How to: Click a single blog post and display it on a separate page

Question

So I am creating a blog for a school project, what I want to do is to be able the click on a blog post on the start page and be redirected to a separate page where only that blog post is displayed. What I have tried is this:

START PAGE LINK:

<a class="blog_title_link" href="singlepost?ID=' . $row['postID'] . '">

AND THE OTHER PAGE:

$pID = -1;
if ($_GET['ID'] != NULL) {
    $pID = $_GET['ID'];
} else {
    echo "ID failed.";
    //Redirect to startpage...
}

$con = mysqli_connect("localhost","root","hejsan123","MyStrength");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}    

$result = mysqli_query($con,"SELECT * FROM posts WHERE postID=") . $pID;

while($row = mysqli_fetch_array($result))
{
    echo '<h1 class="blogTitle"><a class="blog_title_link">' 
   . $row['title'] . '</a></h1><img src="yo.jpg" /><p class="content_box_p">' 
   . $text . '</p>';
   echo "<br>";
}               
mysqli_close($con);

So I have the $pID to check if there actually is a ID, and then I try to get it from the URL so that I can use it in the SQL to choose which information I want from the database. What happens rigth now is

404 not found.

Hope you understand! Thanks!

Answer 1

the 404 Not Found error could it be because of the link:

<a class="blog_title_link" href="singlepost?ID=' . $row['postID'] . '">
                                 __________^___      

Maybe you need singlepost.php?ID...?

And then this line too should be:

$result = mysqli_query($con,"SELECT * FROM posts WHERE postID=" . $pID); 
//                                                       ___not^___but^