Avoid "undefined index" error when the variable does not exist

Question

I have a panel.php in which I load my different pages by the include method. As there is a single panel, the message display system must be included here, and there will not always an error id to be displayed.

I have added the following code to my panel, calling function "envioMensaje" in case there is getting any variable 'msg':

<?php
if ($_GET) {
    if ($_GET['msg']) {
        $id_mensaje=$_GET['msg'];
        $mensaje=envioMensaje($id_mensaje);
        echo $mensaje;
    }
}
?>

This works fine when a message id is sent, but when not I get a "Undefined index: msg in..." error. I have tried also:

if ($_GET) {
    if ($_GET['msg']) {
        $id_mensaje=$_GET['msg'];
        $mensaje=envioMensaje($id_mensaje);
        echo $mensaje;
    } else {
        echo"";
    }
}

And some other variations with no result. Why is it always looking for 'msg'? Could it be because on that same document I have this other conditional asking for $_GET? (i need it for the content to be loaded):

if (!$_GET) {
    include('config/shortcuts.php');
} else {    
    $directorio = $_GET['directory'];
    include('config/'.$directorio);
}

Community · Accepted Answer · 2023-11-17 19:24:28Z

4

Use isset() to check if a variable exists without throwing a PHP notice.

if (isset($_GET['msg'])) {
    //then use $_GET['msg']
}

edited Nov 17, 2023 at 19:24

CommunityBot

11 silver badge

answered Feb 23, 2014 at 20:09

deviantfan

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1 Comment

TecBrat Over a year ago

That's the answer I was going to give. You beat me to it. +1

Kamehameha · Accepted Answer · 2014-02-23 20:10:21Z

1

Check for

if(isset($_GET['msg'])){...}

answered Feb 23, 2014 at 20:10

Kamehameha

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