php 代码运行错误

Question

嗯，这里出了点问题。我输入我的数据库中存在的用户名和密码。在这种情况下它应该回显这个字符串

need to activate

但它呼应了这个

You need to reg

初始化文件

<?php
    //error_reporting(0);
    session_start();

    require 'dbconnect.php';//this works okay so i wouldn't post this file code
    require 'users.php';

    $errors = array();
?>

用户.php

        <?php

        function user_exists($username){
            $username = mysql_real_escape_string($username);

            $query = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE 'username' = '$username'");

            if (!$query) {
                die('Could not query:' . mysql_error());
            }

            return (mysql_result($query, 0) == 1) ? true : false;

        }

        function user_active($username){

            $username = mysql_real_escape_string($username);

            $querytoo = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE 'username' = '$username' AND 'active' = 1");

            if (!$querytoo) {
                die('Could not query:' . mysql_error());
            }

            return (mysql_result($querytoo , 0) == 1) ? true : false;

        }
    ?>

登录.php

<?php
include 'init.php';


if(empty($_POST) === false){
    $username = $_POST['username'];
    $password = $_POST['password'];

    if(empty($username) === true || empty($password) === true){
        $errors[]='You need to enter a username and password';
    } 
    elseif(user_exists($username) === false){
        $errors[]='You need to reg';
    } 
    elseif(user_active($username) === false){
        $errors[]='need to activate';
    } 
    else {
    //
    }

    print_r($errors);
}



?>

部分html

<form action="login.php" method = "post">
                    <ul id="login">
                        <li>
                            username:<br>
                            <input type="text" name="username" size="30" value=""/></li>
                        <li>password:<br>
                            <input type="password" name="password" size="30" value=""/></li>
                        <li><input type = "submit" value ="Log in"></li>
                        <li>
                            <a href="register.php"> Register</a>
                        </li>
                    </ul>

附言查询文本工作正常，我在 mysql 中检查过。在 php 代码中，当我在此处键入 `` 而不是 ''

SELECT COUNT('user_id') FROM users WHERE 'username' = '$username' AND 'active' = 1

出现

'Could not query:'

东西

而且我尝试了 elseif 和 else if 东西，所以我认为没有问题

Answer 1

你的 SQL 是错误的。您正在将静态字符串与您的值进行比较，而不是列名:

    SELECT COUNT('user_id') FROM users WHERE 'username' = '$username' AND 'active' = 1

应该是

SELECT COUNT(user_id) FROM users WHERE username = '$username' AND active = 1

请注意，我从列名称中删除了 '。使用反引号也是有效的:

SELECT COUNT(`user_id`) FROM users WHERE `username` = '$username' AND `active` = 1