Javascript连接两个JSON数组

我有一个包含两个数组的JSON文件， 我希望能够从FEED对象的第二个数组中获得ACLIST中“Rcvr”下的接收方名称

关系是ACLIST Rcvr到FEEDS id，然后返回名称

样本输出：

var输出={[“rcvr”：2，“name”NZMB1“]}

我相信这可以用concat完成？但是我没有运气

提要：

"feeds": [{
  "id": 1,
  "name": "NZWP1",
  "polarPlot": false
}, {
  "id": 2,
  "name": "NZMB1",
  "polarPlot": false
}, {
  "id": 3,
  "name": "PUB_VRS",
  "polarPlot": false
}, {
  "id": 4,
  "name": "NZAP1",
  "polarPlot": true
}, {
  "id": 7,
  "name": "PUB_IN",
  "polarPlot": false
}, {
  "id": 9,
  "name": "MLAT",
  "polarPlot": false
}, {
  "id": 10,
  "name": "ADSBEX",
  "polarPlot": false
}, {
  "id": 11,
  "name": "NZSI/AU",
  "polarPlot": false
}, {
  "id": 13,
  "name": "MLATH",
  "polarPlot": false
}, {
  "id": 14,
  "name": "VRS1",
  "polarPlot": false
}, {
  "id": 15,
  "name": "VRS3",
  "polarPlot": false
}, {
  "id": 18,
  "name": "Pub_Main",
  "polarPlot": false
}, {
  "id": 19,
  "name": "PRIV_IN",
  "polarPlot": false
}, {
  "id": 20,
  "name": "PUB_Pi",
  "polarPlot": false
}, {
  "id": 21,
  "name": "Sat_Feed",
  "polarPlot": false
}, {
  "id": 5,
  "name": "Merged Feed",
  "polarPlot": false
}, {
  "id": 6,
  "name": "Merged Feed 2",
  "polarPlot": false
}, {
  "id": 12,
  "name": "NZ",
  "polarPlot": false
}]

列表：

"acList": [{
  "Id": 11363733,
  "Rcvr": 2,
  "HasSig": false,
  "Icao": "AD6595",
  "Bad": false,
  "Reg": "N962WN",
  "FSeen": "\/Date(1533936304145)\/",
  "TSecs": 1,
  "CMsgs": 2,
  "AltT": 0,
  "Call": "SWA1088",
  "Tisb": false,
  "Spd": 163.0,
  "Trak": 287.0,
  "TrkH": false,
  "Type": "B737",
  "Mdl": "Boeing 737NG 7H4/W",
  "Man": "Boeing",
  "CNum": "36963",
  "From": "MCO Orlando, United States",
  "To": "OKC Will Rogers World, Oklahoma City, United States",
  "Stops": ["PIT Pittsburgh, United States", "BWI Baltimore/Washington International Thurgood Marshal, Baltimore, United States", "PBI Palm Beach, West Palm Beach, United States", "TPA Tampa, United States", "STL Lambert St Louis, United States"],
  "Op": "Southwest Airlines",
  "OpIcao": "SWA",
  "Sqk": "",
  "Vsi": 1536,
  "VsiT": 0,
  "WTC": 2,
  "Species": 1,
  "Engines": "2",
  "EngType": 3,
  "EngMount": 0,
  "Mil": false,
  "Cou": "United States",
  "HasPic": false,
  "Interested": false,
  "FlightsCount": 413,
  "SpdTyp": 0,
  "CallSus": false,
  "Trt": 2,
  "Year": "2011"
}]

你需要一个地图和查找。不过，您仍然需要定义关系。。。没有说明连接是什么


let res=feed.map（feed=>{
让lookup=acList.find（ac=>{
//当ac与馈电相关时，TODO返回true
})
返回{
name:feed.name，
rcvr:lookup.rcvr
}
})
我终于想出办法去做了

var feedslist = [];
for (i = 0; i < feeds.length; i++){
    if (feeds[i].id == p.Rcvr) {
        feedslist.push(feeds[i].name);
    } else if (feedslist != feeds[i].name) {
        feedslist.splice(i, 1);
    }
}

$(" #rcvr").text(feedslist);

}

var feedslist=[]；
对于（i=0；i
到目前为止您尝试了什么？您所说的“我没有运气”是什么意思？预期的结果是什么？您当前得到了什么？在转换为JSON之前加入数组会更容易：
let arr1=[1，2]；设arr2=[3,4]；设arr3=[…arr1，…arr2]；//[1,2,3,4]
甚至不清楚两者之间的关系，以匹配它们。请证明否决票的合理性。我真的很想知道为什么这不能解决这个问题。谢谢，谢谢，不是我，所以不知道为什么。