javascript - jsCall 将值返回到 jquery ajax post 请求之外

Question

我想从 ajax 函数获取 jquery ajax post 请求值到外部。我的代码是这样的，它返回 undefined 作为控制台输出。应该如何解决

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;
}

<小时/>

已编辑

我更改了代码以同步类型 ajax post 请求并检查输出。但它并没有改变。这是我的代码

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;
}

Answer 1

您可以使用Deferred jQuery

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;
}