Display record with image link

[gilanib]

Greetings!
I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code.
ourfranchises.php code:

<?php 
include('fconnection.php');

$sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error());
$row3 = mysql_fetch_array($sql); 
?>
<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>

----------------------------------------------------------------------

franchiseDetails.php code:

<?php
$id = $_GET['id'];
					
$query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1";

$result = mysql_query($query);

if (!$result) {
echo "NO RECORD FOUND";
} else {

while($row3 = mysql_fetch_array($result)):
?>

Manager Name: <?php echo $row3['manager_name'];?>

Please help me. Any help would be appreciated.

Thanks.

[Karthik_pranas]

In this line

<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>">

pass your record id instead of city as follows

<a href="franchiseDetails.php?id=<?php echo $row3['id'];?>">

[diafol]

I want to display each franchise record from database whenever a user click on franchise link.

DO we assume you mean to display data from ONE record based on the city parameter in the url?

[gilanib]

In this line

<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>">

pass your record id instead of city as follows

<a href="franchiseDetails.php?id=<?php echo $row3['id'];?>">

Dear it display one single id for all image links. See the result on http://lanceronlinejobs.com/our_franchises.php

[gilanib]

DO we assume you mean to display data from ONE record based on the city parameter in the url?

Yes dear. Please help me out.

[diafol]

<?php 
include('fconnection.php');
 
$sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error());
$row3 = mysql_fetch_array($sql); 
?>
<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>

This will just print one city link (the first one) as you don't have a while loop. No text is shown with this link, just an image. Tooltip (alt) - try using 'title' attribute too.

Is this what you want? Seems wrong to query the whole tbl_franchise table and then just use the first record.

==============

<?php
$id = $_GET['id'];
 
$query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1";
 
$result = mysql_query($query);
 
if (!$result) {
echo "NO RECORD FOUND";
} else {
 
while($row3 = mysql_fetch_array($result)):
?>
 
Manager Name: <?php echo $row3['manager_name'];?>

I'm assuming this is coming from the previous script link. Again this makes little sense as you're ORDERING BY id, but limiting it to one id, which I assume is unique. Odd.

For this recordset, you one have one record, but then go on to loop over it with a 'while'. This makes no sense.

You seem to have mixed up the purpose of using a loop.

BTW: your get variable should be $_GET not $_GET

I recommend you have a look at the php.net site and learn the basics of loops (for, while, do)

[marases]

Hi, well first thing your passing the variable city to the franchiseDetails.php file, and asking for the id variable...

So just change this and it should work.

<a href="franchiseDetails.php?id=<?php echo $row3['id'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>

Thanks,
Marais