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Resolving a PHP Notice Error

+1 vote

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

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?

posted Sep 17, 2013 by Deepak Dasgupta

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1 Answer

0 votes

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

( !isset( @$sortbypreference ) )
answer Sep 17, 2013 by Salil Agrawal
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