我有多个数组,其中有多种数据类型,例如:
var student= [{
id : 1,
name : 'name1',
year: 2016,
dist_id: 251,
zone_id: 25106
},
{
id : 1,
name : 'name2',
year: 2018,
dist_id: 252,
zone_id: 25212
},];
var dist= [{
id : 251,
name : 'dist1'
},
{
id : 252,
name : 'dist2'
}];
var zone= [{
id : 25106,
name : 'zone1'
},
{
id : 25212,
name : 'zone2'
}];
我想创建一个将所有数据组合为一个数组,以便数组中的字段如下所示:
var merge = [{
id: 1,
name : 'name1',
year: 2016,
distname : 'dist1',
zonename: 'zone1',
},
{
id: 2,
name : 'name2',
year: 2018,
distname : 'dist2',
zonename: 'zone2',
}];
先谢谢了。
答案 0 :(得分:2)
您可以使用Map
作为距离和区域,并将值映射到新对象中。
var student = [{ id: 1, name: 'name1', year: 2016, dist_id: 251, zone_id: 25106 }, { id: 1, name: 'name2', year: 2018, dist_id: 252, zone_id: 25212 }],
dist = [{ id: 251, name: 'dist1' }, { id: 252, name: 'dist2' }],
zone = [{ id: 25106, name: 'zone1' }, { id: 25212, name: 'zone2' }],
distMap = new Map(dist.map(({ id, name: distname }) => [id, { distname }])),
zoneMap = new Map(zone.map(({ id, name: zonename }) => [id, { zonename }])),
merged = student.map(({ id, name, year, dist_id, zone_id }) => Object.assign(
{ id, name, year },
distMap.get(dist_id),
zoneMap.get(zone_id)
));
console.log(merged);
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答案 1 :(得分:1)
您可以将map()
数组student
到所需的merge
数组中,对于每次映射迭代,您都将在dist
和zone
数组中搜索分别在dist_id
和zone_id
上匹配的项目,并将的name
合并到映射结果中:
var student= [
{ id : 1, name : 'name1', year: 2016, dist_id: 251, zone_id: 25106 },
{ id : 1, name : 'name2', year: 2018, dist_id: 252, zone_id: 25212 } ];
var dist= [
{ id : 251, name : 'dist1' }, { id : 252, name : 'dist2' }];
var zone= [
{ id : 25106, name : 'zone1' }, { id : 25212, name : 'zone2' } ];
// Perform a mapping over the student array to aquire merge array in required
// format, with required distname/zonename data
var merge = student.map((s) => {
// Search dist and zone arrays for items that match of dist_id/zone_id
// by filtering and mapping these arrays to find distname and zonename
// for this student
const distname = dist.filter(d => d.id === s.dist_id).map(d => d.name)[0];
const zonename = zone.filter(z => z.id === s.zone_id).map(z => z.name)[0];
return {
id : s.id,
name : s.name,
year : s.year,
distname : distname,
zonename : zonename
}
});
console.log('required merge array:', merge)