如何返回属性与数组匹配的对象数组？

Question

I have an array like

array = [
  { name: "john", tag: ["tag1", "tag2"] },
  { name: "doe", tag: ["tag2"] },
  { name: "jane", tag: ["tag2", "tag3"] }
];

I want to get a new array of objects which contain both "tag2" and "tag3", but not only "tag2" or both "tag1" and "tag2".

Result should be:

newArray = [{ name: "jane", tag: ["tag2", "tag3"] }];

I tried to do it using this process:

tags = ["tag2", "tag3"];
newArray = [];
tags.forEach(t => {
  array.forEach(data => {
    data.tag.forEach(item => {
      if (item === t) {
        newArray.push(data);
      }
    });
  });
});

But I get all the items instead.

Answer 1

If I understood you correctly, you want to search through your top-level array, to find all items whose tag property is an array exactly matching ['tag2', 'tag3'] .

You can achieve this by filtering your array based on the above condition.

Here's one approach:

 const array = [ { name: 'john', tag: ['tag1', 'tag2'] }, { name: 'doe', tag: ['tag2'] }, { name: 'jane', tag: ['tag2', 'tag3'] } ]; const tagsToMatchOn = ['tag2', 'tag3']; // find all elements who's tag property exactly matches // the above tags (in presence, not necessarily in order) const newArray = array.filter(item => ( item.tag.length === tagsToMatchOn.length && tagsToMatchOn.every(t => item.tag.includes(t)) )); console.log(newArray); 

If instead, you wanted to find all items whose tag property is an array including all of ['tag2', 'tag3' ] but can also include other tags, you can try something like this:

 const array = [ { name: 'john', tag: ['tag1', 'tag2'] }, { name: 'doe', tag: ['tag2'] }, { name: 'jane', tag: ['tag2', 'tag3'] } ]; const tagsToMatchOn = ['tag2', 'tag3']; // find all elements who's tag property includes // all of the above tags but can also contain others const newArray = array.filter(item => tagsToMatchOn.every(t => item.tag.includes(t)) ); console.log(newArray); 

Answer 2

This mightn't be the most elegant solution but it does return what you want.

array = [{name:'john',
          tag: ['tag1','tag2'] 
         },
         {name:'doe',
          tag: ['tag2'] 
         },
         {name:'jane',
          tag: ['tag2','tag3'] 
         }
        ];

const newArray = [];
for (let index = 0; index < array.length; index++) {
    if(array[index].tag[0] === 'tag2' && array[index].tag[1] === 'tag3') {
        newArray.push(array[index])
    }
}

or if you want to be a bit more es6:

array.forEach(element => {
  if(element.tag[0] === 'tag2' && element.tag[1] === 'tag3') {
    newArray.push(element)
  }
});

Answer 3

You can do it like this

With the help of filter and every .

What basically i am doing here is first i am lopping through each element of arr (using filter). by using every i am checking does the tag property to element contains all the tags we require. if it does than we include in our final output if not than we don't

 let arr = [{name:'john', tag: ['tag1','tag2'] }, {name:'doe', tag: ['tag2'] }, {name:'jane', tag: ['tag2','tag3'] } ]; let tags = ['tag2','tag3']; let op = arr.filter(e=> tags.every(el=> e.tag.includes(el))); console.log(op);