我希望以第二个json中匹配stop_id的方式合并两个json文件，并将其嵌套为第一个json中的子文档。
这就是我的意思：
first.json ：这是一个包含火车站点站点的json文件。每个都有一个stop_id字段。
[
{
"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
}]
second.json ：这包含通过stop_id
[
{
"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
}]
我希望以一种方式合并这些文档，在second.json中匹配stop_id s嵌套在first.json中的stop_id文档下面。因此，例如，合并的最终结果将如下所示：
merged.json
[{
"stop_id": 70021,
"stop_name": "CALTRAIN - 22ND ST STATION",
"stop_lat": 37.757692,
"stop_lon": -122.392318,
"zone_id": 3329,
"trip": [{
"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
}]
}, {
"stop_id": 70022,
"stop_name": "CALTRAIN - 22ND ST STATION",
"stop_lat": 37.757692,
"stop_lon": -122.392318,
"zone_id": 3329,
"trip": [{
"trip_id": "RTD8449096",
"arrival_time": "07:33:00",
"departure_time": "07:33:00",
"stop_id": 70022,
"stop_sequence": 1
}]
}, {
"stop_id": 70151,
"stop_name": "CALTRAIN - ATHERTON STATION",
"stop_lat": 37.464458,
"stop_lon": -122.198152,
"zone_id": 3331
}]
对于大型数据集，是否可以通过javascript或任何其他方法进行此类合并？
答案 0 :(得分：1)
您可以尝试这样的事情：
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>");
答案 1 :(得分：1)
在普通的Javascript中，您可以使用临时对象来引用目标项，并在另一个循环中推送所有行程。
此解决方案假设stop_id在两个数组中都匹配。
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>');
答案 2 :(得分：1)
你可以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);
答案 3 :(得分：1)
如果浏览器支持不是问题，Array.prototype.find可能是此案例的最佳解决方案。
for (let item1 of arr1) {
item1.trip = arr2.find(item2 => item2.stopid === item1.stopid);
}
答案 4 :(得分：1)
假设第一个JSON样本存储在Array.prototype.map()中，第二个JSON样本存储在stops中，则使用trips进行一次单行调换。 （警告：它需要一个对象扩展函数。我使用了jQuery＆＃39; $.extend()，但任何对象扩展方法都可行，例如Underscore / Lodash＆＃39; s _.extend()或Node.js＆＃39 ; s util._extend。）
var merged = stops.map(function(stop){ return $.extend({}, stop, {trip: trips.filter(function(trip) { return trip.stop_id === stop.stop_id; })}); })
如果你喜欢ES6's arrow functions，你可以让它更漂亮一点：
var merged = stops.map(stop=>$.extend({}, stop, {trip: trips.filter(trip => trip.stop_id === stop.stop_id)}))
这两个都会保持原始数组不变，但是如果您只想将旅行合并到停靠点并且您不关心保留原始数据，则可以删除$.extend()的第一个参数;这样效率更高（特别是对于非常大的数据集），您不必担心将其保存在变量中：
stops.map(stop=>$.extend(stop, {trip: trips.filter(trip => trip.stop_id === stop.stop_id)}))
答案 5 :(得分：0)
map的电台数组中，{p> trains超过filter和stop_id。
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);