Function not returning data as expected - php

Question

I've got a function that should return me a set of links based on an user id. What the function does momentarily is that it returns me just one link instead a set of links based on an user id. The function looks like this:
function retrieve_image_link($user_id)
{
$query = mysql_query("SELECT `image_link` FROM `imgit_images` WHERE user_id = '" . intval($user_id) . "'");
while ($row = mysql_fetch_assoc($query))
{
$link = $row['image_link'] . '<br />';
}
mysql_free_result($query);
return $link;
}
So this is the code that should get me multiple links instead of just one. Where is the problem that the query returns a string instead of an array?
Please help!

Answer 1

you need to apply links to an array like $array[] = "link";
function retrieve_image_link($user_id)
{
$query = mysql_query("SELECT `image_link` FROM `imgit_images` WHERE user_id = '" . intval($user_id) . "'");
while ($row = mysql_fetch_assoc($query))
{
//add bracktes to $link to return an array
$link[] = $row['image_link'] . '<br />';
}
mysql_free_result($query);
return $link;
}
or use .= opperator if you want all the links in one big string
function retrieve_image_link($user_id)
{
$query = mysql_query("SELECT `image_link` FROM `imgit_images` WHERE user_id = '" . intval($user_id) . "'");
while ($row = mysql_fetch_assoc($query))
{
//add . to = to append string to $link
$link .= $row['image_link'] . '<br />';
}
mysql_free_result($query);
return $link;
}

Answer 2

It seems the problem is here
$link = $row['image_link'] . '<br />';
At the end of the loop, $link should have last value instead of all the values. You have to append each link or create an array to push each link.

Answer 3

$link = $row['image_link'] . '<br />';
You're overwriting the value of $link in the loop. Add [] to make it an array:
function retrieve_image_link($user_id)
{
$query = mysql_query("SELECT `image_link` FROM `imgit_images` WHERE user_id = '" . intval($user_id) . "'");
$link = array( );
while ($row = mysql_fetch_assoc($query))
{
$link[] = $row['image_link'] . '<br />';
}
mysql_free_result($query);
return $link;
}

Answer 4

use .=
function retrieve_image_link($user_id)
{
$query = mysql_query("SELECT `image_link` FROM `imgit_images` WHERE user_id = '" . intval($user_id) . "'");
while ($row = mysql_fetch_assoc($query))
{
$link .= $row['image_link'] . '<br />';
}
mysql_free_result($query);
return $link;
}