quant state for value in PHP array?

Asked Sep 21 '18 at 15:31

Active Sep 21 '18 at 15:31

Viewed 24 times

i have a code like this:

$a = array();
$i = 0;
$a['p'] = !isset($a['p']) ? 0 : $a['p'];
$b = ($i==0 && isset($a['p']) && $a['p'] == 'test') ? 'test' : $i+1;

$i = 1;
$a['p'] = !isset($a['p']) ? 0 : $a['p'];
$c = ($i==0 && isset($a['p']) && $a['p'] == 'test') ? 'test' : $i+1;

echo $a.' '.$b;

I`m waiting for 1 in $b, but in fact i have a "test" in $b and 2 in $c How it is possible ?

PS. PHP 5.4.16 @ CentOS 7

asked Sep 21 '18 at 15:31

questor

cos $a['p'] != 'test' and $a['p'] not a string at all – questor Sep 21 '18 at 15:36
it happens because php uses integer comparison, ie `0 == intval('test')` – Iłya Bursov Sep 21 '18 at 15:37
1

`$a['p']` is 0 and when you compare strings and ints some unexpected conversions happen. – u_mulder Sep 21 '18 at 15:37
understood, thanks all for answers, in this case we have to use === for comparison – questor Sep 21 '18 at 15:40

quant state for value in PHP array?

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