使用AJAX / jQuery设置$_SESSION | PHP会话不起作用?
发布时间:2020-12-16 02:51:26 所属栏目:百科 来源:网络整理
导读:我正在为我的网站编写一个登录脚本.我编写了登录脚本,并通过jQuery通过 AJAX调用将表单绑定到它. 这是表单调用的php: ?PHP # Make sure form data was passed to the script IF (isset($_POST) isset($_POST['username']) isset($_POST['password'])){ # Co
我正在为我的网站编写一个登录脚本.我编写了登录脚本,并通过jQuery通过
AJAX调用将表单绑定到它.
这是表单调用的php: <?PHP # Make sure form data was passed to the script IF (isset($_POST) && isset($_POST['username']) && isset($_POST['password'])){ # Connect to the database REQUIRE('../../../../my_db.php'); # Define variables $given_username = $_POST['username']; $given_password = $_POST['password']; $hashed_password = md5($given_password); $matched_username = ""; $matched_password = ""; # See if there is matching info in the database $sql = 'SELECT username,pass FROM users WHERE username="'.$given_username.'" AND pass = "'.$hashed_password.'"'; $result = mysql_query($sql); WHILE($row = mysql_fetch_assoc($result)){ $matched_username = $row['username']; $matched_password = $row['pass']; }; # If there was a match IF ($matched_username != "" && $matched_password != ""){ # Double check the values match IF ($given_username == $matched_username && $hashed_password == $matched_password){ # If there is only one result returned $session_sql = 'SELECT * FROM users WHERE username="'.$matched_username.'" AND pass = "'.$matched_password.'"'; $session_result = mysql_query($session_sql); IF(count(mysql_fetch_assoc($session_result)) != 0 && count(mysql_fetch_assoc($session_result)) < 2){ # If they do,start a session if(!isset($_SESSION)) { session_start(); session_regenerate_id(); }; # Set our session values WHILE($session_row = mysql_fetch_assoc($session_result)){ $_SESSION['id'] = $session_row['id']; $_SESSION['last_login'] = $session_row['last_login']; $_SESSION['username'] = $session_row['username']; $_SESSION['signup_date'] = $session_row['signup_date']; }; # Set users last login date and time to this login $update_sql = 'UPDATE users SET last_login = NOW WHERE username="'.$matched_username.'" AND pass = "'.$matched_password.'"'; $update = mysql_query($update_sql); echo json_encode(array("success"=>"user logged in","session"=>$_SESSION)); }ELSE echo json_encode(array("error"=>"More than one user with the same information. What did you do?!")); }ELSE echo json_encode(array("error"=>"invalid login provided")); }ELSE echo json_encode(array("error"=>"invalid login provided")); }ELSE echo json_encode(array("error"=>"you must supply a username and password")); ?> 但是如果我做console.log(result.session)我得到[],这让我觉得通过ajax设置会话变量是不可行的,或者会话本身不能正常工作. 我没有从这段代码中得到任何错误. 有人能指出我正确的方向吗? 我不认为我可以访问php.ini,但我记得很久以前你必须设置会话在一个文件中运行,但对于我的生活我找不到一个例子. 解决方法
确保在脚本的开头有session_start().如果在进入session_start()之前发出任何通知或任何内容,那么您将收到如下错误:
此外,您的脚本对SQL注入是开放的.确保您正确地转义用户名和密码!注意:您最好的选择是使用prepared statements. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |