statement evaluate to true?
if ($opts['width'] == 'fs' || $opts['height'] == 'fs' || $opts['ratio'] == 'fs') {
var_dump($opts); // result of this see bellow
}
Result of var_dump($opts)
inside (!) if-statement:
array(3) {
'width' => int(200)
'height' => int(0)
'ratio' => int(0)
}
How is this possible? None of the array's values is (stirng) fs?
Because 0 == 'fs'
. See this conversion table .
PHP has the ===
operator to compare both value and type.
For a more extensive table: Type-juggling and (strict) greater/lesser-than comparisons in PHP
链接地址: http://www.djcxy.com/p/58516.html上一篇: 字符串'w32'== 0在php中评估为true。 是吧?
下一篇: 声明评估为真?