考虑这个对象,有两个NL语言通道和一个EN语言通道:
[
{
"name": "De Redactie",
"channels": [
{
"name": "headlines",
"pubDate": "2017-05-15 09:15:00",
"language": "nl",
"items": [
]
},
{
"name": "headlines English",
"pubDate": "2017-05-14 18:05:00",
"language": "en",
"items": [
]
},
{
"name": "politiek",
"pubDate": "2017-05-14 20:11:00",
"language": "nl",
"items": [
]
}
]
}
]
我如何划分它们以便我能得到这个结果:
[
{
"name": "De Redactie",
"channels": [
{
"name": "headlines",
"pubDate": "2017-05-15 09:15:00",
"language": "nl",
"items": [
]
},
{
"name": "politiek",
"pubDate": "2017-05-14 20:11:00",
"language": "nl",
"items": [
]
}
]
}
{
"name": "De Redactie",
"channels": [
{
"name": "headlines English",
"pubDate": "2017-05-14 18:05:00",
"language": "en",
"items": [
]
}
]
}
]
请注意,这是dummyData。实际数据可以包含一种语言的x量和第二种或第三种或第四种的y量......
我试过在lodash文档中查找正确的函数组合。还尝试了各种复杂的forEach结构,但无法绕过它。
最好使用lodash或打字稿的解决方案,因为我在Angular 4中工作。
答案 0 :(得分:3)
使用Array#map迭代数组。对于每个对象,使用带有destructuring的object rest提取频道数组。使用Array#reduce对频道进行迭代,并使用相同语言将频道分组到Map。通过传播Map's values iterator来转换回数组。
通过映射对象并将组指定为对象的channels
prop来创建对象数组。通过传播到Array#concat:
const data = [{"name":"De Redactie","channels":[{"name":"headlines","pubDate":"2017-05-15 09:15:00","language":"nl","items":[]},{"name":"headlines English","pubDate":"2017-05-14 18:05:00","language":"en","items":[]},{"name":"politiek","pubDate":"2017-05-14 20:11:00","language":"nl","items":[]}]}];
const result = [].concat(...data.map(({ channels, ...rest }) => {
const channelGroups = [...channels.reduce((m, channel) => {
m.has(channel.language) || m.set(channel.language, []);
m.get(channel.language).push(channel);
return m;
}, new Map()).values()];
return channelGroups.map((channels) => ({
...rest,
channels
}));
}));
console.log(result);
答案 1 :(得分:1)
const res = _.chain(arr)
.flatMap(item => _.map( // get array of channels with parent name
item.channels,
channel => _.assign({}, channel, { parentName: item.name })
))
.groupBy('language') // group channels by language
.values() // get channel arrays for each lang
.map(langArrs => ({ // set finished structure
name: _.first(langArrs).parentName, // get name for lang channels
channels: _.omit(langArrs, ['parentName']) // set channels without parent name
}))
.value();