PHP 5.3.1 传递引用问题

问题描述

好吧，这是一个奇怪的问题，所以请耐心听我解释.

Ok, this is a weird problem, so please bear with me as I explain.

我们将开发服务器从 PHP 5.2.5 升级到 5.3.1.

We upgraded our dev servers from PHP 5.2.5 to 5.3.1.

在切换后加载我们的代码，我们开始收到如下错误:

Loading up our code after the switch, we start getting errors like:

警告:mysqli_stmt::bind_param() 的参数 2 应为参考，值在/home/spot/trunk/system/core/Database.class.php 第 105 行中给出

提到的(105)行如下:

the line mentioned (105) is as follows:

call_user_func_array(Array($stmt, 'bind_param'), $passArray);

我们将该行更改为以下内容:

we changed the line to the following:

call_user_func_array(Array($stmt, 'bind_param'), &$passArray);

此时(因为allow_call_time_pass_reference)关闭，php抛出这个:

at this point (because allow_call_time_pass_reference) is turned off, php throws this:

已弃用:/home/spot/trunk/system/core/Database.class.php 中的第 105 行已弃用调用时传递引用

在尝试解决此问题一段时间后，我崩溃了并将 allow_call_time_pass_reference 设置为开启.

After trying to fix this for some time, I broke down and set allow_call_time_pass_reference to on.

这摆脱了 Deprecated 警告，但现在 Warning: Parameter 2 to mysqli_stmt::bind_param() 预计将成为参考 警告每次都抛出，有或没有引用.

That got rid of the Deprecated warning, but now the Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference warning is throwing every time, with or without the referencing.

我对如何解决这个问题毫无头绪.如果目标方法是我自己的，我只会在 func 声明中引用传入的变量，但它是一个(相对)本机方法 (mysqli).

I have zero clue how to fix this. If the target method was my own, I would just reference the incoming vars in the func declaration, but it's a (relatively) native method (mysqli).

有人遇到过这种情况吗?我该如何解决?

Has anyone experienced this? How can I get around it?

谢谢.

推荐答案

您正在传递一个元素数组 ($passArray).传递的数组内的第二个项目需要是一个引用，因为那确实是您传递给函数的项目列表.

You are passing an array of elements ($passArray). The second item inside the passed array needs to be a reference, since that is really the list of items you are passing to the function.

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