在登录php表单上工作 [英] working on a login php form

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; 
    } 
?>  

推荐答案

用户名）
{

username) {

用户名 = sanitize（

username = sanitize(

用户名）;
return （mysql_result（mysql_query（ SELECT COUNT（'id'）FROM'ex练'WHERE'login'='

username); return (mysql_result(mysql_query("SELECT COUNT ('id') FROM 'practice' WHERE 'login' = '

这篇关于在登录php表单上工作的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！