此问题仅供参考。如何在PHP中检查变量是否已设置且is_null
我想知道如果有一种方法来检查,如果一个变量is_set,如果是检查是否被设置为null
基础上PHP type comparison tables不好像这是可能的。
例如,假设我们有这样的代码:
<?php
$x = null;
if(isset($x) && is_null($x)){
echo '$x is set to NULL';
}else{
echo '$x was never initialized';
}
echo "\n";
if(isset($y) && is_null($y)){
echo '$y is set to NULL';
}else{
echo '$y was never initialized';
}
echo "\n";
if(is_null($z)){
echo '$z is set to NULL';
}else{
echo '$z was never initialized';
}
?>
我期望的页面显示:
$x is set to NULL
$y was never initialized
$z is set to NULL <it should give an E_NOTICE>
但我正在逐渐
$x was never initialized
$y was never initialized
$z is set to NULL
我正在分配一个值,该值为NULL,意味着'is_null($ x)'应该返回true以及'isset($ x)',因为变量是SET且值为NULL – Fabrizio