使用jQuery延迟使用包含$ .each的多个ajax

Question

我正在尝试为以下方案创建一个javascript对象

一项调查采访了多人关于他们在几顿饭中消费的食物.该对象需要嵌套如下: -

case={}
case[x].Interview={}
case[x].Interview[y].meals={}
case[x].Interview[y].meals[z].Food=[]

我通过以下代码实现了这一点

var $caseoffset=0
loadcases()
function loadcases() {
    $.ajax({
        url: "functions.php",data: {offset: $caseoffset,method: "getCase"},method: "post",dataType: 'json',
        success: function(result) {
            cases = result;
            loadinterview(cases[$caseoffset].fldCaseID)         
        }
    })  
}

function loadinterview($CaseID) {
    $.ajax({
        url: "functions.php",
        data: {method: "getinterview",caseid: $CaseID}, method: "post",dataType: 'json',
        success: function(result) {
            thiscase=cases[$caseoffset]
            thiscase.interviewcount=result.length
            thiscase.interviews={}

            $.each(result,function(key,val){
                thiscase.interviews[val.fldInterviewID]=val
                loadmeals(val.fldInterviewID)
            })  
        }    
    })
}
function loadmeals($InterviewID) {
    $.ajax({
        url: "functions.php",
        data: {method: "getmeal",InterviewID: $InterviewID},method: "post",dataType: 'json',
        success: function(result) {
            thiscase.interviews[parseInt($InterviewID)].mealcount = result.length
            thiscase.interviews[parseInt($InterviewID)].meals={}

            $.each(result, function(key, val) {

                thiscase.interviews[parseInt($InterviewID)].meals[parseInt(val.fldMealHistoryID)] = val
                getfoodinmeal($InterviewID, val.fldMealHistoryID)
            })
        }
    })
}

function getfoodinmeal($interviewid, $mealid) {
    $.ajax({
        url: "functions.php",data: {method: "getfoodinmeal",mealid: $mealid},
        method: "post",

        dataType: 'json',

        success: function(result){
            foodinmeal = [];

            $.each(result, function(key, val) {
                foodinmeal.push(val.fldFoodID)
            })

            thiscase.interviews[$interviewid].meals[$mealid].food = foodinmeal

        }
    })
}

问题是,一旦编制了每个采访者消耗的所有食物,我想进行一些计算.如何创建延迟语句来解决这个问题.

Answer 1

从jQuery 1.5开始,$.ajax()返回一个jqXHR实现Promise接口.

这意味着您可以与它结合使用Promise.all()(https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/all)

这样的事情应该有效:

Promise.all([
    $.ajax({
        url: "a.php"
    }),
    $.ajax({
        url: "b.php"
    }),
    $.ajax({
        url: "c.php"
    }),
]).then(function() {
    // the 3 $ajax() call are finished
})