Question

以下是我的对象数据

var testObj ={
    "AAA": {
    "number":{
        "123":{},
        "5435":{},
        "5466":{}
    }
  },
  "BBB":{
     "number":{
        "54656":{},
        "111":{},
        "4656":{}
    }
  },
  "CCC":{
    "number":{
        "214543":{},
        "32432":{},
        "12435":{}
    }
  },
  "DDD":{
    "number":{
        "343":{},
        "123213":{},
        "111":{}
    }
  }
}

如何过滤此对象并获取一个编号为“111”的新对象？

var testObj ={
  "BBB":{
     "number":{
        "54656":{},
        "111":{},
        "4656":{}
    }
  },
  "DDD":{
    "number":{
        "343":{},
        "123213":{},
        "111":{}
    }
  }
}

Answer 1

您可以使用Object.keys()和reduce()并使用hasOwnProperty()检查密钥

＆＃13;

var testObj = {"AAA":{"number":{"123":{},"5435":{},"5466":{}}},"BBB":{"number":{"111":{},"4656":{},"54656":{}}},"CCC":{"number":{"12435":{},"32432":{},"214543":{}}},"DDD":{"number":{"111":{},"343":{},"123213":{}}}}

var result = Object.keys(testObj).reduce(function(r, key) {
  if (testObj[key].number.hasOwnProperty('111')) {
    r[key] = testObj[key]
  }
  return r;
}, {})

console.log(result)

＆＃13;

Answer 2

您可以使用Object.keys和Array.prototype.reduce获取所有对象键，使用自定义谓词过滤掉（在这种情况下，是否设置了foo.number.111）并构建对象结果：

＆＃13;

const testObj = {
  "AAA": {
    "number": {
      "123": {},
      "5435": {},
      "5466": {}
    }
  },
  "BBB": {
    "number": {
      "54656": {},
      "111": {},
      "4656": {}
    }
  },
  "CCC": {
    "number": {
      "214543": {},
      "32432": {},
      "12435": {}
    }
  },
  "DDD": {
    "number": {
      "343": {},
      "123213": {},
      "111": {}
    }
  }
};

const result = Object.keys(testObj).filter(key => {
  return testObj[key].number['111'];
}).reduce((p, key) => {
  p[key] = testObj[key];
  return p;
}, {});

console.log(result);

＆＃13;

Answer 3

您可以将Object.keys()，for..of循环，JSON.stringify()，RegExp.prorotype.test()与RegExp <property name including quotes followed by ":">，delete一起使用。要保留原始对象，请使用Object.assign()创建具有原始对象属性的新对象，从新对象中迭代和delete个对象。

var testObj ={
    "AAA": {
    "number":{
        "123":{},
        "5435":{},
        "5466":{}
    }
  },
  "BBB":{
     "number":{
        "54656":{},
        "111":{},
        "4656":{}
    }
  },
  "CCC":{
    "number":{
        "214543":{},
        "32432":{},
        "12435":{}
    }
  },
  "DDD":{
    "number":{
        "343":{},
        "123213":{},
        "111":{}
    }
  }
}

let prop = `"111"`, re = new RegExp(prop + ":");

for (let o of Object.keys(testObj)) 
    if (!re.test(JSON.stringify(testObj[o]))) 
      delete testObj[o];

console.log(testObj);

Answer 4

您可以执行以下操作;

var testObj ={
    "AAA": {
    "number":{
        "123":{},
        "5435":{},
        "5466":{}
    }
  },
  "BBB":{
     "number":{
        "54656":{},
        "111":{},
        "4656":{}
    }
  },
  "CCC":{
    "number":{
        "214543":{},
        "32432":{},
        "12435":{}
    }
  },
  "DDD":{
    "number":{
        "343":{},
        "123213":{},
        "111":{}
    }
  }
},
filtered = Object.keys(testObj).filter(k => Object.keys(testObj[k].number).includes("111"))
                 .reduce((p,k) => (p[k] = Object.assign({},testObj[k]),p),{});
console.log(filtered);

如何过滤对象并获取新对象？

4 个答案: