PHP mysqli查询不起作用答案

【问题标题】：PHP mysqli query isnt workingPHP mysqli查询不起作用
【发布时间】：2012-10-26 21:11:04
【问题描述】：

当用户输入他们的详细信息时，他们点击登录但它不起作用，我与数据库的连接很好，但这个文件不起作用，任何帮助将不胜感激，谢谢

include '../connection.php'; //used to include connection file that is 1 level higher in the directory 

$username = $_REQUEST['username'];
$password = $_REQUEST['password'];

$fquery = 'SELECT Username FROM login LIMIT 0, 30 ';
$squery = 'SELECT Password FROM login LIMIT 0, 30 ';

$username_query = mysqli_query($dbc, $fquery);
$password_query = mysqli_query($dbc, $squery);

$username_row = mysqli_fetch_array($username_query);
$password_row = mysqli_fetch_array($password_query);

if($username == $username_row && $password == $password_row) {
    echo 'username and password correct';
}


?>

【问题讨论】：

“不工作”是什么意思？一个空白屏幕？错误信息？返回了错误的数据？
另外，您可能想查看查询中的逻辑；您正在查看表中是否存在用户名和密码，但不一定是同一个帐户....
请注意，这永远不会奏效。您假设您的用户/密码永远不会超过 30 个，然后将请求变量与数组进行比较，并不能确保用户实际上与密码匹配。
@andrewsi 当我点击输入时，表单什么也不显示，它只是保持不变
您的逻辑中似乎缺少一些部分。您是否真的单步调试过您的代码以查看每个变量是什么？或者至少您的查询是什么以及它们返回什么？

标签： php database login mysqli

【解决方案1】：

<?php

include '../connection.php'; //used to include connection file that is 1 level higher in the directory 

$username = $_REQUEST['username'];
$password = $_REQUEST['password'];

$query = 'SELECT Username FROM login WHERE Username = ? AND Password = ?';

/* set a default value to check against */
$valid_user = '';

/* use prepared statement */
$stmt = mysqli_stmt_init($dbc);
if (mysqli_stmt_prepare($stmt, $query)) {
    /* set question marks equal to values */
    mysqli_stmt_bind_param($stmt, 'ss', $username, $password);
    mysqli_stmt_execute($stmt);

    /* get the valid username only if query is successful */
    mysqli_stmt_bind_result($stmt, $valid_user);
    mysqli_stmt_fetch($stmt);

    /* close the statment */
    mysqli_stmt_close($stmt);
}


/* check if default was overwritten */
if($valid_user != '') {
    echo 'username and password correct';
}
?>

试试这个，应该会完成你想做的事情。

【讨论】：

【解决方案2】：

     $username_query = mysqli_query($dbc, $fquery);
     $password_query = mysqli_query($dbc, $squery);       
     $username_row   = $username_query->fetch_array(MYSQLI_ASSOC);
     $password_row   = $password_query->fetch_array(MYSQLI_ASSOC);        

     if($username == $username_row['username'] && $password == $password_row['Password']) {
       echo 'username and password correct';
     }

【讨论】：

【解决方案3】：

$username = mysqli_real_escape_string($dbc, $_REQUEST['username']);
$password = mysqli_real_escape_string($dbc, $_REQUEST['password']);
$query = "SELECT * FROM login WHERE Username = '$username' AND Password = '$password' LIMIT 1";

if(mysqli_num_rows($query) > 0)
    echo 'username and password correct';

【讨论】：