小编Cor*_*rey的帖子

我正在为我的Web应用程序创建一个登录脚本,并且我正在尝试使用它$count = mysqli_stmt_num_rows($stmt);来查找从sql select语句返回的行数,因此我可以决定是否应该启动会话.

问题是,$ count始终为0,即使我输入与我的数据库中的数据匹配的有效用户名和密码.我已经测试了select语句,它运行正常.没有给出错误,语法,SQL或其他内容,所以我对于发生什么事情感到有些困惑.

码:

<?php

    $link = mysqli_connect("localhost", "****", "****", "****");

    //check connection
    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }

    // username and password sent from form 
    $myusername=$_POST['myusername'];   
    $mypassword=$_POST['mypassword']; 

// Move to MySQL(i) as MySQL is now obslete and use Prepare statment for protecting against SQL Injection in better and easier way
    $stmt = mysqli_prepare($link, 'SELECT username, password FROM `users` WHERE  `username` =  ? AND  `password` =  ?');

    /* bind parameters for markers …