假設有已經存在的學生字典數組，學生有姓名和所在教室號，先查詢符合條件的學校，便利學生字典數組，創建學生對象賦值姓名、學校和班級，但是班級需要先查詢教室的位置來確定，這樣就出現循環查詢了？這樣的問題該怎麼解決呢？

var studentArr = new Array({'name': 'a','room':'101'},{'name': 'b','room':'102'},{'name': 'c','room':'103'},{'name': 'd','room':'104'});
var objects = new Array();
var schoolQuery = new AV.Query(Shcool);
schoolQuery.equalTo('name','**高中');
schoolQuery.find().then(function(schoolReuslts){
    for (var i = 0; i < studentArr.count; i ++){
        var student = studentArr[i];
        var object = new Student();

        object.set('name',student['name']);
        object.set('room',student['room']);
        object.set('school',schoolReuslts[0]);

        var classQuery = new AV.Query(Class);
        classQuery.equalTo('school',schoolReuslts[0]);
        classQuery.equalTo('room',student['room']);
        classQuery.find().then(function(classResults){
                object.set('class',classResults[0]);
                objects.push(object);
            }, function(error){
                console.log(error);
            });
        }
    return AV.Object.saveAll(objects);
}).then(function(objects){
    //全部保存成功    
}
}).catch(function(error) {
    console.log(error);