passing data using hyperlink in PHP

Question

I'm trying to pass the path of an image ['guest'] from one page to another using a link. (I'm storing URLs of images in database) Can't seem to get the image to display, which is a bigger image of 'url'. I'm doing it this way so that I can have a larger image displayed in the target page (does_this_work.php) plus adding some other bits on the page too.

I'm still learning and can;t seem to see what I'm doing wrong. Any help appreciated,

<?php 

$host=""; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name=""; // Table name 

// Connect to server and select databse.
mysql_connect($host, $username, $password)or die("cannot connect"); 
mysql_select_db($db_name) or die("cannot select DB");


$photo=mysql_query("SELECT * FROM `images` ORDER BY (ID = 11) DESC, RAND() LIMIT 7");

while($get_photo=mysql_fetch_array($photo)){ ?>

<div style="width:300px;">


<a href="does_this_work.php?big_image=$get_photo['guest']>" target=""><img src="<?     
echo $get_photo['url']; ?>" title="">


 </div>

 <? } ?>

I then use the following code to try and display the array data in the target file

<?php

echo "this is the page where you get a larger picture of image on previous page, plus     
further info";

$big_image = $_GET['guest'];

echo $big_image;

?>

while there are many other things wrong with this: $_GET['big_image'] not $_GET['guest'] — user557846
– user557846, Commented Feb 29, 2012 at 22:11

BenOfTheNorth · Accepted Answer · 2012-02-29 22:10:59Z

1

You're missing an opening tag in here (And a closing semi colon, but that's not as problematic here):

<a href="does_this_work.php?big_image=$get_photo['guest'] ?>"

Change to:

<a href="does_this_work.php?big_image=<?= $get_photo['guest']; ?>"

answered Feb 29, 2012 at 22:10

BenOfTheNorth

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Comments

Anton · Accepted Answer · 2012-02-29 22:32:49Z

There are several errors in your code. First of all, this is how you use $_GET and $_POST:

ex. site.php?argument=value

To retrieve the value of argument, you need this code in site.php:

//The variable must not necessarily be $value
$value = $_GET['argument'];
//Alt.
$value = $_POST['argument'];

Secondly (like the other answers tell you) you are missing a php-tag here:

<a href="does_this_work.php?big_image=$get_photo['guest']>" target=""><img src="<? echo $get_photo['url']; ?>" title="">

Instead it should be:

<a href="does_this_work.php?big_image=<?php echo $get_photo['guest']; ?>" target=""><img src="<?php echo $get_photo['url']; ?>" title="">

Now, in order to make that compatible with your second code, you need to change the argument sent to guest like this:

<a href="does_this_work.php?guest=<?php echo $get_photo['guest']; ?>" target=""><img src="<?php echo $get_photo['url']; ?>" title="">

OR change the $_GET['guest']; to $_GET['big_image'];

I think I got it all right..

thanks for that. Have made changes but still not getting the image displayed on does_this_work.php. Have tried putting
Does $big_image; include the <img>-tag? If not, have you added it to the echo? (Talking about your second script now.)
thanks for that. Can't tell you how long I've been on that for!cheers

Ken Wheeler · Accepted Answer · 2012-02-29 22:10:45Z

0

<a href="does_this_work.php?big_image=$get_photo['guest'] ?>" target=""><img src="<?     
echo $get_photo['url']; ?>" title="">

You are missing your php starting tags. This should read:

<a href="does_this_work.php?big_image=<? $get_photo['guest'] ?>" target=""><img src="<?     
echo $get_photo['url']; ?>" title="">

answered Feb 29, 2012 at 22:10

Ken Wheeler

1,95813 silver badges12 bronze badges

3 Comments

Wrikken Over a year ago

<? $get_photo['guest'] ?>, what do you expect that to 'do'?

Ken Wheeler Over a year ago

<?= Oops. Yeah that usually helps.

user1022772 Over a year ago

Cheers for your help on that.

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