cant display two image links from mysql php - php

Question

function find_image_by_id() {
global $connection;
$query = "SELECT * ";
$query .= "FROM images ";
$query .= "WHERE page_id={$_GET["page"]}";
$image_set = mysqli_query($connection, $query);
confirm_query($image_set);
return $image_set;
}
function display_image_by_id(){
$current_image = find_image_by_id();
while($image=mysqli_fetch_assoc($current_image)){
$output = "<div class=\"images\">";
$output .= "<img src=\"images/";
$output .= $image["ilink"];
$output .= "\" width=\"72\" height=\"72\" />";
$output .= $image["phone_name"];
$output .= "</div><br />";
}
mysqli_free_result($current_image);
return $output;
}
This is the code I'm using to show the images stored as links in mysql
and the images are in a folder. But what happens after this code is executed only the second
value is displayed. I want both value/ images to be displayed.

Answer 1

Try something like that-
All you need to do is just initialize this variable outside the loop.
$output =''; //initialize before
SO your function look like this -
function display_image_by_id(){
$current_image = find_image_by_id();
$output =''; //initialize before
while($image=mysqli_fetch_assoc($current_image)){
$output .= "<div class=\"images\">";
$output .= "<img src=\"images/";
$output .= $image["ilink"];
$output .= "\" width=\"72\" height=\"72\" />";
$output .= $image["phone_name"];
$output .= "</div><br />";
}
mysqli_free_result($current_image);
return $output;
}