如何比较和过滤es6中两个对象数组中不匹配的数组

Question

在我的 Angular 应用程序中，我试图只获取从 UI 修改的更新数组。

在这里，我有两个数组，其中一个 role_name 已更改。我想比较和过滤掉修改过的对象，只返回那些修改过的数组。（在这种情况下，role_name值被修改）

我试过在 ES6 的帮助下过滤掉，但它不起作用。

有人可以帮助我，我在哪里失踪？

var arr1 = [ {
  "num" : null,
  "role_name" : "ABC",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "1",
  "user_name" : "Rywan"
},
{
  "num" : null,
  "role_name" : "DEF",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "2",
  "user_name" : "adecg"
},
{
  "num" : null,
  "role_name" : "GHJ",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "3",
  "user_name" : "dde"
},
{
  "num" : null,
  "role_name" : "RRT",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "4",
  "user_name" : "kumar"
},
{
  "num" : null,
  "role_name" : "SFR",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "5",
  "user_name" : "SASI"
}
];

var arr2 = [ {
  "num" : null,
  "role_name" : "ABC",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "1",
  "user_name" : "Rywan"
},
{
  "num" : null,
  "role_name" : "ROLE_CHANGED",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "2",
  "user_name" : "adecg"
},
{
  "num" : null,
  "role_name" : "GHJ",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "3",
  "user_name" : "dde"
},
{
  "num" : null,
  "role_name" : "RRT",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "4",
  "user_name" : "kumar"
},
{
  "num" : null,
  "role_name" : "SFR",
  "profile" : "ff",
  "user" : "1234",
  "rn" : "5",
  "user_name" : "SASI"
}
];

   let filteredData;
     filteredData = arr1.filter(function(o1){
      //filter out (!) items in arr2
       return arr2.some(function(o2){
           return o1.role_name !== o2.role_name;  
      });
   });
   
   console.log(filteredData);

输出应该是

[
{
      "num" : null,
      "role_name" : "ROLE_CHANGED",
      "profile" : "ff",
      "user" : "1234",
      "rn" : "2",
      "user_name" : "adecg"
    }
]

Answer 1

您的逻辑有点缺陷，因为您.some()将始终返回 true ，因为总有一个元素会有所不同。您想要做的是实际检查是否有任何角色名称相同。当没有找到匹配项（即.some()返回 false）时，您就知道 role_name 已更改：

const filteredData = arr2.filter((o1, i) => {
  return !arr1.some((o2) => {
    return o1.role_name === o2.role_name;
  });
});

const filteredData = arr2.filter((o1, i) => {
  return !arr1.some((o2) => {
    return o1.role_name === o2.role_name;
  });
});