Ajax的响应与我认为的不一样 [英] Ajax response doesn't equal what I think it should

问题描述

我正在尝试调用 displayUsers 函数，如果响应等于" loggedIn "(响应来自php中针对ajax请求的echo语句).它总是直接跳转到else语句，并且不执行 displayUsers().但是，当我警报响应时，它会显示loginIn.

I'm trying to call a displayUsers function, if response equals "loggedIn" (response is coming from echo statement in php for ajax request). It always jumps straight to the else statement and doesn't execute displayUsers(). However, when I alert response it displays loggedIn.

这是我的代码:

   function ajaxRequest(url, method, data, asynch, responseHandler) {
    var request = new XMLHttpRequest();
    request.open(method, url, asynch);

    if (method == "POST") {
        request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
    }
    request.onreadystatechange = function() {
        if (request.readyState == 4) {
            if (request.status == 200) {
                responseHandler(request.responseText);
            }
        }
    }

    request.send(data);
}

//loginCheck
    function loginCheck() {
        var username = document.getElementById("usernameLogin").value;
        var password = document.getElementById("passwordLogin").value;
        var data="usernameLoginAttempt="+username+"&passwordLoginAttempt="+password;
        ajaxRequest("../PHP/CODE/login_check.php", "POST", data, true, loginCheckResponse);
    }

    function loginCheckResponse(response) {
    //check response, if it is "loggedIn" then call show users function
    alert(response);
    if (response == "loggedIn") {
        displayUsers();
    } else {
        alert("Login Failed. Please try again.")

    }

}

推荐答案

// response is an object which you get from ajex.
// You have not written how you call loginCheckResponse()
// call like loginCheckResponse(response.<variable which you return from service page>)
function loginCheckResponse(response)
{
    //check response, if it is "loggedIn" then call show users function
    alert(response);
    if (response == "loggedIn") {
        displayUsers();
    } else {
        alert("Login Failed. Please try again.")
    }

}

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