PHP:只能通过引用传递变量

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问题描述

我在第 57 行收到此错误:$password = str_replace($key, $value, $password, 1);

I am getting this error on line 57: $password = str_replace($key, $value, $password, 1);

据我所知，我只是传入变量.以下是更多背景信息:

As far as I can tell, I am only passing in variables. Here is some more context:

$replace_count = 0;
foreach($replacables as $key => $value)
{
    if($replace_count >= 2)
        break;
    if(strpos($password, $key) !== false)
    {
        $password = str_replace($key, $value, $password, 1);
        $replace_count++;
    }

}

推荐答案

您不能传递 1 的常量，解决方法是将其设置为变量.

You can't pass a constant of 1, a fix is to set it to a variable as so.

变化:

$password = str_replace($key, $value, $password, 1);

到:

$var = 1
$password = str_replace($key, $value, $password, $var);

更新: 更改为在方法调用之外根据评论中的反馈声明变量.

UPDATE: Changed to declare variable outside of the method call from feedback in comments.

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