PHP-检查变量是否未定义

问题描述：

考虑此jquery语句

Consider this jquery statment

isTouch = document.createTouch !== undefined

我想知道我们是否在PHP中有类似的语句，不是isset()，而是从字面上检查未定义的值，例如:

I would like to know if we have a similar statement in PHP, not being isset() but literally checking for an undefined value, something like:

$isTouch != ""

PHP中是否存在与上述类似的东西?

Is there something similar as the above in PHP?

答

您可以使用-

$isTouch = isset($variable);

如果定义了$variable，则将存储true，否则将存储false.如果需要反演，只需删除!.

It will store true if $variable is defined else false. If need the invers the simply remove the !.

注意:如果var存在并且具有非NULL的值，则返回TRUE，否则返回FALSE.

Note : Returns TRUE if var exists and has value other than NULL, FALSE otherwise.

或者如果您要检查false，0等，请使用empty()-

Or if you want to check for false, 0 etc also then use empty() -

$isTouch = empty($variable);

empty()适用于-