PHP if(value !== (value OR value)){}

Question

I'm trying to create an if() statement to catch any unwanted tampering or coding errors.

All parts are strings.

I've tried correct and incorrect values in the array, but it seams to default to true every time.. outputting the error.

if ($array['field'] !== ('b' || 'c')){
    echo 'An error has been encounterd <br/>';
    exit();
}

Any ideas? thanks

Quentin · Accepted Answer · 2014-02-12 10:27:24Z

5

You're comparing $array['field'] to the result of 'b' || 'c'.

You need to compare it to each of the things you want to match against.

You have to say "If the thing is not 'b' and it is also not 'c'".

if ($array['field'] !== 'b' && $array['field'] !== 'c'){

answered Feb 12, 2014 at 10:27

Quentin

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1 Comment

Brilliant! Big green tick is coming your way when the system allows

Zaheer Ahmed · Accepted Answer · 2014-02-12 10:26:58Z

2

try, this way:

if ($array['field'] !== 'b' && $array['field'] !==  'c'){

answered Feb 12, 2014 at 10:26

Zaheer Ahmed

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That's never going to be true. That shouldn't be an ||.

cornelb · Accepted Answer · 2014-02-12 10:27:49Z

1

if (!in_array($array['field'], array('b', 'c'))) {
    // ...
}

answered Feb 12, 2014 at 10:27

cornelb

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