Question

我有Ajax登录提交，工作得很好。现在我需要在成功时将$ user_id发送回登录页面。但无法弄清楚如何。

以下是我所拥有的。

这是php页面

            <?
            if (!securePage($_SERVER['PHP_SELF'])){die();}

            //Prevent the user visiting the logged in page if he/she is already logged in
            if(isUserLoggedIn()) { header("Location: account.php"); die(); }

            //Forms posted
            if(!empty($_POST))
            {
                $errors = array();
                $username = sanitize(trim($_POST["user"]));
                $password = trim($_POST["password"]);

                //Perform some validation
                //Feel free to edit / change as required
                if($username == "")
                {
                    $response['success'] = false;
                }
                if($password == "")
                {
                    $response['success'] = false;
                }

                if(count($errors) == 0)
                {
                    //A security note here, never tell the user which credential was incorrect
                    if(!usernameExists($username))
                    {
                        $response['success'] = false;
                    }
                    else
                    {
                        $userdetails = fetchUserDetails($username);
                        //See if the user's account is activated
                        if($userdetails["active"]==0)
                        {
                            $response['success'] = false;
                        }
                        else
                        {
                            //Hash the password and use the salt from the database to compare the password.
                            $entered_pass = generateHash($password,$userdetails["password"]);

                            if($entered_pass != $userdetails["password"])
                            {
                                //Again, we know the password is at fault here, but lets not give away the combination incase of someone bruteforcing
                                $response['success'] = false;
                            }
                            else
                            {
                                //Passwords match! we're good to go'

                                $response['success'] = true;
                            }

                        }
                    }
                }
            }
            //$user_id = $loggedInUser->user_id;
            echo json_encode($response);
            ?>

这是调用php页面的ajax。还有我需要从php页面检索ID的地方。

            <script type="text/javascript">


                //login ajax to send over user and pass LS
                function handleLogin() {
                    var form = $("#loginForm");
                    //disable the button so we can't resubmit while we wait
                    $("#submitButton",form).attr("disabled","disabled");
                    var e = $("#user", form).val();
                    var p = $("#password", form).val();

                    console.log("click");
                    if(e != "" && p != "") {
                        var str = form.serialize();
                        //McDOn(str);
                        $.ajax({
                               type: 'POST',
                               url: 'http://vsag.actualizevps.com/loginmobile.php',
                               crossDomain: true,
                               data:  {user: e, password :p},
                               dataType: 'json',
                               async: false,
                               success: function (response){
                               //alert ("response");
                               if (response.success) {
                               //alert("you're logged in");
                               window.localStorage["user"] = e;
                               //window.localStorage["password"] = md5(p);
                               //window.localStorage["UID"] = data.uid;
                               window.location = "create.html";
                               }
                               else {

                               alert("Your login failed");
                               //window.location("main.html");
                               location.reload();
                               }


                               },
                               error: function(error){
                               //alert(response.success);
                               alert('Could not connect to the database' + error);
                               window.location = "main.html";
                               }
                               });
                    }
                    else {
                        //if the email and password is empty
                        alert("You must enter user and password");

                    }
                    return false;
                }
            </script>

Answer 1

此时$ response值只是true或false，你可以返回一个数组：

$response = array("Success" => true, "UserId" => $user_id);

和你的AJAX响应，响应变量

response.UserId

将包含用户ID

返回用户ID以及Ajax Success响应

1 个答案: