Javascript用另一个元素数组过滤一个数组

Question

I want to implement a tag filter for a search and get an array of entries (variable: entries) like this with many entries:

"entries":[
      {
         "id":1,
         "details":"text",
         "tags":[
            {
               "id":9,
               "label":"Label9"
            },
            {
               "id":6,
               "label":"Label6"
            },
         ],
         "date":"Mar 8, 2018 2:45:30 PM"
      }]

I want to filter this array with another array of tags (variable: tags) like this:

"tags":[
            {
               "id":6,
               "label":"Label6"
            }

At the end I need an array of those entries which contain all tags in the tags array. ]

I wrote this code but something is wrong. It compares the id of each tag.

 const entries = [{ "id": 1, "details": "text", "tags": [{ "id": 9, "label": "Label9" }, { "id": 6, "label": "Label6" }, ], "date": "Mar 8, 2018 2:45:30 PM" }] const tags = [{ "id": 6, "label": "Label6" }] function containSearchTag(tags) { return entries.filter(function(el) { for (let i = 0; i < el.tags.length; i++) { for (let j = 0; j < tags.length; j++) { return el.tags[i].id === tags[j].id; } } }); } console.log(containSearchTag(tags)); 

Answer 1

You could use a Set and filter the array by checking the tags array with the set.

 var entries = [{ id: 1, details: "text", tags: [{ id: 9, label: "Label9" }, { id: 6, label: "Label6" }], date: "Mar 8, 2018 2:45:30 PM" }], tags = [{ id: 6, label: "Label6" }], ids = new Set(tags.map(({ id }) => id)), result = entries.filter(({ tags }) => tags.some(({ id }) => ids.has(id))); console.log(result); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 2

You can use of Array.some to make it easier; like :

 const entries = [{ id: 1, details: 'text', tags: [{ id: 9, label: 'Label9' }, { id: 6, label: 'Label6', }, ], date: 'Mar 8, 2018 2:45:30 PM', }, { id: 1, details: 'text', tags: [{ id: 9, label: 'Label9', }, { id: 12, label: 'Label12', }, ], date: 'Mar 8, 2018 2:45:30 PM', }]; function filterByTag(tags) { return entries.filter(x => x.tags.some(y => tags.some(z => z.id === y.id))); } console.log(filterByTag([{ id: 6, label: 'Label6', }]).length); console.log(filterByTag([{ id: 9, label: 'Label6', }]).length); console.log(filterByTag([{ id: 17, label: 'Label6', }]).length); 

Answer 3

Use Array.filter , Array.every and Array.some

 let entries = [{"id":1,"details":"text","tags":[{"id":9,"label":"Label9"},{"id":6,"label":"Label6"},],"date":"Mar 8, 2018 2:45:30 PM"}]; let tag = [{"id":6,"label":"Label6"}]; // Create an array of unique id's let ids = tag.map(({id}) => id); // filter only those entries which have all the tag ids specified in tag array let result = entries.filter(({tags}) => ids.every((id) => tags.some((tag) => tag.id === id))); console.log(result); 

Answer 4

Try This I hope it will help you

 var data = JSON.parse('{"entries":[{"id":1,"details":"text","tags":[{"id":9,"label":"Label9"},{"id":6,"label":"Label6"}],"date":"Mar 8, 2018 2:45:30 PM"}]}'); let tag = JSON.parse('[{"id":6,"label":"Label6"}]'); data.entries.forEach((element)=>{ element.tags.forEach((value)=>{ tag.forEach((find)=>{ if(find.id === value.id) { console.log("tag is avaliable "); console.log(value); } }); }); }); 

Answer 5

An optimized solution can be to create a map of all the ids of tags present corresponding to each entry and than search for all the tag ids of the tag array. Search in map can be done in O(1) time. which can help in reducing the overall time complexity.

 var entries = [{"id":1,"details":"text","tags":[{"id":9,"label":"Label9"},{"id":6,"label":"Label6"}],"date":"Mar 8, 2018 2:45:30 PM"}]; var tags =[{"id":6, "label":"Label6"}]; var result = entries.filter((obj) => { var tagMap = obj.tags.reduce((a, o) =>{ a[o.id] = true; return a;}, {}); var bool = true; tags.forEach((o) => {var x = tagMap[o.id] ? true : false; bool = bool && x;}) return bool; }); console.log(result);