Javascript / Typescript - 从对象中删除

Question

如何删除包含filter1的所有对象？

const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
}

Answer 1

易。

1. 要删除对象，您可以使用delete API。
2. 要检查对象是否具有特定密钥，您可以使用 hasOwnProperty API。

你在这里可以做的是。首先循环obj中的所有键，然后检查其value（一个对象）是否具有filter1属性。如果他们那么delete它就离开了。

示例：

const obj = {
    0: {
        'filter1': {
            'key1': {
                'email1': 'email1'
            }
        }
    },
    1: {
        'filter1': {
            'key12': {
                'email12': 'email12'
            }
        }
    },
    2: {
        'filter2': {
            'key2': {
                'email2': 'email2'
            }
        }
    }
};

for (var key in obj) {
    var objectInKey = obj[key];
    if (objectInKey.hasOwnProperty("filter1")) {
        delete obj[key]
    }
}

console.log(JSON.stringify(obj, null, 4));

console.log(obj[0]); // <-- Should be undefined.
console.log(obj[2]); // <-- Should not be undefined as obj wasn't deleted.

Answer 2

您可以使用reduce获取新对象：

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

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const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.keys(obj).reduce((newObj, key) => {
  if (!obj[key].filter1) {
    newObj[key] = obj[key];
  }

  return newObj;
}, {});

console.log(newObject);

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或@zerkms说，使用entries：

const entries = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

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const obj = {
  0: {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  1: {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  2: {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
};

const newObject = Object.entries(obj).reduce((newObj, [key, val]) => {
  if (val.filter1) {
    return newObj;
  }

  return {
    ...newObj,
    [key]: val,
  }
}, {});

console.log(newObject);

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Answer 3

您可以使用Object.entries获取一组键值对，然后使用Array#reduce过滤数据并将其转换回对象格式。

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const obj = {
  0: { 'filter1': { 'key1':  { 'email1':  'email1'  } } },
  1: { 'filter1': { 'key12': { 'email12': 'email12' } } },
  2: { 'filter2': { 'key2':  { 'email2':  'email2'  } } }
};

console.log(Object.entries(obj).reduce((m, [k, v]) => (v.filter1 || (m[k] = v), m), {}));

// { "2": { "filter2": { "key2": { "email2": "email2" } } } }

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进一步阅读：

• Object.entries ^(mdn)
• Array#reduce ^(mdn)

Answer 4

要修改对象到位（而不是复制它或创建数组），我只使用一个简单的循环和删除：

let k = Object.keys(obj);
for (let i = 0; i < k.length ; i++ ) {
    if (obj[k[i]].hasOwnProperty('filter1')) delete obj[k[i]];
}

Answer 5

您可以使用filter，

obj = obj.filter(o => o.filter1 == undefined);

请参阅filter参考here。

<强> P.S。此示例中的obj为var而不是const ，请参阅下面的代码段

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var obj = [
  {
    'filter1': {
      'key1': {
        'email1': 'email1'
      }
    }
  },
  {
    'filter1': {
      'key12': {
        'email12': 'email12'
      }
    }
  },
  {
    'filter2': {
      'key2': {
        'email2': 'email2'
      }
    }
  }
];

console.log(obj); // initial obj
obj = obj.filter(o => o.filter1 == undefined);
console.log(obj); // element with `filter1` property is now deleted

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