Question

好吧，基本上我正在使用youtube上的注册和登录教程。哪个使用旧版本的PHP，我试图更新代码，但是当PHP查询数据库时，它不会返回我期望它返回的内容。

基本上我通过将$ username设置为＆＃39; euan＆＃39;来测试查询的工作原理。它存在于数据库中，但出于某种原因，它不会返回“euan”＃e;存在。

的init.php

<?php 
    session_start();
    error_reporting(0);
    require 'database/connect.php'; echo '[connect loaded]'; echo ' ';
    require 'functions/users.php'; echo '[users loaded]'; echo ' ';
    require 'functions/general.php'; echo '[general loaded]'; echo ' ';
    $errors = array();
?>

Connect.php

<?php 
    $connection_error = 'We\'re experiencing connection issues. Come back later.';
    $con = mysqli_connect('localhost','root','') or die($connection_error);
    mysqli_select_db($con,'forum') or die($connection_error);
?>

Login.php

<?php
    include 'core/init.php';
    $username = "euan";
    echo $username; echo ' '; echo ' '; echo ' ';
    if (user_exists($username) === true) {
        echo '[Yeah that exists]';
    } else {
        echo '[Nah thats not there]'; echo ' ';
        echo $username; echo ' ';
    }
    if (empty($_POST) === false) {
        $username = $_POST['username'];
        $password = $_POST['password'];

        if (empty($username) === true || empty($password) === true)  {
            $errors[] = 'You need to enter a username and password';
        } else if (user_exists($username) === false) {
            $errors[] = 'We can\'t find that username. Have you registered?';
        }
    }
?>

users.php

<?php
    function user_exists($username) {
        echo $username;
        //$username = sanitize($username);
        $query = mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = $username");
        return(mysqli_query($query, 0) == 1) ? true : false;
    }
?>

数据库图片:(由于代表无法嵌入）

http://i.imgur.com/tl5SujK.png

http://i.imgur.com/g1niTB3.png

Answer 1

在您的users.php

中，$username应在'引号内，如下所示，以便您的查询看起来像where username='euan'

mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");

并更改函数返回，使用mysqli_affected_rows($con)==1

这可以解决您的问题，$query也不是必需的

<?php
function user_exists($username) {
    echo $username;
    //$username = sanitize($username);
    mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");
    return(mysqli_affected_rows($con) == 1) ? true : false;
}
?>

PHP没有从MySQL查询中返回正确的信息

1 个答案: