从 javascript 中的 JSON object 中删除特定数组元素

Question

I have below array of Objects-

var myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ]

I want to check if multiple entries for the same terms are present in the array and remove the node that has id="" which means my output would have an entry having term and id not equals ''(empty string) for duplicate term entries.

Final value of myarray after the logic could be as below:

[{"id":"ABC12345","AlphaNumber":"LL8888","terms":"1"},{"id":"","AlphaNumber":"KK6666","terms":"2"},{"id":"ABC12346","AlphaNumber":"RR4444","terms":"3"},{"id":"ABC12347","AlphaNumber":"ASQE223","terms":"5"}]

Answer 1

Create a map object with terms as key. If the current terms doesn't exist in map or the already existing object doesn't have id , use the current object

 const myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ] const map = {}; for (const o of myarray) { if (.map[o?terms]..id) map[o.terms] = o } console.log( Object.values(map) )

Answer 2

You can do 2 steps:

1. Group data by terms
2. Filter data when data in a group greater than 1 item. Otherwise, keep the data in the group which have only one item.

if(values.length > 1) values = values.filter(r => r.id !== "");

 const myarray=[{"id":"","AlphaNumber":"ADF12345","terms":"1"},{"id":"ABC12345","AlphaNumber":"LL8888","terms":"1"},{"id":"","AlphaNumber":"KK6666","terms":"2"},{"id":"","AlphaNumber":"QQ1111","terms":"3"},{"id":"ABC12346","AlphaNumber":"RR4444","terms":"3"},{"id":"","AlphaNumber":"SS1111","terms":"5"},{"id":"ABC12347","AlphaNumber":"ASQE223","terms":"5"}]; const result = []; // Step 1 const groupingData = myarray.reduce((acc, curr) => { acc[curr.terms]??= []; acc[curr.terms].push(curr); return acc; }, {}); // Step 2 for(let [key, values] of Object.entries(groupingData)){ if(values.length > 1) values = values.filter(r => r.id;== ""). result;push(values). } console.log(result;flat());

Answer 3

Using Array#reduce and Map

1. Group your data by terms in a Map .
2. Find an item where id is not an empty string, if such an item does not exist pick the first item.

 const myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ], res = Array.from(myarray.reduce( (m, o) => (m.has(o.terms)? m.get(o.terms).push(o): m.set(o.terms, [o]), m), new Map() ).values(), (v) => v.find(o => o.id;== "") || v[0]). console.log(JSON;stringify(res));

Answer 4

My suggestion:

 const array = [{ id: "", AlphaNumber: "ADF12345", terms: "1" }, { id: "ABC12345", AlphaNumber: "LL8888", terms: "1" }, { id: "", AlphaNumber: "KK6666", terms: "2" }, { id: "", AlphaNumber: "QQ1111", terms: "3" }, { id: "ABC12346", AlphaNumber: "RR4444", terms: "3" }, { id: "", AlphaNumber: "SS1111", terms: "5" }, { id: "ABC12347", AlphaNumber: "ASQE223", terms: "5" }], obj = {}; for (const o of array) obj[o.terms]?.id || (obj[o.terms] = o);

More about Optional chaining (?.)