如何使用ES6 HOF Javascript在数组中查找重复的对象（所有键值对应该相同）

debugcn 发表于 Dev

Nakul Nagariya

var records = [ 
   { 
      "defaultContact":true,
      "contactName":"testContactName",
      "mobileNumber":"900000000000",
      "mobileDialCode":"+91 IN",
      "faxNumber":"123",
      "faxDialCode":"+91 IN",
      "emailId":"[email protected]"
   },
   { 
      "defaultContact":false,
      "contactName":"xyz",
      "mobileNumber":"900000001000",
      "mobileDialCode":"+91 IN",
      "faxNumber":"123",
      "faxDialCode":"+91 IN",
      "emailId":"[email protected]"
   },
   { 
      "defaultContact":false,
      "contactName":"asdasd",
      "mobileNumber":"123",
      "mobileDialCode":"+91 IN",
      "faxNumber":"",
      "faxDialCode":"",
      "emailId":""
   },
   { 
      "contactName":"asdasd",
      "defaultContact":false,
      "emailId":"",
      "faxDialCode":"",
      "faxNumber":"",
      "mobileDialCode":"+91 IN",
      "mobileNumber":"123"
   }
];

上面是一个对象数组，我已经使用两个for循环完成了此操作，但这看起来并不好，有人可以建议如何使用ES6高阶函数来完成此操作。

这里重复表示每个属性与每个属性完全匹配时。

以下是我的操作方法：

let duplicateRecords = [];
    for (let i = 0; i < records.length; i++) {
      for (let j = i + 1; j < records.length; j++) {
        if (
          records[i].contactName === records[j].contactName &&
          records[i].emailId === records[j].emailId &&
          records[i].faxDialCode === records[j].faxDialCode &&
          records[i].faxNumber === records[j].faxNumber &&
          records[i].mobileDialCode === records[j].mobileDialCode &&
          records[i].mobileNumber === records[j].mobileNumber
        ) {
          duplicateRecords = [records[j]];
        }
      }
    }

任何帮助，将不胜感激。

亚历山大·库哈连科（Alexander Kukharenko）

在您的情况下，使用循环for比使用高阶函数更合适。但是您可以使用Object.keys和Array.filter使您的代码更加通用和简短。

for (let i = 0; i < records.length; i++) {
   const keys = Object.keys(records[i]);
     for (let j = i + 1; j < records.length; j++) {
         const isDublicated = !Boolean(keys.filter(key => records[i][key] !== records[j][key]).length);
         isDublicated ? dublicatedRecords.push(records[j]) : null;
     }
}

或使用高阶函数：

const dublicatedRecords = records.reduce((acc, record, index) => {
  const keys = Object.keys(record);
  records.forEach((otherRecord, otherIndex) => {
      const isDublicated = !Boolean(keys.filter(key => record[key] !== otherRecord[key]).length);
      isDublicated && index !== otherIndex ? acc.push(otherRecord) : null;
  })
  return acc;
}, []);

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