php 不能在写上下文中使用函数返回值？
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Can't use function return value in write context?
提问by Sophie Mackeral
I have this code:
我有这个代码：
public function __construct($Directory = null)
{
if ($Directory === null) {
trigger_error("Directory Must Be Set!", E_USER_ERROR);
}
if (isset($Directory)) {
if (!empty(trim($Directory))) { //Error Line
echo "test";
}
}
}
(The echo is for my debugging purposes.)
（回声是为了我的调试目的。）
I get returned with the fatal error:
我返回致命错误：
Can't use function return value in write context
不能在写上下文中使用函数返回值
According to PHP storm this returns:
根据 PHP 风暴，这将返回：
Variable Expected
可变预期
But using the code directly from this question:
但是直接使用这个问题的代码：
White spaces throwing off HTML form validation
This is the correct syntax, as i've used it in the past... But, in this situation this is throwing an error. Why is this?
这是正确的语法，因为我过去使用过它......但是，在这种情况下，这会引发错误。为什么是这样？
回答by Ken Herbert
From the PHP docs on the emptyfunction:
来自该empty函数的 PHP 文档：
Note: Prior to PHP 5.5, empty() only supports variables; anything else will result in a parse error. In other words, the following will not work: empty(trim($name)). Instead, use trim($name) == false.
注意：PHP 5.5 之前，empty() 只支持变量；其他任何事情都会导致解析错误。换句话说，以下将不起作用：empty(trim($name))。相反，使用trim($name) == false。
So you will have to split that line into something like the following:
因此，您必须将该行拆分为如下所示的内容：
$trimDir = trim($Directory);
if(!empty($trimDir))
