不能在 mysqli_stmt 对象上使用 call_user_func_array

2021-12-25 00:00:00 php mysqli

我正在为 MySQLi 编写一个包装类.在那里，我正在编写一个函数来接受查询和可变数量的参数，我可以在其中调用 mysqli_stmt::bind_param.代码如下:

I am writing a wrapper class for MySQLi. In there, I am writing a function to accept a query and variable number of parameters where I can invoke mysqli_stmt::bind_param. Here is the code:

<?php
    class DbHelper {
        ....
        public function Execute($query, $params){
            $this->open(); # Opens a connection to the database using MySQLi API
            $stmt = $this->mysqli->prepare($query);
            try{
                $result = call_user_func_array(array($stmt, 'bind_param'), $params);
            }
            catch(Exception $ex){
                # Handle Exception
            }
        }
        ....
    }
?>

这是我调用函数的方式:

Here's how I am calling the function:

<?php
    $db = new DbHelper();
    $params = array('i', $stateID);
    $result = $db->Execute('SELECT * FROM mst_cities WHERE State_ID = ?', $params);    
?>

当我运行代码时，我收到如下警告:
警告:mysqli_stmt::bind_param() 的参数 2 应为引用，值在......中给出.

When I run the code, I get an warning as this:
Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given in......

我该怎么办?

推荐答案

来自 文档:

注意:

将 mysqli_stmt_bind_param() 与 call_user_func_array() 结合使用时必须小心.请注意，mysqli_stmt_bind_param() 需要通过引用传递参数，而 call_user_func_array() 可以接受可以表示引用或值的变量列表作为参数.

Care must be taken when using mysqli_stmt_bind_param() in conjunction with call_user_func_array(). Note that mysqli_stmt_bind_param() requires parameters to be passed by reference, whereas call_user_func_array() can accept as a parameter a list of variables that can represent references or values.

评论中有一些解决方法，例如 看到这个:

There are some workarounds in the comments, for example see this:

call_user_func_array(array($stmt, 'bind_param'), refValues($params));

function refValues($arr)
{
    if (strnatcmp(phpversion(),'5.3') >= 0) //Reference is required for PHP 5.3+
    {
        $refs = array();
        foreach($arr as $key => $value)
            $refs[$key] = &$arr[$key];
        return $refs;
    }
    return $arr;
}

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