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if $variable !=('term1','term2','term3')

Posted on 2011-03-17
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Last Modified: 2012-05-11
if $variable !=('term1','term2','term3')

is this the correct syntax
0
Comment
Question by:rgb192
  • 3
  • 3
  • 2
  • +1
9 Comments
 
LVL 8

Expert Comment

by:Rik-Legger
ID: 35159226
if ($variable != 'term1' OR $variable != 'term2' OR $variable != 'term3') {
    // 
}

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Expert Comment

by:enachemc
ID: 35159241
if ($variable != 'term1' || $variable != 'term2' || $variable != 'term3') {
    //
}

|| is or
&& is and
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Author Comment

by:rgb192
ID: 35159258
i have 100 terms
can I do a list or array
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LVL 8

Expert Comment

by:Rik-Legger
ID: 35159284
$array = array(
    'term1',
    'term2',
);

if (! in_array($variable, $array)) {
    //
}

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LVL 12

Expert Comment

by:enachemc
ID: 35159285
$a = array('1.10', 12.4, 1.13);

if (in_array('12.4', $a, true)) {
    echo "'12.4' found with strict check\n";
}


http://php.net/manual/en/function.in-array.php
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LVL 12

Expert Comment

by:Avinash Zala
ID: 35162723
will go with enachemc, +1 for this...
0
 

Author Comment

by:rgb192
ID: 35167686
is there a way to do a for loop

because I would have to repeat myself in the if...
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Accepted Solution

by:
Rik-Legger earned 500 total points
ID: 35167783
You mean like this?

<?php 

$array = array(
    'term1',
    'term2',
);

foreach ($array as $value) {
	if ($value != $variable) {
		//
	}
}

?>

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Author Closing Comment

by:rgb192
ID: 35245430
thanks
0

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