这个if语句如何评估为true？

Question

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?

Answer 1

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