我有很多客户,我只想提取每个客户的经度和纬度,并且我想得到这样的客户
latLong=[{lat:123,long:322},{lat:111,long:333},{lat:33,long:11}]
这是我的客户数组,我只想从每个对象中提取经纬度,然后将它们放入对象数组中。
"customers": [
{
"id": "5",
"aname": "Sts Customers",
"type": "2",
"address": "house no e 4",
"area_id": "7",
"cell_no": "03334488",
"opn_type": "Debit",
"opngbl": "2564",
"lat": "33.7997",
"longi": "73.04052769999998",
"company_id": "1",
"acct_no": ""
},
{
"id": "14",
"aname": "New Customer",
"type": "2",
"address": "house no 4",
"area_id": "8",
"cell_no": "7878",
"opn_type": "Credit",
"opngbl": "2541",
"lat": "33.7997",
"longi": "73.04052769999998",
"company_id": "1",
"acct_no": ""
},
{
"id": "15",
"aname": "one more customer",
"type": "2",
"address": "jjkhklj",
"area_id": "8",
"cell_no": "8876987",
"opn_type": "Credit",
"opngbl": "454",
"lat": "33.7997",
"longi": "73.04052769999998",
"company_id": "1",
"acct_no": ""
}
],
答案 0 :(得分:0)
您可以执行以下操作。
const source = {
customers: [
{
id: '5',
aname: 'Sts Customers',
type: '2',
address: 'house no e 4',
area_id: '7',
cell_no: '03334488',
opn_type: 'Debit',
opngbl: '2564',
lat: '33.7997',
longi: '73.04052769999998',
company_id: '1',
acct_no: '',
},
{
id: '14',
aname: 'New Customer',
type: '2',
address: 'house no 4',
area_id: '8',
cell_no: '7878',
opn_type: 'Credit',
opngbl: '2541',
lat: '33.7997',
longi: '73.04052769999998',
company_id: '1',
acct_no: '',
},
{
id: '15',
aname: 'one more customer',
type: '2',
address: 'jjkhklj',
area_id: '8',
cell_no: '8876987',
opn_type: 'Credit',
opngbl: '454',
lat: '33.7997',
longi: '73.04052769999998',
company_id: '1',
acct_no: '',
},
],
};
const result = source.customers.map(({ lat, longi: long }) => ({ lat, long }));
console.log(result);
基本上,我们可以使用map
遍历source.customers
数组。
然后我们将传递给回调的customer
对象解构以仅选择lat
和longi
,将longi
重命名为long
。
最后,我们返回一个仅包含我们感兴趣的键值对的对象。