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Editorial Team
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Asked: 2026-05-25T20:41:26+00:00

I’m using this line: ‘.((isset($imgRight) && in_array(1, $imgRight)) ? ‘checked=checked’ : ”).’ and, in

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I’m using this line:

'.((isset($imgRight) && in_array(1, $imgRight)) ? 'checked="checked"' : '').'

and, in some cases, $imgRight can be false. That’s why there’s isset(), but it still fails.

What do I need to do to avoid this warning?

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1 Answer

  1. Editorial Team

    Just because something is false doesn’t mean it’s not set:

    $foo = false;
isset($foo); //true

    You can just use:

    ($imgRight && in_array(1, $imgRight)) ? 'checked="checked"' : '')

    or to be very safe (if imgRgiht might be null, or some non-falsey value that isn’t an array):

    ((!empty($imgRight) && is_array($imgRight) && in_array(1, $imgRight)) ? 'checked="checked"' : '')
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