本文介绍了在登录php表单上工作的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这是我的错误..
警告:mysql_result()要求参数1为资源,在E:\ XAMPP \\\ htdocs \中给出布尔值第6行的basicphpedit\core\functions\users.php
数组([0] =>我们找不到该用户名,你注册了吗?)
here is my error..
Warning: mysql_result() expects parameter 1 to be resource, boolean given in E:\XAMPP\htdocs\basicphpedit\core\functions\users.php on line 6
Array ( [0] => We can't find that username, have you registed? )
<?php
function user_exists($username)
{
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT ('id') FROM 'practice' WHERE 'login' = '$username'"), 0) == 1) ? TRUE : FALSE;
} // "SELECT COUNT ('id') FROM 'practice' WHERE 'login' = '$username'"), 0) == 1) ? TRUE : FALSE;
function user_active($username)
{
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT ('id') FROM 'practice' WHERE 'login' = '$username' AND 'active' = 1"),0)===1)?TRUE:FALSE;
}
function id_from_username($username)
{
$username = sanitize($username);
return mysql_result(mysql_query("SELECT 'id' FROM 'practice' WHERE 'login'='$username'"),0,'id');
}
function login($username, $password)
{
$user_id = id_from_username($username);
$username = sanitize($username);
//$password = md5($password);
return (mysql_result(mysql_query("SELECT COUNT('id') FROM 'practice' WHERE 'login'='$username' AND 'password'='$password'"),0)==1) ?$user_id : FALSE;
}
?>
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