将json作为子文档合并到另一个json中

您可以array.forEach。

first.forEach(function (val, index, theArray) {
   val.trip = [];
   second.forEach(function (val2, index, theArray) {
       if(val2.stop_id === val.stop_id){
           val.trip.push(val2);
       }
   });    
});
console.log(first);

一把可以摆弄的小提琴。https://jsfiddle.net/u3etdqrz/3/

您可以尝试以下操作：

var stnList = [{"stop_id": 70021,"stop_name": "CALTRAIN - 22ND ST STATION",  "stop_lat": 37.757692,"stop_lon": -122.392318,"zone_id": 3329}, {  "stop_id": 70022,  "stop_name": "CALTRAIN - 22ND ST STATION",  "stop_lat": 37.757692,  "stop_lon": -122.392318,  "zone_id": 3329}, {  "stop_id": 70151,  "stop_name": "CALTRAIN - ATHERTON STATION",  "stop_lat": 37.464458,  "stop_lon": -122.198152,  "zone_id": 3331}];
var travelList = [{  "trip_id": "RTD8997283",  "arrival_time": "05:40:00",  "departure_time": "05:40:00",  "stop_id": 70021,  "stop_sequence": 1}, {  "trip_id": "RTD8997283",  "arrival_time": "05:52:00",  "departure_time": "05:52:00",  "stop_id": 70021,  "stop_sequence": 2}, {  "trip_id": "RTD8449096",  "arrival_time": "07:33:00",  "departure_time": "07:33:00",  "stop_id": 70022,  "stop_sequence": 1}]
travelList.forEach(function(t) {
  // Find oobject
  var _stn = stnList.find(function(stn) {
    return stn.stop_id === t.stop_id
  });
  
  // check if trip property is defined or not
  if (!_stn.trip) _stn.trip = [];
  _stn.trip.push(t);
});
document.write("<pre>" + JSON.stringify(stnList, 0, 4) + "</pre>");

在纯Javascript中，您可以使用一个临时对象作为对目标项的引用，并将所有行程推送到另一个循环中。

此解决方案假定stop_id在两个阵列中都匹配。

function merge(array1, array2) {
    var o = {};
    array1.forEach(function (a) {
        o[a.stop_id] = a;
    });
    array2.forEach(function (a) {
        o[a.stop_id].trip = o[a.stop_id].trip || [];
        o[a.stop_id].trip.push(a);
    });
}
var array1 = [{ "stop_id": 70021, "stop_name": "CALTRAIN - 22ND ST STATION", "stop_lat": 37.757692, "stop_lon": -122.392318, "zone_id": 3329 }, { "stop_id": 70022, "stop_name": "CALTRAIN - 22ND ST STATION", "stop_lat": 37.757692, "stop_lon": -122.392318, "zone_id": 3329 }, { "stop_id": 70151, "stop_name": "CALTRAIN - ATHERTON STATION", "stop_lat": 37.464458, "stop_lon": -122.198152, "zone_id": 3331 }],
    array2 = [{ "trip_id": "RTD8997283", "arrival_time": "05:40:00", "departure_time": "05:40:00", "stop_id": 70021, "stop_sequence": 1 }, { "trip_id": "RTD8997283", "arrival_time": "05:52:00", "departure_time": "05:52:00", "stop_id": 70021, "stop_sequence": 2 }, { "trip_id": "RTD8449096", "arrival_time": "07:33:00", "departure_time": "07:33:00", "stop_id": 70022, "stop_sequence": 1 }];
merge(array1, array2);
document.write('<pre>' + JSON.stringify(array1, 0, 4) + '</pre>');

如果浏览器支持不是问题，Array.prototype.find可能是这种情况的最佳解决方案。

for (let item1 of arr1) {
  item1.trip = arr2.find(item2 => item2.stopid === item1.stopid);
}

这里有一个使用Array.prototype.map()的小线性，假设第一个JSON样本存储在stops中，第二个存储在trips中。（注意：它需要一个对象扩展函数。我使用了jQuery的$.extend()，但任何对象扩展方法都可以，例如Undercore/Lodash的_.extend()或Node.js的util._extend。）

var merged = stops.map(function(stop){ return $.extend({}, stop, {trip: trips.filter(function(trip) { return trip.stop_id === stop.stop_id; })}); })

如果你喜欢ES6的箭头功能，你可以把它做得更漂亮一点：

var merged = stops.map(stop=>$.extend({}, stop, {trip: trips.filter(trip => trip.stop_id === stop.stop_id)}))

这两种方法都会使原始数组保持不变，但如果您只是想将行程合并到站点中，而不关心保留原始数组，则可以删除$.extend()的第一个参数；这更有效（尤其是对于非常大的数据集），并且您不必担心将其保存在一个变量中：

stops.map(stop=>$.extend(stop, {trip: trips.filter(trip => trip.stop_id === stop.stop_id)}))

与stop_id匹配的站点阵列中的trains和filter上的map。

function getStations(stations, id) {
  return stations.filter(function(obj) {
    return obj.stop_id === id;
  });
}
function mergeInfo(trains, stations) {
  return trains.map(function(obj) {
    var filtered = getStations(stations, obj.stop_id);
    if (filtered.length) obj.trip = filtered;
    return obj;
  });
}
var output = mergeInfo(trains, stations);

演示