Resolving a PHP Notice Error

[Ron Piggott]

I am wanting to establish a default sort by preference when the user hasn’t specified one. I setup to test this with:

<?php

if ( !is_set( $sort_by_preference ) ) {

$sort_by_preference = "government_wording";

}

?>

But I am receiving a Notice error:

Notice: Undefined variable: sort_by_preference in GIFI_codes.php on line 11- Line 11 is “if ( !is_set( $sort_by_preference ) ) {“What is the correct way to test this without triggering a Notice error?Ron
Ron Piggott



www.TheVerseOfTheDay.info

[Daniel Brown]

On Tue, Sep 17, 2013 at 3:38 PM, Ron Piggott

Post by Ron Piggott
<?php
if ( !is_set( $sort_by_preference ) ) {

Did you create a function is_set(), or did you mean to use the
construct isset()?

[Sebastian Krebs]

Post by Ron Piggott
I am wanting to establish a default sort by preference when the user
<?php
if ( !is_set( $sort_by_preference ) ) {
$sort_by_preference = "government_wording";
}
?>
Notice: Undefined variable: sort_by_preference in GIFI_codes.php on line
11- Line 11 is “if ( !is_set( $sort_by_preference ) ) {“What is the correct
way to test this without triggering a Notice error?Ron
Ron Piggott

Because actually you are looking for isset() (not is_set()).

Post by Ron Piggott
www.TheVerseOfTheDay.info