问题描述

我正在尝试创建一个小型 SQL 查询类.

I'm trying to create a small SQL query class.

这是我的班级，但我不知道为什么，我有这个错误:严格的标准:在第 52 行中只应通过引用传递变量

Here is my Class but i don't why, I've this error : Strict Standards: Only variables should be passed by reference in line 52

第 52 行是:

if (!$stmt->bind_param($param[$i][0], mysqli_real_escape_string($this->mysqli, $param[$i][1]))) {

我的代码(我开始了):

My code (i'm beginning) :

<?php
class Sql{

    private $db;
    private $user;
    private $pwd;
    private $url;

    private $param;

    private $mysqli;

    function __construct($db, $user, $pwd, $url){
        $this->db = $db;
        $this->user = $user;
        $this->pwd = $pwd;
        $this->url = $url;


    }

    /**
     * mysqli::connection()
     * 
     * @return 
     */
    public function connection()
    {
        try{
            $this->mysqli = new mysqli($this->db, $this->user, $this->pwd, $this->url);
        }catch(Exception $e){
            throw new Exception("Impossible de se connecter à la base " . $this->db);
        }
    }

    public function select($query, $param, $debug=false){

        $this->connection();

        $r = $this->InitialiseResult("select");

        if (!($stmt = $this->mysqli->prepare($query))) {
            echo "Echec de la préparation : (" . $this->mysqli->errno . ") " . $this->mysqli->error;
        }

        //Param
        for($i=0;$i<sizeof($param);$i++){
            if (!$stmt->bind_param($param[$i][0], mysqli_real_escape_string($this->mysqli, $param[$i][1]))) {
                echo "Echec lors du liage des paramètres : (" . $stmt->errno . ") " . $stmt->error;
            }
        }

        if (!$stmt->execute()) {
            echo "Echec lors de l'exécution : (" . $stmt->errno . ") " . $stmt->error;
        }

        if (!($res = $stmt->get_result())) {
            echo "Echec lors de la récupération du jeu de résultats : (" . $stmt->errno . ") " . $stmt->error;
        }else{

            $r["state"] = true;
            $r["rows"] = $res->fetch_assoc();
            $r["num_rows"] = $res->num_rows;

            if($debug)
                var_dump($r);

        }

        return $r;

    }


    /**
     * mysqli::InitialiseResult()
     *
     * @param mixed $p
     * @return
     */
    public function InitialiseResult($p)
    {
        $r = array(); //on écrase
        $r["state"] = false;

        switch($p){
            case "select":

                $r["rows"] = array();
                $r["num_rows"] = 0;
                break;

        }

        return $r;
    }
}
?>

我尝试将 $param 放在一个属性中并使用它是 mysqli_real_escape_string() 但错误仍然存在.

I've try to put $param in a property and use that is mysqli_real_escape_string() but the error is still there.

有什么想法吗?

推荐答案

$stmt->bind_param() 要求所有的params都通过引用传递，所以不能传递函数的返回值直接(而不是先将其分配给变量).但是，正如评论中已经提到的，您根本不需要转义参数，这是使用准备好的语句的优势之一.

$stmt->bind_param() requires all params to be passed by reference, so you can't pass function's return value directly (without assigning it to a variable first, that is). But, as was already mentioned in the comments, you don't need to escape the parameters at all, that's one of the advantages of using prepared statements.

这篇关于mysqli : 严格的标准:只应通过引用传递变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！