在JavaScript中按顺序加载AJAX集合
Load AJAX collections in order in JavaScript

由于您只需要所有可用的结果，并且每个请求都不依赖于先前的请求，因此可以使用jQuery.when

let students = null;
let teachers = null;
let schools = null;

$(document).ready(function() {
  $.when(
      getSchools(),
      getTeachers()
  ).done(function(shoolResults, teacherResults) {
      window.schools = shoolResults;
      window.teachers = teacherResults;
      handleTeachers();
      getStudents();
  });


  function getSchools() {
    return $.ajax({
      type: 'GET',
      url: schoolsUrl
    });
  }

  function getTeachers() {
    return $.ajax({
      type: 'GET',
      url: teachersUrl
    });
  }

  function handleTeachers() {
   $.each(teachers, function (key, item) { 
     item.school = schools[item.schoolId].name;
    });
  }
});

正如其他人已经建议的那样，您可以链接请求。 我对您的代码进行了一些更改。

1. 添加了严格模式 ，有助于防止错误
2. IFFE中包装了该代码，以防止全球污染

如果所有api都属于同一服务器，则可以在服务器端处理所有这些数据并返回一个已填充的json。 这样，您的服务器将在构造此json上做一些额外的工作，但另一方面，您将只发出一个ajax请求而不是3。这将更快地工作，并且可以将json缓存一段时间

第一个解决方案的代码

(function () {

'use strict';
const studentsUrl = 'api/students';
const teachersUrl = 'api/teachers';
const schoolsUrl = 'api/schools';

let students = null;
let teachers = null;
let schools = null;

var scoolData = {
    schools: null,
    teachers: null,
    students: null
};

$(document).ready(function () {
    getSchools().then(function (schools) {
        scoolData.schools = schools;
        getTeachers().then(function (teachers) {
            scoolData.teachers = teachers;
            $.each(scoolData.teachers, function (key, item) {
                item.school = scoolData.schools[item.schoolId].name;
            });
        });
    });
});

function getSchools() {
    return $.get(schoolsUrl);
}

function getTeachers() {
    return $.get(teachersUrl,
        function (result) {
            scoolData.teachers = result;

            // >>> SHOULD BE SURE, SCHOOLS already loaded!!!
            $.each(teachers, function (key, item) {
                item.school = scoolData.schools[item.schoolId].name;
            });
        });
}
})();

如果您希望他们按顺序排列（尽管我不确定我为什么会这样，因为您无论如何都会检索所有的学校/老师/学生），只需执行此操作即可。

注意 ：在以下示例中， get*函数是虚拟变量。 相反，只需返回来自它们的$.get调用的结果：

 function getSchools() { return Promise.resolve({1: {name: 'school1'}}); } function getTeachers() { return Promise.resolve({1: {name: 'teacher1', schoolId: 1}}); } function getStudents() { return Promise.resolve({1: {name: 'student1', teacherId: 1}}); } (async () => { const schools = await getSchools(); const teachers = await getTeachers(); const students = await getStudents(); // Alternative for the $.each code Object.values(teachers).forEach(teacher => teacher.school = schools[teacher.schoolId].name); Object.values(students).forEach(student => student.teacher = teachers[student.teacherId].name); console.log(schools, teachers, students); })(); 

另一个注意事项：这是ES8代码，如果您需要支持较旧的浏览器并且不能使用像Babel这样的编译器，我将发布一个非异步/等待版本。

不依赖于ES8的代码 ：

 function getSchools() { return Promise.resolve({1: {name: 'school1'}}); } function getTeachers() { return Promise.resolve({1: {name: 'teacher1', schoolId: 1}}); } function getStudents() { return Promise.resolve({1: {name: 'student1', teacherId: 1}}); } let schools = null, teachers = null, students = null; getSchools().then(_schools => { schools = _schools; return getTeachers(); }).then(_teachers => { teachers = _teachers; return getStudents(); }).then(_students => { students = _students; for (var _ in teachers) { teachers[_].school = schools[teachers[_].schoolId].name; } for (var _ in students) { students[_].teacher = teachers[students[_].teacherId].name } console.log(schools, teachers, students); }); 

调用getTeachers(); 当getSchools(); 返回成功或完成，成功是首选，因为如果有错误，则运行完整。

我认为您正在寻找这个。

 getSchools().done(function(data){
      var someId = data.findThatId;
      getTeachers(someId);
 }); 

您将需要从Ajax调用返回的数据获得data中done 。

您可以异步加载它们，但必须等到两个调用完成。

为此，请在ajax调用之前添加return并将结果合并到ready函数中（而不是在教师调用的成功处理程序中）：

let schoolsPromise = getSchools();
let teachersPromise = getTeachers();
$.when(schoolsPromise, teachersPromise)
  .then((schools, teachers) => {
    $.each(teachers, (key, item) => {
      item.school = schools[item.schoolId].name;
    });
  });