我从包含currentDataResult和pastDataResult的api返回一个对象数组。
[
{
"sectionCurrentDataResult": [
{
"section_id": 14785,
"subdivision_name": "Stratton Woods",
},
{
"section_id": 14790,
"subdivision_name": "Stratton Woods",
},
{
"section_id": 14791,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14792,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14781,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14786,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14787,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14788,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14782,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14783,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 14784,
"subdivision_name": "Stratton Woods"
},
{
"section_id": 5326,
"subdivision_id": 1439,
"subdivision_name": "Stratton Woods"
}
]
},
{
"sectionPastDataResult": [
{
"section_id": 5326,
"price_min": 177,
"price_max": 235
},
{
"section_id": 14785,
"price_min": 190,
"price_max": 220
},
{
"section_id": 14786,
"price_min": 238,
"price_max": 292
},
{
"section_id": 14788,
"price_min": 186,
"price_max": 205
},
{
"section_id": 14790,
"price_min": 150,
"price_max": 269
},
{
"section_id": 14783,
"price_min": 150,
"price_max": 260
},
{
"section_id": 14787,
"price_min": 90,
"price_max": 90
},
{
"section_id": 14792,
"price_min": 177,
"price_max": 235
},
{
"section_id": 14791,
"price_min": 145,
"price_max": 221
},
{
"section_id": 14784,
"price_min": 148,
"price_max": 186
},
{
"section_id": 14781,
"price_min": 155,
"price_max": 200
},
{
"section_id": 14782,
"price_min": 150,
"price_max": 170
}
]
}
]
我需要将匹配的pastDataObject（by section_id）对象作为嵌套数组推送到currentDataResult对象中。这就是它需要的样子
"sectionCurrentDataResult": [
{
"section_id": 14785,
"subdivision_name": "Stratton Woods",
"sectionHistory":[{
"section_id": 14785,
"price_min": 190,
"price_max": 220
}]
},
{
"section_id": 14790,
"subdivision_name": "Stratton Woods",
"sectionHistory":[{
"section_id": 14790,
"price_min": 150,
"price_max": 269
}]
},
etc....
]
我创建了一个服务，它接收当前和过去的数据结果，并重新排序过去的数据结果以匹配当前。我需要帮助的是将过去的数据数组推送到当前数据对象。现在它错误地将整个过去的数据数组推送到当前数据数组中的第一个对象。我已经设置了一个带有我的代码的plunker。
app.controller('MainCtrl', function($scope,bigEnchilada,inputHistorySvc) {
for (var i = 0; i < bigEnchilada[0].sectionCurrentDataResult.length; i++) {
bigEnchilada[0].sectionCurrentDataResult[i].sectionHistory = inputHistorySvc.historyInputs(bigEnchilada);
}
$scope.sections = bigEnchilada[0].sectionCurrentDataResult;
});
答案 0 :(得分：2)
这就是你要找的东西吗？
var app = angular.module('angularjs-starter', []);
app.controller('MainCtrl', function($scope,bigEnchilada,inputHistorySvc) {
var list = inputHistorySvc.historyInputs(bigEnchilada);
for (var i = 0; i < bigEnchilada[0].sectionCurrentDataResult.length; i++){
for (var j = 0; j < list.length; j++ ) {
if(bigEnchilada[0].sectionCurrentDataResult[i].section_id == list[j].section_id){
bigEnchilada[0].sectionCurrentDataResult[i].sectionHistory = list[j];
}
}
}
$scope.sections = bigEnchilada[0].sectionCurrentDataResult;
});
输出：
[
{
"section_id": 14785,
"subdivision_name": "Stratton Woods",
"sectionHistory": {
"section_id": 14785,
"price_min": 190,
"price_max": 220
}
},
{
"section_id": 14790,
"subdivision_name": "Stratton Woods",
"sectionHistory": {
"section_id": 14790,
"price_min": 150,
"price_max": 269
}
},
{
"section_id": 14791,
"subdivision_name": "Stratton Woods",
"sectionHistory": {
