php if variable1 is not in array or if variable1 is not variable2 then echo

Question

The first part before || works on a standalone basis, but the second added (OR) second condition doesn't perform.

eg. $category="Cycle", or ($category) Cycle is not ($primaryName) Running I would expect to echo Running;

eg $category="Surf" echo nothing;

eg $category="Tech" and $primaryName="Tech" echo nothing;

eg $category="Tech" and &primaryName="Clean" echo "Clean";

What am I doing wrong here?

<?php $categorycheck = array("Surf", "Sail", "Driving", "Flying");

if ((! in_array($category, $categorycheck)) || ($category != $primaryName))
{
    echo '<li>' . $primaryName . '</li>';
}
?>

It looks like you need AND instead of OR inside your if statement. — Mutale
– Mutale, Commented Feb 5, 2014 at 21:10
2 times -if not- and -or or and- or got me confused. Need a coffee! Thanks Mutale! — Peter van der Lely
– Peter van der Lely, Commented Feb 5, 2014 at 21:16

Relu Motica · Accepted Answer · 2014-02-05 21:15:12Z

2

Replace || with &&, otherwise it will evaluate as true every time $category is not an element of $categorycheck.

answered Feb 5, 2014 at 21:15

Relu Motica

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Comments

Peter van der Lely · Accepted Answer · 2014-02-05 21:18:43Z

2

thanks to Mutale, the working code is:

<?php $categorycheck = array("Surf", "Sail", "Driving", "Flying");

if ((! in_array($category, $categorycheck)) && ($category != $primaryName))
{
echo '<li>' . $primaryName . '</li>';
}
?>

answered Feb 5, 2014 at 21:18

Peter van der Lely

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