警告：mysql_result（）期望参数1是资源，布尔在第6行给出/home/roscapex/public_html/core/functions/users.php

Question

我试图做一个登录/注册系统，我不断收到错误： 警告：请求mysql_query（）预计参数2是资源，在整数给出/home/roscapex/public_html/core/functions/users.php第8行

警告：mysql_result（）预计至少2个参数，1中给出/home/roscapex/public_html/core/functions/users.php第8行

我的代码是：

<?php 
function user_exists($username) { 

$username = sanitize($username); 

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

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

} 
?> 

注意这可能是重复的，但我尝试了所有其他相关问题。

Answer 1

代码中有多处错误。让我注释它：

<?php 
function user_exists($username) { 

$username = sanitize($username); 

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'"); // $query now contains the result of the query (a Resource) 

// Here you try to run the resource *again* trough mysql_query, with with 2 parameters (which results in the warning). I am not quite sure what you intended. 
return (mysql_result(mysql_query($query, 0) == 1) ? true : false); 

} 
?> 

我修好了。

<?php 
function user_exists($username) { 
    $username = sanitize($username); // This hopefully santizes the username. Better use mysql_real_escape_string($username) 

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

    $result = mysql_query($query) or die(mysql_error()); // Check if query was successful 
    $row = mysql_fetch_array($result); // fetch the first row 
    return ($row[0] > 1); // If there is one or more users with this name, return true. 

} 
// omit the closing ?>, because it causes trouble, when there are any chars behind it (headers alreay sent error etc.)