按照说明我必须在虚线之间插入代码。 假设刷新并看到'存在' 我有正确的数据库,表,一切我非常肯定但是 我看不存在。 我应该继续遵循指南还是我搞砸了?
如果我正确地遵循它,是否有人有保证可以使用的指南?
我的login.php
<?php
include 'core/init.php';
-----------------------------------------
if (user_exists('alex') === true) {
echo "exists";
}
die('no exist');
-------------------------------------------
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 not find that user';
}
}
?>
我有init.php
<?php
session_start();
error_reporting(0);
require 'database/connect.php';
require 'functions/general.php';
require 'functions/users.php';
$errors = array();
?>
我有users.php
<?php
function user_exists($username) {
$username = sanitize($username);
$query = mysql_query("SELECT COUNT('user_id') FROM 'users' WHERE 'username' = '$username'");
return (mysql_result($query, 0) == 1) ? true : false;
}
?>
i have general.php
<?php
function sanitize($data) {
return mysql_real_escape_string($data);
}
?>
我用:
<form action="login.php" method="post">
注意:如果有人熟悉phpacademy,我正在关注他的指南
答案 0 :(得分:1)
您的die
始终被调用,您应该将代码更改为:
if (user_exists('alex') === true) {
echo "exists";
}
else{
die('no exist');
}