如何为对象添加值并删除当前值?
How to add value to object and remove current value?
我有这样的数据:
users = [{
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "HARD"
}, {
"emp_id": 2,
"user": {
"emp_full_name": "test2",
"emp_email": "test2@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "EASY"
}, {
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 4,
"name": "Javascript",
"hashtag_group_id": 1
},
"difficulty": "HARD"
}]
我想将 hashtag
添加到具有相同 emp_id
的对象中。如果 emp_id
有多个数据,则应删除包含 emp_id
和单个 hashtag
数据的数据。
所以基本上这就是我的预期:
[{
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": [{
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
}, {
"id": 4,
"name": "Javascript",
"hashtag_group_id": 1
}],
"difficulty": "HARD"
}, {
"emp_id": 2,
"user": {
"emp_full_name": "test2",
"emp_email": "test2@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "EASY"
}]
如何这样转换数据?
我不知道如何解决这个问题,我尝试使用 filter()
和 map()
以及一些验证条件,但无法正常工作。
您可以创建一个由 emp_id
键控的地图并通过该键收集用户。当已有条目时,使用 [].concat
扩展 hashtag
。如果它还不是数组,这将创建一个数组。
const users = [{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "test@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "HARD"},{"emp_id": 2,"user": {"emp_full_name": "test2","emp_email": "test2@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "EASY"},{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "test@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 4,"name": "Javascript","hashtag_group_id": 1},"difficulty": "HARD"}];
const map = new Map;
for (const user of users) {
const match = map.get(user.emp_id);
if (match) match.hashtag = [].concat(match.hashtag, user.hashtag);
else map.set(user.emp_id, {...user});
}
const result = [...map.values()];
console.log(result);
您可以将 .reduce()
与 .findIndex()
一起使用。试试这个
let users = '[{"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "test@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "HARD"}, {"emp_id": 2, "user": {"emp_full_name": "test2", "emp_email": "test2@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "EASY"}, {"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "test@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 4, "name": "Javascript", "hashtag_group_id": 1 }, "difficulty": "HARD"} ]';
users = JSON.parse(users)
users = users.reduce((arr, o) => {
let idx = arr.findIndex(({emp_id}) => emp_id === o.emp_id);
if( idx !== -1 ) arr[idx].hashtag = [].concat(arr[idx].hashtag, o.hashtag)
else arr.push(o)
return arr;
}, []);
console.log(users)
我有这样的数据:
users = [{
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "HARD"
}, {
"emp_id": 2,
"user": {
"emp_full_name": "test2",
"emp_email": "test2@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "EASY"
}, {
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 4,
"name": "Javascript",
"hashtag_group_id": 1
},
"difficulty": "HARD"
}]
我想将 hashtag
添加到具有相同 emp_id
的对象中。如果 emp_id
有多个数据,则应删除包含 emp_id
和单个 hashtag
数据的数据。
所以基本上这就是我的预期:
[{
"emp_id": 1,
"user": {
"emp_full_name": "Test",
"emp_email": "test@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": [{
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
}, {
"id": 4,
"name": "Javascript",
"hashtag_group_id": 1
}],
"difficulty": "HARD"
}, {
"emp_id": 2,
"user": {
"emp_full_name": "test2",
"emp_email": "test2@gmail.com",
"emp_phone_no": null,
"preferred_work_type": null
},
"hashtag": {
"id": 1,
"name": "NodeJs",
"hashtag_group_id": 1
},
"difficulty": "EASY"
}]
如何这样转换数据?
我不知道如何解决这个问题,我尝试使用 filter()
和 map()
以及一些验证条件,但无法正常工作。
您可以创建一个由 emp_id
键控的地图并通过该键收集用户。当已有条目时,使用 [].concat
扩展 hashtag
。如果它还不是数组,这将创建一个数组。
const users = [{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "test@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "HARD"},{"emp_id": 2,"user": {"emp_full_name": "test2","emp_email": "test2@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "EASY"},{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "test@gmail.com","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 4,"name": "Javascript","hashtag_group_id": 1},"difficulty": "HARD"}];
const map = new Map;
for (const user of users) {
const match = map.get(user.emp_id);
if (match) match.hashtag = [].concat(match.hashtag, user.hashtag);
else map.set(user.emp_id, {...user});
}
const result = [...map.values()];
console.log(result);
您可以将 .reduce()
与 .findIndex()
一起使用。试试这个
let users = '[{"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "test@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "HARD"}, {"emp_id": 2, "user": {"emp_full_name": "test2", "emp_email": "test2@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "EASY"}, {"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "test@gmail.com", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 4, "name": "Javascript", "hashtag_group_id": 1 }, "difficulty": "HARD"} ]';
users = JSON.parse(users)
users = users.reduce((arr, o) => {
let idx = arr.findIndex(({emp_id}) => emp_id === o.emp_id);
if( idx !== -1 ) arr[idx].hashtag = [].concat(arr[idx].hashtag, o.hashtag)
else arr.push(o)
return arr;
}, []);
console.log(users)