"section_id": 14791,
"price_min": 145,
"price_max": 221
}
},
等
答案 1 :(得分：1)
var inputHistory = inputHistorySvc.historyInputs(bigEnchilada);
for (var i = 0; i < bigEnchilada[0].sectionCurrentDataResult.length; i = i + 1) {
for (x = 0; x < inputHistory.length; x = x + 1) {
if (bigEnchilada[0].sectionCurrentDataResult[i].section_id === inputHistory[x].section_id) {
bigEnchilada[0].sectionCurrentDataResult[i].sectionHistory = inputHistory[x];
}
}
}
$scope.sections = bigEnchilada[0].sectionCurrentDataResult;
答案 2 :(得分：1)
您可以使用临时对象和两个独立的循环。一个用于获取所有引用，第二个用于将数据分配给引用。此解决方案中Big O：O(n)。
var data = [{ "sectionCurrentDataResult": [{ "section_id": 14785, "subdivision_name": "Stratton Woods", }, { "section_id": 14790, "subdivision_name": "Stratton Woods", }, { "section_id": 14791, "subdivision_name": "Stratton Woods" }, { "section_id": 14792, "subdivision_name": "Stratton Woods" }, { "section_id": 14781, "subdivision_name": "Stratton Woods" }, { "section_id": 14786, "subdivision_name": "Stratton Woods" }, { "section_id": 14787, "subdivision_name": "Stratton Woods" }, { "section_id": 14788, "subdivision_name": "Stratton Woods" }, { "section_id": 14782, "subdivision_name": "Stratton Woods" }, { "section_id": 14783, "subdivision_name": "Stratton Woods" }, { "section_id": 14784, "subdivision_name": "Stratton Woods" }, { "section_id": 5326, "subdivision_id": 1439, "subdivision_name": "Stratton Woods" }] }, { "sectionPastDataResult": [{ "section_id": 5326, "price_min": 177, "price_max": 235 }, { "section_id": 14785, "price_min": 190, "price_max": 220 }, { "section_id": 14786, "price_min": 238, "price_max": 292 }, { "section_id": 14788, "price_min": 186, "price_max": 205 }, { "section_id": 14790, "price_min": 150, "price_max": 269 }, { "section_id": 14783, "price_min": 150, "price_max": 260 }, { "section_id": 14787, "price_min": 90, "price_max": 90 }, { "section_id": 14792, "price_min": 177, "price_max": 235 }, { "section_id": 14791, "price_min": 145, "price_max": 221 }, { "section_id": 14784, "price_min": 148, "price_max": 186 }, { "section_id": 14781, "price_min": 155, "price_max": 200 }, { "section_id": 14782, "price_min": 150, "price_max": 170 }] }];
void function () {
var o = {};
data[0].sectionCurrentDataResult.forEach(function (a) {
o[a.section_id] = a;
});
data[1].sectionPastDataResult.forEach(function (a) {
o[a.section_id].sectionHistory = [a];
});
}();
document.write('<pre>' + JSON.stringify(data, 0, 4) + '</pre>');
答案 3 :(得分：1)
如果你打破这个json，你可以实际避免嵌套循环，按section_id对它们进行排序，然后在一个循环中创建所需的对象。您还可以避开if -else循环和equality check
// Get sorted array of sectionCurrentDataResult
var array1 = a[0]["sectionCurrentDataResult"].sort(function(x,y){
return x.section_id>y.section_id? 1 : x.section_id<y.section_id? -1 :0;
})
// Get sorted array of sectionPastDataResult
var array2= a[1]["sectionPastDataResult"].sort(function(x,y){
return x.section_id>y.section_id? 1 : x.section_id<y.section_id? -1 :0;
})
// Will be populated with merged data
var sectionCurrentDataResult=[]
for(var a = 0;a<array1.length;a++){
var sectionHistory=[];
sectionHistory.push({
"section_id":array2[a].section_id,
"price_min":array2[a].price_min,
"price_max":array2[a].price_max
})
sectionCurrentDataResult.push({
"section_id":array1[a].section_id,
"subdivision_name":array1[a].subdivision_name,
"sectionHistory":sectionHistory
})
}
console.log(sectionCurrentDataResult);