if语句不适用于php中的false [英] The if statement doesn't work for false in php
问题描述
当试图熟悉PHP中的if语句时，发生了这种情况.
第一次我在下面尝试此代码.
When trying to get familiar with if statement in PHP, this happened.
First time i tried this code below.
if(true) {echo 'true';} else {echo 'false';}
当条件为true时，输出为true.同样，当条件为false(if(false))时，它将回显false.
And the output was true when the condition is true. Again, when the condition is false (if(false)) it echos false.
但是我尝试了同样的方法，同时使用变量作为条件，同时更改了变量的值.
But i tried the same, using a variable as the condition, while changing the value of the variable.
$con='false';
if($con){echo 'true';} else{echo 'false';}
在这种情况下，即使变量值为false或true，输出也为true.同时，当使用1和0代替true和false时，if statement可以正常工作.为什么会这样?
At this situation the output is true even when the variable value is false or true. At the same time, the if statement working fine when 1 and 0 is used instead true and false. Why is this happening?
推荐答案
PHP在if表达式中做了一些偷偷摸摸的事情.以下值被认为是FALSE:
PHP does some sneaky things in the if expression. The following values are considered FALSE:
- 布尔值FALSE本身
- 整数0(零)
- 浮点数0.0(零)
- 空字符串和字符串"0"
- 具有零个元素的数组
- 具有零成员变量的对象(仅PHP 4)
- 特殊类型NULL(包括未设置的变量)
- 从空标签创建的SimpleXML对象
其他所有值都被视为TRUE(包括任何资源).
Every other value is considered TRUE (including any resource).
您实际上传递的是一个字符串，该字符串表示单词false，而不是值false.因为不在上面的列表中，所以实际上被认为是正确的！
You're actually passing a string that says the word false, rather than the value false itself. Because that isn't in the above list, it is actually considered true!
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