Javascript 不等待 Promise 在函数中解析

Question

我的js代码需要等到ajax完成才能加载数据。

if (district_search_value[0] == true) {
        var zone_search_value = checkZone(district_search_value[1], datas[i]['recipient_zone']);
        console.log(zone_search_value);
    }

checkZone 功能：

   function checkZone(district_id, zone_name) {
    var value = [];
    value[0] = false;
    value[1] = 0;
    find_zones_from_district_id().then(function(data) {
        // Run this when your request was successful
        console.log(data);
        var zones = JSON.parse(data);
            //console.log(zones);
            for (zones_row = 0; zones_row < zones.length; zones_row++) {
                if (zone_name == zones[zones_row]['zone_name']) {
                    value[0] = true;
                    value[1] = zones[zones_row]['id'];
                }
            }
        return value;
    }).catch(function(err) {
        // Run this when promise was rejected via reject()
        console.log(err);
    })
}

主要的promise函数：

function find_zones_from_district_id(district_id) {
    return new Promise(function (resolve, reject) {
        $.ajax({
            type: 'POST',
            url: 'ajaxData_for_user.php',
            data: {import_excel_find_zones_from_district_id: 'district_id'},
            success: function (data) {
                resolve(data) // Resolve promise and go to then()
            },
            error: function (err) {
                reject(err) // Reject the promise and go to catch()
            }
        });
    });

}

javascript 不等待 promise 解决的原因是什么？ 我也尝试过 jquery 中的 $.when ，但这也不会等待 ajax 加载数据。

Answer 1

两件事需要修复，第三件事是可选的

1. 在使用 .then() 记录结果之前等待承诺解决

   if (district_search_value[0] == true) {
     checkZone(district_search_value[1], datas[I]['recipient_zone'])
     .then(zone_search_value => {
       console.log(zone_search_value);
     }
   }
   

2. checkZone 必须返回它从find_zones_from_district_id得到的承诺

   function checkZone(district_id, zone_name) {
     var value = [];
     value[0] = false;
     value[1] = 0;
     return find_zones_from_district_id().then(function(data) {
       // ...
   

3. 正如评论者指出的那样，没有必要包装 ajax 调用...

   function find_zones_from_district_id(district_id) {
     return $.ajax({
       type: 'POST',
       url: 'ajaxData_for_user.php',
       data: {import_excel_find_zones_from_district_id: 'district_id'} // did you want to pass a string here or district_id parameter?
     })
   }