Home Backend Development PHP Tutorial empty跟isset的解决之道 - PHP变量是否定义的判断

empty跟isset的解决之道 - PHP变量是否定义的判断

Jun 13, 2016 pm 01:07 PM
empty isset null return undefined

empty和isset的解决之道 - PHP变量是否定义的判断

当访问一个未定义的变量时,PHP会产生警告

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使用之前可以用empty();或 isset()来判断变量是否定义

$arr['k1']; 尽管$arr是未定义的,还是可以用用这两个函数来判断

如果使用 如果使用if判断,将会产生? Undefined variable 的警告

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一个变量的未定义与和将变量设为null是不同的,用unset()会将变量置为未定义

?

函数中无return或return;调用此函数得到结果为NULL

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