clicking images for their detail page - php

Question

I'm trying to enable a user to go to a post detail page. They can find these images after a search. But I always get this:
Warning: Illegal string offset 'postID' in C:\xampp\htdocs\eindwerk2\index.php on line 106
Warning: Illegal string offset 'postID' in C:\xampp\htdocs\eindwerk2\index.php on line 180
this is the php code. The seach function works btw.
$_SESSION['KEYWORD'] = array();
$error = '';
//$allResults2 = array();
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 INNER JOIN user ON posts.postUser = user WHERE postTags = :postTags";
$pdoQuery = "SELECT posts.postID, posts.postUser, posts.postImageUrl, posts.postDescription, user.firstname, user.lastname, user.userID FROM posts INNER
JOIN user ON posts.postUser=user.email WHERE postTags = :postTags;";
$pdoResult = $pdoConnect->prepare($pdoQuery);
$pdoExec = $pdoResult->execute(array(":postTags" => $postTags));
if(!empty($_POST)){
while ($row = $pdoResult->fetch(PDO::FETCH_ASSOC)) {
$_SESSION['KEYWORD'][] = $row['postImageUrl']['postID'];
//$poster = $row['postID'];
//$allResults2[] = $row['postUser'];
//echo $postTags;
}
}
else {
echo 'no';
}
if(count($_SESSION['KEYWORD']) === 0) {
$error = "Sorry no results!";
}
This is the html
<div class="search">
<?php
foreach ($_SESSION['KEYWORD'] as $imageLink){
$postid = $imageLink['postID'];
echo '';
}
/*foreach ($allResults2 as $imageUser){
echo '$imageUser';
}*/
?>
<div class="searchError"><?php echo $error; ?></div>
</div>

Answer 1

Your error is from this line of code:
$_SESSION['KEYWORD'][] = $row['postImageUrl']['postID'];
This isn't legal syntax in PHP. You can reproduce it with:
<?php
$test = array(
test1 => "Hello",
test2 => "hello2"
);
echo $test["test1"]["test2"];
?>
When you try to execute this code, you get:
PHP Warning: Illegal string offset 'test2' in
Sadly, I'm not that sure what you try to achieve with this line of code. Anyway, your SQL result is a one-dimensional array. Your code acts like you work with a two dimensional array. You can only use $row["postImageUrl"] or $row["postID"] you you can't use both directly.
If you describe a bit better what you try to do with this line, I can help you. But actually I have no idea what your thoughts were.