew") or die(mysql_error());
$result = mysql_query("SELECT * FROM releases
WHERE products_id=products_id")
or die(mysql_error());
if ($result){
$row = mysql_fetch_array($result);
echo $_GET['products_id'];
}
?>

Answer 1

If I'm understanding you correctly, your first script needs to be updated to:
echo '';
That will turn your image into a link. At that point your second script should be able to use:
_GET['id']
to retrieve this value and use it in your query. Thusly:
$result = mysql_query("SELECT * FROM releases
WHERE products_id=" . mysql_real_escape_string($_GET['id']));
Now if you really want to products_id in the URL instead of id, just change it 'id' to products_id in the echo line in the first script and all of the $_GET uses in the second script.

Answer 2

Try changing this line:
echo '<img class="releaseimg" src="'.$row['products_image'].'">';
To this one:
echo "<a href='detailed.php?products_id=".$row['products_id']."'><img class='releaseimg' src='".$row['products_image']."'></a>";

Answer 3

Hmm try using $row['products_id'] instead of $_GET['products_id']; $_GET is used when accessing a webpage and the reason that $row['products_image'] worked is because you accessed the database results.

Answer 4

You have misplaced the use of products_id...
Your code should like something like this...
<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("crnew") or die(mysql_error());
$result = mysql_query("SELECT * FROM releases
WHERE products_id='".intval($_GET['id'])."'")
or die(mysql_error());
if ($result){
$row = mysql_fetch_array($result);
echo $row['products_id'];
}
?>