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A tragedy for php arrays? _PHP Tutorial

WBOY
WBOYOriginal
2016-07-21 15:29:52845browse

Copy code The code is as follows:

$a=1;
$b=2;
$t = array(
array('a', 'string', $field['a']), // Name
if($a==$b){array('b','1')}
);
print_r($t);
exit;

1. Because the array is long. Didn't notice
2. The development code did not clear the cache in time. . . . The tragedy begins here
The following is the correct usage
$a==$b?array('sDesc','1'):array('sDesc','2')
Copy code The code is as follows:

$a=1;
$b=2;
$t = array(
array('a', 'string', $field['a']), // name
$a==$b?array('b','1'):array('b',' 2')
);
print_r($t);
exit;

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/323351.htmlTechArticleCopy the code The code is as follows: $a=1; $b=2; $t = array( array('a ', 'string', $field['a']), // name if($a==$b){array('b','1')} ); print_r($t); exit; 1 , due to the long length of the array. No note...
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