Want to post multiple search results - php

Question

I tried to make a search function, when your search a tag, its supposed to give you the images with that tag. When i just echo in the php code (line 101), it gives all the links. However, when i try to post it in the html, it only gives one result back.
php code
$postTags = "";
if (isset($_POST['Find'])) {
try {
$pdoConnect = new PDO("mysql:host=localhost;dbname=imdterest", "root", "");
} catch (Exception $exc) {
echo $exc->getMessage();
exit();
}
$postTags = $_POST['naam'];
$pdoQuery = "SELECT * FROM posts WHERE postTags = :tags";
$pdoResult = $pdoConnect->prepare($pdoQuery);
$pdoExec = $pdoResult->execute(array(":tags" => $postTags));
while ($row = $pdoResult->fetch(PDO::FETCH_ASSOC)) {
$postTags = $row['postImageUrl'];
//echo $postTags;
}
}
html code
<div class="search">
<img src="<?php echo $postTags; ?>">
</div>

Answer 1

the problem is that in your php code you have WHILE loop and you echo link for all found rows. That is why it prints all results. But also you don't store the values but overwrite it every while iteration. In your HTML you don't have while loop and you just print overwritten variable (probably last result).
You can either store the results in an array or move while loop to HTML location.
Storing into array:
$allResults = array();
while ($row = $pdoResult->fetch(PDO::FETCH_ASSOC)) {
$allResults[] = $row['postImageUrl'];
}
and then somewhere in your html code
<div class="search">
<?php
foreach ($allResults as $imageLink){
echo '<img src="' . $imageLink . '">";
}
?>
</div>

Answer 2

The variable $postTags is getting overwritten and will hold only last value of MySQL result set. Create an array and loop over it.
$postTags = array();
while ($row = $pdoResult->fetch(PDO::FETCH_ASSOC)) {
$postTags[] = $row['postImageUrl'];
}
In HTML:
foreach($postTags as $val)
{ ?>
<div class="search">
<img src="<?php echo $val; ?>">
</div>
<?php
} ?>

Answer 3

Change your while loop like this, As you are not using array next data will over write the previous and all you are left with one image link that is last image link. hence to prevent overriding use array.
$postTags=[];
while ($row = $pdoResult->fetch(PDO::FETCH_ASSOC)) {
$postTags[] = $row['postImageUrl'];
//echo $postTags;
}
And html code like this , As now you have multiple images you need to display all and for that you need to iterate and foreach is one method to iterate through array.
<?php
foreach ($postTags as $postTag) {
?>
<div class="search">
<img src="<?php echo $postTag; ?>">
</div>
<?php }?>