Output a series of images from a db into a variable? - php

Question

I have a number of images in my db and they all correspond to the same project ID.
I want them to be displayed next to each other in the browser.
But the following code is output only the latest image but not all of them:
//get all the images in this project and add them to the content variable for output
if (!empty($FormProjectID)) {
$DBQuery3 = mysqli_query($dblink, "SELECT * FROM images WHERE project_id = '$FormProjectID'");
if (mysqli_num_rows($DBQuery3) < 1) {
$content = '
<p>This project is empty. Upload some files to get started.</p>
';
} else {
while($row = mysqli_fetch_array($DBQuery3)) {
$DBImageID = $row['image_id'];
$DBProjectID = $row['project_id'];
$DBImageName = $row['image_name'];
$DBImageDescription = $row['image_description'];
$DBDateCreated = $row['date_created'];
$DBLinkToFile = $row['link_to_file'];
$DBGivenName = $row['given_name'];
//if the image was given a name by the user, display it
//otherwise display the generated name
if (strlen($DBGivenName) > 1) {
$FileName = $DBGivenName;
} else {
$FileName = $DBImageName;
}
$content = '
<div class="image">
<img src="'.$DBLinkToFile.'" alt="'.$FileName.'" title="'.$FileName.'"/>
</div>
';
}
}
How do I get all the image so that in the end the html looks like this:
<div class="image">
<img src="'.$DBLinkToFile.'" alt="'.$FileName.'" title="'.$FileName.'"/>
</div>
<div class="image">
<img src="'.$DBLinkToFile.'" alt="'.$FileName.'" title="'.$FileName.'"/>
</div>
<div class="image">
<img src="'.$DBLinkToFile.'" alt="'.$FileName.'" title="'.$FileName.'"/>
</div>
Later on in my html page I have:
<?php echo $content; ?>

Answer 1

change $content = ' to $content .= '
PHP String Operators
Note: set the $content variable to empty string or null before your while loop

Answer 2

What you're looking for is string concatenation (see string operators)
to fix your problem you'll need to first initialise an empty string, do this before your while loop
} else {
$content = '';
while($row = mysqli_fetch_array($DBQuery3)) {
and then use the concatenation operator . to append each image
$content .= '
<div class="image">
<img src="'.$DBLinkToFile.'" alt="'.$FileName.'" title="'.$FileName.'"/>
</div>
';