Question

给出以下示例：

public function replaceMyText($search, $replace, &$content)
{
    $newContent = str_replace($search, $replace, $content, $count = 1)
    $content = $newContent;
}

这会导致Warning只能通过引用传递变量吗？如果是这样，我无法完全理解原因。

在将其传递给$content函数之前，应该将str_replace分配给另一个变量吗？

Answer 1

编辑

请尝试以下代码：

function replaceMyText($search, $replace, $content)
{
    $newContent = str_replace($search, $replace, $content,$count=1);
    return $newContent;
}

$searchValue = "test";
$replaceWith = "magic trick";
$actualContent = "This is a test.";

$replaced = replaceMyText($searchValue,$replaceWith,$actualContent);

echo $replaced;

Answer 2

<?php
function replaceMyText($search, $replace, &$content)
{
    $newContent = str_replace($search, $replace, $content, $count = 1);
    $content = $newContent;
}

replaceMyText("123", "456", "123456");

在不使用变量的情况下使用此功能会导致致命错误

Fatal error: Only variables can be passed by reference in /usercode/file.php on line 8

因为

不应通过引用传递其他表达式，因为结果是不确定的。

来自http://php.net/manual/en/language.references.pass.php

您只能以此使用

$a = "123456";
replaceMyText("123", "456", $a);
echo $a;

对不起，我英语不好。希望它能对您有所帮助。

通过引用传递主题的Str_replace

2 个答案: