合并两个对象数组并覆盖键值

Question

I am trying to merge data from json that comes in array of objects. I was using the underscore solution from here merge two json object based on key value in javascript , but it turns out it doesnt override existing items which I need to do as well now.

The result should be all items of array 1 in the same order, overriden by array 2 where id = id. Items in array 2 that does not exist in array 1 should be pushed to the end of the result.

First array:

[
 {id: 8, category: "A"}
 {id: 2, category: "D"}
 {id: 5, category: "C"}
 {id: 9, category: "B"}
]

Second array:

[
 {id: 1, category: "X"}
 {id: 2, category: "Y"}
]

Expected result:

[
 {id: 8, category: "A"}
 {id: 2, category: "Y"}
 {id: 5, category: "C"}
 {id: 9, category: "B"}
 {id: 1, category: "X"}
]

Answer 1

Use filter , find and concat

Given that

var arr1 = [
 {id: 8, category: "A"},
 {id: 2, category: "D"},
 {id: 5, category: "C"},
 {id: 9, category: "B"}
];
var arr2 = [
 {id: 12, category: "X"},
 {id: 2, category: "Y"}
];

If the order is not important

var output = arr2.concat( 
        arr1.filter( s => 
            !arr2.find( t => t.id == s.id ) 
        )//end filter 
);//end concat

Demo

 var arr1 = [{ id: 8, category: "A" }, { id: 2, category: "D" }, { id: 5, category: "C" }, { id: 9, category: "B" } ]; var arr2 = [{ id: 12, category: "X" }, { id: 2, category: "Y" } ]; var output = arr2.concat( arr1.filter(s => !arr2.find(t => t.id == s.id) ) //end filter ); //end concat console.log(output);

If the order is important

var output = arr1.map( 
       s => arr2.find( 
           t => t.id == s.id ) || s 
).concat( //end map of arr1
      arr2.filter( 
           s => !arr1.find( t => t.id == s.id ) 
      ) //end filter
);//end concat

Demo

 var arr1 = [{ id: 8, category: "A" }, { id: 2, category: "D" }, { id: 5, category: "C" }, { id: 9, category: "B" } ]; var arr2 = [{ id: 12, category: "X" }, { id: 2, category: "Y" } ]; var output = arr1.map( s => arr2.find( t => t.id == s.id) || s ).concat( //end map of arr1 arr2.filter( s => !arr1.find(t => t.id == s.id) ) //end filter ); //end concat console.log(output);

Answer 2

You could use a Map as closure and store the index of the result array for this id .

 var first = [{ id: 8, category: "A" }, { id: 2, category: "D" }, { id: 5, category: "C" }, { id: 9, category: "B" }], second = [{ id: 12, category: "X" }, { id: 2, category: "Y" }], result = [first, second].reduce((m => (r, a) => { a.forEach(o => { if (m.has(o.id)) { r[m.get(o.id)] = o; return; } m.set(o.id, r.push(o) - 1); }); return r; })(new Map), []); console.log(result);

Answer 3

You can set a loop on your secondArray and check each object with id value against object with id of firstArray . If you find a match then simply replace the object else push the object:

 var firstArray = [ {id: 8, category: "A"}, {id: 2, category: "D"}, {id: 5, category: "C"}, {id: 9, category: "B"} ]; var secondArray = [ {id: 12, category: "X"}, {id: 2, category: "Y"} ]; secondArray.forEach((obj)=>{ var match = false; for(var i=0; i<firstArray.length; i++){ if(firstArray[i].id === obj.id){ match = true; firstArray[i] = obj; break; } } if(!match){ firstArray.push(obj); } }); console.log(firstArray);

Answer 4

You can use a forEach to iterate through second array. For each object with the same id in the first array, update the category otherwise push in the new array.

 const first = [{id: 8, category: "A"},{id: 2, category: "D"},{id: 5, category: "C"},{id: 9, category: "B"}], second = [{id: 12, category: "X"},{id: 2, category: "Y"}], merged = [...first]; second.forEach(o => { let obj = first.find(({id,category}) => id === o.id); obj ? obj.category = o.category : merged.push({...o}); }); console.log(merged);

Answer 5

我认为减少更好

first.reduce((res, item) => res.filter(i => i.id !== item.id).concat(item), second);

Answer 6

Using underscore I managed to come up with this answer my own question. It is probably not the most efficent

const newarr = _.map(arr1, obj1 => {
  const r = _.find(arr2, obj2 => {
    return obj1[match] === obj2[match]
  })
  if (typeof r === 'undefined') {
    return obj1
  } else {
    return r
  }
})
_.each(arr2, obj => {
  if (_.indexOf(arr1, _.findWhere(arr1, {id: obj.id})) === -1) { 
   newarr.push(obj) 
  }
})

Answer 7

这应该有效：

const newArr = second.reduce((res, item) => res.filter(i => i.id !== item.id).concat(item), first);