我正在尝试将几个JSON对象合并为一个，但我很难搞清楚如何获得所需的结果。
我不擅长创建复杂的for循环，这阻止了我成功。
示例area1.json
{
data: [
{
date: "2018031300",
p: [
{
lon: -5.555,
precip-intensity: 0,
wind-dir: 90.25001,
temperature: 3.7924042,
wind-speed: 8.692518,
weather-symbol: 3,
pressure-sealevel: 100247.5
},
{
lon: -4.444,
precip-intensity: 0,
wind-dir: 89.87501,
temperature: 3.1674042,
wind-speed: 8.731581,
weather-symbol: 3,
pressure-sealevel: 100238.5
}
]
},
{
date: "2018031301",
p: [
{
lon: -5.555,
precip-intensity: 0.00390625,
wind-dir: 97.500015,
temperature: 3.6734467,
wind-speed: 8.874923,
weather-symbol: 3,
pressure-sealevel: 100265.625
},
{
lon: -4.444,
precip-intensity: 0.00390625,
wind-dir: 96.000015,
temperature: 3.1109467,
wind-speed: 9.066329,
weather-symbol: 3,
pressure-sealevel: 100260.625
}
]
}
]
}
示例area2.json
{
data: [
{
date: "2018031300",
p: [
{
lon: -7.777,
precip-intensity: 0.0078125,
wind-dir: 96.25001,
temperature: 4.792404,
wind-speed: 10.200331,
weather-symbol: 3,
pressure-sealevel: 100237
},
{
lon: -6.666,
precip-intensity: 0,
wind-dir: 86.50001,
temperature: 4.167404,
wind-speed: 11.106581,
weather-symbol: 3,
pressure-sealevel: 100243
}
]
},
{
date: "2018031301",
p: [
{
lon: -7.777,
precip-intensity: 0.03125,
wind-dir: 92.625015,
temperature: 4.7359467,
wind-speed: 11.066329,
weather-symbol: 3,
pressure-sealevel: 100242.125
},
{
lon: -6.666,
precip-intensity: 0.0078125,
wind-dir: 90.750015,
temperature: 4.2359467,
wind-speed: 11.24211,
weather-symbol: 3,
pressure-sealevel: 100269.625
}
]
}
]
}
这是理想的结果
结合area3.json
{
data: [
{
date: "2018031300",
p: [
{
lon: -7.777,
precip-intensity: 0.0078125,
wind-dir: 96.25001,
temperature: 4.792404,
wind-speed: 10.200331,
weather-symbol: 3,
pressure-sealevel: 100237
},
{
lon: -6.666,
precip-intensity: 0,
wind-dir: 86.50001,
temperature: 4.167404,
wind-speed: 11.106581,
weather-symbol: 3,
pressure-sealevel: 100243
},
{
lon: -5.555,
precip-intensity: 0,
wind-dir: 90.25001,
temperature: 3.7924042,
wind-speed: 8.692518,
weather-symbol: 3,
pressure-sealevel: 100247.5
},
{
lon: -4.444,
precip-intensity: 0,
wind-dir: 89.87501,
temperature: 3.1674042,
wind-speed: 8.731581,
weather-symbol: 3,
pressure-sealevel: 100238.5
}
]
},
{
date: "2018031301",
p: [
{
lon: -7.777,
precip-intensity: 0.03125,
wind-dir: 92.625015,
temperature: 4.7359467,
wind-speed: 11.066329,
weather-symbol: 3,
pressure-sealevel: 100242.125
},
{
lon: -6.666,
precip-intensity: 0.0078125,
wind-dir: 90.750015,
temperature: 4.2359467,
wind-speed: 11.24211,
weather-symbol: 3,
pressure-sealevel: 100269.625
},
{
lon: -5.555,
precip-intensity: 0.00390625,
wind-dir: 97.500015,
temperature: 3.6734467,
wind-speed: 8.874923,
weather-symbol: 3,
pressure-sealevel: 100265.625
},
{
lon: -4.444,
precip-intensity: 0.00390625,
wind-dir: 96.000015,
temperature: 3.1109467,
wind-speed: 9.066329,
weather-symbol: 3,
pressure-sealevel: 100260.625
}
]
}
]
}
这个帖子的答案几乎就是...... Merge JSON Data by Key
我试过这段代码......
function mergeJson(target) {
for (var argi = 1; argi < arguments.length; argi++) {
var source = arguments[argi];
for (var key in source) {
if (!(key in target)) {
target[key] = [];
}
for (var i = 0; i < source[key].length; i++) {
target[key].push(source[key][i]);
}
}
}
return target;
}
var finalJson = mergeJson({}, area1, area2);
这给了我两个阵列......
答案 0 :(得分：3)
您复制粘贴了执行不同合并的代码段。给定的合并确实：
{ a: [1, 2] } + { a: [3] } = { a: [1, 2, 3] }
您的数据格式要求更复杂的合并：
{ data: [ { key: "a", values: [1, 2] } ] } +
{ data: [ { key: "a", values: [3] } ] } =
{ data: [ { key: "a", values: [1, 2, 3] }
放开第一个街区的逻辑，你会得到：
{ data: [ { key: "a", values: [1, 2] },
{ key: "a", values: [3] } ] }
正如你所说，这不是你想要的。
此比较应明确指出您需要找到一种方法来收集类似key值的数据（格式为date），然后才能开始合并。
这是一个示例实现，包括使用您提供的代码段时的预处理任务。代码在评论中解释。
var area1={data:[{date:"2018031300",p:[{lon:-5.555},{lon:-4.444}]},{date:"2018031301",p:[{lon:-5.555},{lon:-4.444}]}]},area2={data:[{date:"2018031300",p:[{lon:-7.777},{lon:-6.666}]},{date:"2018031301",p:[{lon:-7.777},{lon:-6.666}]}]};
const mergeAreas = (area1, area2) => {
// Skip the part of the structure that's always the same:
const vals1 = area1.data;
const vals2 = area2.data;
// to { [date]: points }
const valObj1 = arrayToObject(x => x.date, x => x.p, vals1);
const valObj2 = arrayToObject(x => x.date, x => x.p, vals2);
// Merge using the strategy you provided:
const mergedObj = mergeJson({}, valObj1, valObj2);
// Go back to the original format: [ { date, p } ]
const mergedData = objToArray(([date, p]) => ({ date, p }), mergedObj);
// Wrap it back up in the base structure: { data: [ ] }
return {
data: mergedData
}
};
// Takes an array of items and returns a single object.
// The items are stored in a key returned by `getKey`
// The items are represented by a value returned by `getValue`
// When `getKey` returns duplicates, it overwrites the previous entry
function arrayToObject(getKey, getValue, xs) {
return xs.reduce(
(acc, x) => Object.assign(acc, { [getKey(x)]: getValue(x) }),
{}
);
};
// This takes an object and returns an array based on its entries
function objToArray(kvpToItem, obj) {
return Object.entries(obj).map(kvpToItem);
}
// Provided by OP from answer https://stackoverflow.com/a/16302909/3297291
function mergeJson(target) {
for (var argi = 1; argi < arguments.length; argi++) {
var source = arguments[argi];
for (var key in source) {
if (!(key in target)) {
target[key] = [];
}
for (var i = 0; i < source[key].length; i++) {
target[key].push(source[key][i]);
}
}
}
return target;
}
// Run with example data
var finalJson = mergeAreas(area1, area2);
console.log(finalJson);