Issue passing variable using isset()

Question

Would really appreciate it if someone could point me in the right direction.

I am starting to learn PHP again as I have not touched it for years.

Below is a snippet of code I have wrote for a Message System.

<?php


$stmt = $user_home->runQuery("SELECT * FROM tbl_messages WHERE recipientID=:uid");
$stmt->execute(array(":uid"=>$_SESSION['userSession']));
while ($qryMessages = $stmt->fetch(PDO::FETCH_ASSOC))

    {
    $mID = $qryMessages['msgID'];
    $mTitle = $qryMessages['msgTitle'];


        echo "<td><a href='?read&mID=$mID'>" . $mTitle . "</a></td>";
    }

?>

Some HTML here etc....

<?php 

if(isset($_GET['read&mID=' . $mID]))
 {
    echo $mID
 }
?>

I want to be able to show the message on the same page by using isset() but cannot seem to pass the $mID variable to it when its clicked, I keep receiving this error message:

Notice: Undefined variable: mID in D:\xampp\htdocs\development\template\home\messages.php on line 234

Any feedback would be greatly appreciated.

Aniruddha Chakraborty · Accepted Answer · 2016-07-10 16:50:39Z

2

You have to use isset like this

<?php 

if(isset($_GET['mID']))
 {
    echo $_GET['mID'];
 }
?>

PHP manual

answered Jul 10, 2016 at 16:50

Aniruddha Chakraborty

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Comments

Indrasis Datta · Accepted Answer · 2016-07-10 16:50:41Z

1

In the $_GET array, you can specify the query string parameter keys. Here, the query string keys are read and mID. Please note that $mID is the value, so we need not mention it in the isset() function.

You could replace this with:

if (isset($_GET['mID'])) {
   echo $_GET['mID'];                 // This will print the value of $mID
}

answered Jul 10, 2016 at 16:50

Indrasis Datta

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1 Comment

James Over a year ago

Thank you for your explaination, completely see where I've gone wrong. Makes perfect sense to me now. Thanks again

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