我有一个对象数组如下所示,我使用sequelize ORM从我的数据库中读取: 我希望从一个部分获得我的所有视频,但我能用sequelize返回的更好的是:
[{
"id": 2,
"name": "Ru",
"subsection": 1,
"Video": {
"id": 11,
"source": "sourrrccrsss22222",
"videoSubSection": 2
}
},
{
"id": 2,
"name": "Ru",
"subsection": 1,
"Video": {
"id": 12,
"source": "sourrrccrsss111",
"videoSubSection": 2
}
},
{
"id": 1,
"name": "Oc",
"subsection": 1,
"Video": {
"id": 13,
"source": "sourrrcc",
"videoSubSection": 1
}
},
{
"id": 1,
"name": "Oc",
"subsection": 1,
"Video": {
"id": 14,
"source": "sourrrcc",
"videoSubSection": 1
}
}]
有没有办法合并和组合我的数组中的对象来获得这样的东西:
[{
"id": 2,
"name": "Ru",
"subsection": 1,
"Video": [{
"id": 11,
"source": "sourrrccrsss22222",
"videoSubSection": 2
},{
"id": 12,
"source": "sourrrccrsss111",
"videoSubSection": 2
}]
},
{
"id": 1,
"name": "Oc",
"subsection": 1,
"Video": [{
"id": 13,
"source": "sourrrcc",
"videoSubSection": 1
},{
"id": 14,
"source": "sourrrcc",
"videoSubSection": 1
}]
}
接近最多的函数是_.mergeWith(object,sources,customizer),但我遇到的主要问题是我有对象并且需要合并这个对象。
答案 0 :(得分:1)
在普通的Javascript中,您可以将Array#forEach()
与数组的临时对象一起使用。
var data = [{ id: 2, name: "Ru", subsection: 1, Video: { id: 11, source: "sourrrccrsss22222", VideoSubSection: 2 } }, { id: 2, name: "Ru", subsection: 1, Video: { id: 12, source: "sourrrccrsss111", VideoSubSection: 2 } }, { id: 1, name: "Oc", subsection: 1, Video: { id: 13, source: "sourrrcc", VideoSubSection: 1 } }, { id: 1, name: "Oc", subsection: 1, Video: { id: 14, source: "sourrrcc", VideoSubSection: 1 } }],
merged = function (data) {
var r = [], o = {};
data.forEach(function (a) {
if (!(a.id in o)) {
o[a.id] = [];
r.push({ id: a.id, name: a.name, subsection: a.subsection, Video: o[a.id] });
}
o[a.id].push(a.Video);
});
return r;
}(data);
document.write('<pre>' + JSON.stringify(merged, 0, 4) + '</pre>');
答案 1 :(得分:1)
也许尝试transform():
_.transform(data, (result, item) => {
let found;
if ((found = _.find(result, { id: item.id }))) {
found.Video.push(item.Video);
} else {
result.push(_.defaults({ Video: [ item.Video ] }, item));
}
}, []);
使用reduce()也可以在此处使用,但transform()
不那么详细。
答案 2 :(得分:0)
你可以这样做(test
是你的db输出)
var result = [];
var map = [];
_.forEach(test, (o) => {
var temp = _.clone(o);
delete o.Video;
if (!_.some(map, o)) {
result.push(_.extend(o, {Video: [temp.Video]}));
map.push(o);
} else {
var index = _.findIndex(map, o);
result[index].Video.push(temp.Video);
}
});
console.log(result); // outputs what you want.