小编Jah*_*hed的帖子

我有一个简单的PHP错误,我无法找到解决方案.我插入以下代码:

    <?php
session_start();
require_once '../includes/db.php';  

$address1 = $_REQUEST["address1"];
$address2 = $_REQUEST["address2"];
$city = $_REQUEST["city"];
$postcode = $_REQUEST["postcode"];

mysql_query("UPDATE customers SET address1='$address1', address2='$address2', city='$city', postcode='$postcode' WHERE username = '".$_SESSION['username']."')")

or die(mysql_error());

?> 

此代码给出了以下错误:

您的SQL语法有错误; 检查与您的MySQL服务器版本对应的手册,以便在第1行的')'附近使用正确的语法

第1行是以下我认为它与sql语句无关:

    <?php
session_start();
?>

感谢大家!:)

我有一个包含以下字段的mysql表:

id用户名密码角色

管理员的角色编号为1,普通用户的角色编号为2

    <?php
// we must never forget to start the session
session_start();

$errorMessage = '';
if (isset($_POST['username']) && isset($_POST['password'])) {
    include 'library/connect.php';

    $username   = $_POST['username'];
    $password = $_POST['password'];

    // check if the user id and password combination exist in database
    $sql = "SELECT role FROM user WHERE username = '$username' AND password = '$password'";

    $result = mysql_query($sql) or die('Query failed. ' . mysql_error()); 
    $row = mysql_fetch_array($result);

    if (mysql_num_rows($result) == 1 AND $row['role']==2) {
        // the user id and …