Question

数组1：

[{Id:1,Name:'abc'},{Id:5,Name:'bcd'},{Id:3,Name:'xyz'}]

数组2：

[{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:1,Name:'Ajay1'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4']

在重新排列两个数组时，输出应该是：

数组1：

[{Id:1,Name:'abc'},{Id:3,Name:'xyz'},{Id:5,Name:'bcd'}]

数组2：

[{Id:1,Name:'Ajay1'},{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4'}]

OR

[{Id:1,Name:'Ajay1'},{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:4,Name:'Ajay4'},{Id:2,Name:'Ajay2'}]

让我知道上面的最佳方式。两个数组中的现有ID应该是相同的顺序，其中一个数组中的额外ID应该在最后。

Answer 1

我希望这就是你要找的东西

var Arr1 = [{Id:1,Name:'Ajay1'},{Id:5,Name:'Ajay5'},{Id:3,Name:'Ajay3'}];
var Arr2 = [{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:1,Name:'Ajay1'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4'}];

var finalArr = [];
var unMatchedArr = [];
Arr2.forEach(function(item) {
  var inArr = Arr1.filter(function(a1) {
    return a1.Id === item.Id;
  });

  if(inArr.length) {
    finalArr.push(item);
  }

  else {
    unMatchedArr.push(item);
  }

});

var finalArr = finalArr.concat(unMatchedArr);
snippet.log(JSON.stringify(finalArr));

<!-- Script provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

finalArr包含结果

Answer 2

您可以使用下划线sortBy对第一个数组进行排序。

然后合并filter和every以找到剩余的。最后，您可以使用本机concat方法组合两个数组来创建第二个数组，如下所示：

＆＃13;

var arr1 =[{Id:1,Name:'abc'},{Id:5,Name:'bcd'},{Id:3,Name:'xyz'}],
arr2 = [{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:1,Name:'Ajay1'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4'}],
  newArr1 = _.sortBy(arr1, "Id"),
  diff = _.filter(arr2, function(obj) {
    return _.every(arr1, function(elm) {
      return elm.Id != obj.Id;
    });
  }),
  newArr2 = newArr1.concat(diff);

console.log(newArr1,newArr2);
snippet.log(JSON.stringify(newArr1));
snippet.log(JSON.stringify(newArr2));

＆＃13;

<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/underscore.js/1.7.0/underscore-min.js"></script>

＆＃13;

您也可以使用其他方式同时组合reject和some：

＆＃13;

var arr1 = [{Id:1,Name:'Ajay1'},{Id:5,Name:'Ajay5'},{Id:3,Name:'Ajay3'}],
  arr2 = [{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:1,Name:'Ajay1'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4'}],
  newArr1 = _.sortBy(arr1, "Id"),
  diff = _.reject(arr2, function(obj) {
    return _.some(arr1, function(elm) {
      return elm.Id == obj.Id;
    });
  }),
  newArr2 = newArr1.concat(diff);

console.log(newArr1, newArr2);
snippet.log(JSON.stringify(newArr1));
snippet.log(JSON.stringify(newArr2));

＆＃13;

<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/underscore.js/1.7.0/underscore-min.js"></script>

＆＃13;

<强>更新

＆＃13;

var arr1 =[{Id:1,Name:'abc'},{Id:5,Name:'bcd'},{Id:3,Name:'xyz'}],
arr2 = [{Id:3,Name:'Ajay3'},{Id:5,Name:'Ajay5'},{Id:1,Name:'Ajay1'},{Id:2,Name:'Ajay2'},{Id:4,Name:'Ajay4'}],
  newArr1 = _.sortBy(arr1, "Id"),
  matches = _.filter(_.sortBy(arr2, "Id"), function(obj) {
    return _.find(arr1, function(elm) {
      return elm.Id == obj.Id;
    });
  }),
  diff = _.difference(arr2, matches),
  newArr2 = matches.concat(diff);
snippet.log(JSON.stringify(newArr1));
snippet.log(JSON.stringify(newArr2));

＆＃13;

<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/underscore.js/1.7.0/underscore-min.js"></script>

＆＃13;

在javascript中基于id排列两个不等长的数组

2 个答案: