javascript - JQuery ajax 发布到 PHP 服务器无法正常工作 - "SyntaxError: Unexpected token < in JSON at position 0"

Question

这个问题已经有答案了:

"SyntaxError: Unexpected token < in JSON at position 0" (40 个回答)

已关闭 5 年前。

我正在尝试将json数据发送到服务器，并接收json响应，代码如下:

JS:

function login() {

console.log("clicked");

//get values of form into variables
var email = $("#email").val();
var password = $("#password").val();

//create data array
var dataString = {"email": email, "password": password};

console.log(dataString);

//check for blank inputs
if ($.trim(email).length == 0) {
    myApp.alert("Email required", "Login Failed");
} else if ($.trim(password).length == 0) {
    myApp.alert("Password required", "Login Failed");
}

//if form isnt empty, post ajax request to server
if ($.trim(email).length > 0 & $.trim(password).length > 0) {

    console.log("input checked");

    $.support.cors = true;

    //ajax post
    $.ajax({
        type: "POST",
        url: "http://gingr-server.com/login.php",
        data: dataString,
        dataType: 'json',
        crossDomain: true,
        cache: false,

        beforeSend: function() {
            $("#login").val('Connecting...');
            console.log("connecting to server");
        },

        //display success/fail message - put something in data on server
        success: function(data, textString, xhr) {
            if (data.status == "correct") {
                localStorage.login = "true";
                localStorage.email = email;
                localStorage.id = data.id;
                mainView.router.loadPage("swipe.html");
            } else {
                myApp.alert("Incorrect email or password", "Login Failed");
            }
        },

        error: function(xhr, ajaxOptions, errorThrown) {
            console.log(xhr);
            console.log(errorThrown);
            myApp.alert("Unknown error, please try again", "Login Failed");
        },

        complete: function(data) {
            if (data.readyState == "0") {
                console.log("unsent");
            } else if (data.readyState == "4") {
                console.log("done");
            }   
        }

    });
} 
return false;
}

PHP:

<?php
header("access-control-allow-origin: *");
header("access-control-allow-methods: GET, POST, PATCH, PUT, DELETE, OPTIONS");
header("access-control-allow-headers: X-Requested-With, Content-Type");

header('Content-Type: application/json');

//include databse info
require_once "db_config.php";

    $email = $mysqli->real_escape_string(htmlspecialchars($_POST["email"]));
    $password = $mysqli->real_escape_string(htmlspecialchars($_POST["password"]));

    //get hashed password from db
    $result = $mysqli->query("SELECT password, id FROM user_table WHERE 
    email = '$email'");
    $row = $result->fetch_assoc();
    $passwordHash = $row["password"];
    $userID = $row["id"];

    //free memory
    $result->free();

    //verify password for that email
    if (password_verify($password, $passwordHash)) {
        $response = array(  "status" => "correct",
                            "id" => $userID));
    } else {
        $response = array("status" => "incorrect"));
    }

    echo json_encode($response);

    $mysqli->close();
?>

错误信息如下:

语法错误:JSON 中位置 0 处出现意外标记 < 在解析()

对我来说，客户端似乎正在接收 HTML(因此是 < 标签)而不是 json，但我不知道为什么!

Answer 1

找到了!一点也不聪明...

$response = array(  "status" => "correct",
                            "id" => $userID));

发现));造成所有问题...我认为就寝时间