将数组分配给静态函数变量php

I have a replace callback method where I am trying to replace multiple string occurrences with array values accordingly.

我有一个替换回调方法，我尝试用数组值替换多个字符串。

I have passed the $parametersArray to the callback method through the use keyword as follows (the regex matches 3 elements):

我已经通过use关键字(regex匹配3个元素)将$parametersArray传递给回调方法:

$string = 'Welcome Mr MM1, MM2 MM3 to the website';
$parametersArray = array('K', 'A' , 'AD');

$line = preg_replace_callback(
    '(MM[1-9])',
    // anonymous
    function () use ($parametersArray) {
        static $paramArray = $parametersArray;
        return array_shift($paramArray);
    },
    $string
);

I am getting the below error:

我得到了以下错误:

Parse error: syntax error, unexpected '$parametersArray' (T_VARIABLE)

语法错误，意外的'$parametersArray' (T_VARIABLE)

If I set the array to the static variable explicitly, I do not get the error and get the expected behavior.

如果我显式地将数组设置为静态变量，就不会得到错误并得到预期的行为。

Is there a problem with assigning the array as variable directly to the statically defined variable within the function?

是否存在将数组作为变量直接分配给函数中静态定义的变量的问题?

3 个解决方案

#1

You can init static variable like this

可以像这样初始化静态变量

$line = preg_replace_callback(
    '(MM[1-9]{1})',
    // anonymous
    function ($matches) use ($parametersArray) {
        static $paramArray;

        if (null === $paramArray) {
            $paramArray = $parametersArray;
        }

        return array_shift($paramArray);
    },
    $string
);

#2

As per the docs (See example #6): You cannot initialize a static variable with an expression:

根据文档(参见示例#6):不能用表达式初始化静态变量:

<?php
function foo(){
    static $int = 0;          // correct 
    static $int = 1+2;        // wrong  (as it is an expression)
    static $int = sqrt(121);  // wrong  (as it is an expression too)

    $int++;
    echo $int;
}

#3

You can reference the array and use it original values in the closure afterwards. Note that the parameters get initialized at definition not when you call it - so this might work for you:

您可以引用数组，然后在闭包中使用它的原始值。注意，参数在定义时被初始化，而不是当您调用它时——因此这可能对您有用:

<?php

$string = 'Welcome Mr MM1, MM2 MM3 to the website';

$parametersArray = array('K', 'A' , 'AD');
$refArray =& $parametersArray;
$line = preg_replace_callback(
            '(MM[1-9]{1})',
            // anonymous
            function () use (&$refArray, $parametersArray) {
                # stays the same in every call
                # it is a copy after all
                print_r($parametersArray);
                return array_shift($refArray);
            },
            $string
);

echo $line;
# Welcome Mr K, A AD to the website
?>

#1