CODEWITHSUNDEEP
phphtml
user1022772
user1022772

Reputation: 651

passing data using hyperlink in PHP

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;

?>

Upvotes: 1

Views: 943

Answers (3)

Anton
Anton

Reputation: 1515

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..

Upvotes: 0

BenOfTheNorth
BenOfTheNorth

Reputation: 2872

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']; ?>"

Upvotes: 1

Ken Wheeler
Ken Wheeler

Reputation: 1958

<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="">

Upvotes: 0

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