問題描述

我正在為 MySQLi 編寫一個包裝類.在那里，我正在編寫一個函數(shù)來接受查詢和可變數(shù)量的參數(shù)，我可以在其中調(diào)用 mysqli_stmt::bind_param.代碼如下:

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
            }
        }
        ....
    }
?>

這是我調(diào)用函數(shù)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);    
?>

當我運行代碼時，我收到如下警告:
警告:mysqli_stmt::bind_param() 的參數(shù) 2 應(yīng)為引用，值在......中給出.

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......

我該怎么辦?

推薦答案

來自 文檔:

注意:

將 mysqli_stmt_bind_param() 與 call_user_func_array() 結(jié)合使用時必須小心.請注意，mysqli_stmt_bind_param() 需要通過引用傳遞參數(shù)，而 call_user_func_array() 可以接受可以表示引用或值的變量列表作為參數(shù).

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.

評論中有一些解決方法，例如 看到這個:

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;
}

這篇關(guān)于不能在 mysqli_stmt 對象上使用 call_user_func_array的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！