Javascript jsCall返回值到jquery ajax post请求外部

Javascript jsCall返回值到jquery ajax post请求外部,javascript,jquery,ajax,jquery-ajaxq,Javascript,Jquery,Ajax,Jquery Ajaxq,我想从ajax函数中获取jquery ajax post请求值。我的代码是这样的,它返回未定义的作为控制台输出。你应该怎样修理它 function submit() { var outputFromAjax = submitViaPost('administrator/validationForInputValuesOfAddRole'); console.log(outputFromAjax); } function submitViaPost(url) { va

我想从ajax函数中获取jquery ajax post请求值。我的代码是这样的,它返回
未定义的
作为控制台输出。你应该怎样修理它

function submit() {
    var outputFromAjax = submitViaPost('administrator/validationForInputValuesOfAddRole');

    console.log(outputFromAjax);
}

function submitViaPost(url) {

    var formData = $('form').serializeArray();

    var output;

    $.post(urlForPhp + '/' + url, formData, function (outputData) {
        output = outputData;
    });
    return output;
}

已编辑

我将代码更改为sync类型的ajax post请求并检查输出。但它没有改变。这是我的密码

function submit() {
    var outputFromAjax = submitViaPost('administrator/validationForInputValuesOfAddRole');

    console.log(outputFromAjax);
}

function submitViaPost(url) {

    var formData = $('form').serializeArray();

    var output;

    $.ajax({
        url: urlForPhp + '/' + url,
        data: formData,
        dataType: 'JSON',
        async: false,
        method: 'POST',
        success: function (e) {
            output =  e;
        }
    });
    return output;
}

不返回,将其设置为回调
$。post
是异步的

function submit() {
    submitViaPost('administrator/validationForInputValuesOfAddRole', function(out){ //Result comes here
      var outputFromAjax = out;
      console.log(outputFromAjax);
    });
}

function submitViaPost(url , callback) { //Added callback
    var formData = $('form').serializeArray();
    $.ajax({
    url: urlForPhp + '/' + url,
    data: formData,
    dataType: 'JSON',
    async: false,
    method: 'POST',
    success: function (e) {
        callback(e);
    }
 });
}
你可以用


submitViaPost函数立即返回,返回值始终未定义。您必须在post回调使用post的done回调完成后保存该值。那么我是如何对其进行编码的呢?返回值应该在$.post函数中?
function submit() {
    submitViaPost('administrator/validationForInputValuesOfAddRole').then(function (outputFromAjax) {
        console.log(outputFromAjax);
    });
}

function submitViaPost(url) {
    var dfd = jQuery.Deferred();
    var formData = $('form').serializeArray();
    $.post(urlForPhp + '/' + url, formData, function (outputData) {
        dfd.resolve(outputData);;
    });
    return dfd;
}