Question

我有这个JSON对象 myFilters ：

{"filters":
      {"role":"","jobs":[]}
}

我可以使用函数clean（myFilters）从其中正确删除空对象：

function clean(obj) {
  for (var propName in obj) {
    if (
      obj[propName] === null ||
      obj[propName] === undefined ||
      obj[propName] === ""
    ) {
      delete obj[propName];
    }
  }

现在，我的 myFilters 对象变为：

{"filters":
    { "jobs":[] }
}

现在如何从JSON对象中删除空数组和键？

Answer 1

您应该再添加一个条件，例如

function clean(obj) {
  for (var propName in obj) {
    if (
      obj[propName] === null ||
      obj[propName] === undefined ||
      obj[propName] === "" ||
      Array.isArray(obj[propName]) && obj[propName].length === 0
    ) {
      delete obj[propName];
    }
  }
}

Answer 2

您应该先检查属性的类型，然后再通过    typeof

职位属性是一个对象，您可以通过其属性检查其值   长度。如果其 length 等于0，则为空。

function clean(obj) {

    for (var propName in obj) {

        if (typeof (obj[propName]) == 'object') {

            if (obj[propName].length == 0) {
                delete obj[propName];
            }

        } else {

            if (
                obj[propName] === null ||
                obj[propName] === undefined ||
                obj[propName] === ""
            ) {
                delete obj[propName];
            }

        }

    }

}

Answer 3

我喜欢Saveli Tomac's solution，所以我赞成。让我也向您展示原始解决方案的其他缩写。

如前所述，如果要查找空数组，则需要再检查2件事。那么，如何更轻松地检查null，undefined和''值呢？

if (!undefined) { console.log('undefined needs to be deleted') };
if (!null) { console.log('null needs to be deleted') };
if (!'') { console.log(`'' needs to be deleted`) };

检查Array.length是否具有0值也可以缩短，如下所示：

const array1 = [];
const array2 = [1,2,3];
if (!array1.length) { console.log('array1 has 0 length') };
if (!array2.length) { console.log('array2 has 0 length') };

因此，根据这些代码段，您可以像以下内容一样进行其他缩短：

// extended with other types for the demo
let myObject = { "filters": { "role": "", "jobs": [], "nullValue": null, "undefinedIsHere": undefined, "arrayWithValue": [1,2,3], "stringValue": "hello", "numberishere": 123 } };

const clean = (obj) => {
  for (let propName in obj) {
    if (
      !obj[propName] ||
      Array.isArray(obj[propName]) && !obj[propName].length
    ) { delete obj[propName] };
  }
}

clean(myObject.filters);
console.log(myObject);

或使用1️⃣内衬：

// extended with other types for the demo
let myObject = { "filters": { "role": "", "jobs": [], "nullValue": null, "undefinedIsHere": undefined, "arrayWithValue": [1,2,3], "stringValue": "hello", "numberishere": 123 } };

const clean = (obj) => {
  Object.keys(obj).forEach(propName => (!obj[propName] || Array.isArray(obj[propName]) && !obj[propName].length) && delete obj[propName]);
}

clean(myObject.filters);
console.log(myObject);

在此处进一步阅读：

我希望这会有所帮助！

Answer 4

应该是这样的：

function clean(obj) {
  for (var propName in obj) {
    if (obj.hasOwnProperty(propName) && 
      obj[propName] === null ||
      obj[propName] === undefined ||
      obj[propName] === "" || 
      (Array.isArray(obj[propName]) && obj[propName].length <= 0)
    ) {
      delete obj[propName];
    }
  }
}

Answer 5

Saveli的答案应该可以正常工作。这是您可以用来获得相同结果的另一种方法。

const object = {
  "filters": {
    "role": "",
    "jobs": [],
    "foo": undefined,
    "baz": null,
    "bar": {},
    "moreJobs": ['1', '2']
  }
}

const result = {
  filters: Object.keys(object.filters).reduce((acc, key) => {
    if (
      object.filters[key] !== null &&
      object.filters[key] !== undefined &&
      object.filters[key] !== '' &&
      typeof object.filters[key] === 'object' && Object.keys(object.filters[key]).length > 0
    ) {
      acc[key] = object.filters[key];
    }

    return acc;
  }, {})
};

console.log(result);

Answer 6

尝试一下：

var filterObj = {
	"filters": {
		"role": "",
		"jobs": []
	}
};

for (var i in filterObj) {
  for (var j in filterObj[i]) {
  	if ((filterObj[i][j] === null) ||
        (filterObj[i][j] === undefined) ||
        (filterObj[i][j].length === 0)) {
        delete filterObj[i][j];
    }
  }
}

console.log(filterObj);

使用JavaScript从对象中删除空数组

6 个答案: