和錯誤的SQL語法

Question

這裏是MySQL錯誤即時得到：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECY * FORM user WHERE username='' AND password='' LIMIT 1' at line 1 

這裏是我的代碼：

<?php 
session_start(); 
include_once("connect.php"); 

if (isset($_POST['username'])) { 
    $username = $_POST['username']; 
    $password = $_POST['password']; 
    $sql = "SELECY * FORM users WHERE username='".$username."' AND password='".$password."' LIMIT 1"; 
    $res = mysql_query($sql) or die(mysql_error()); 
    if (mysql_num_rows($res) == 1) { 
     $row = mysql_fetch_assoc($res); 
     $_SESSION['uid'] = $row ['id']; 
     $_SESSION['username'] = $row ['username']; 
     header("Location: index.php"); 
     exit(); 
    } else {  
     echo "Invalid login information. Please return to the previous page."; 
     exit(); 
    } 
} 
?> 

有人能幫助我嗎？ :) 我不知道即時通訊工作在這裏工作，找不到錯誤。

Answer 1

SELECY 

是一個拼寫錯誤，請嘗試：

SELECT 

Answer 2

$sql = "SELECY * FORM users 
WHERE username='".$username."' 
AND password='".$password."' 
LIMIT 1"; 

變化SELECY到SELECT和FORM到FROM。

正確的查詢應該是：

$sql = "SELECT * FROM users 
WHERE username='".$username."' 
AND password='".$password."' 
LIMIT 1";