我们可以使用JSON。stringify和JSON。解析以从对象中删除空白属性。

jsObject = JSON.parse(JSON.stringify(jsObject), (key, value) => {
               if (value == null || value == '' || value == [] || value == {})
                   return undefined;
               return value;
           });

使用Nullish合并可用ES2020

const filterNullishPropertiesFromObject = (obj) => {
  const newEntries = Object.entries(obj).filter(([_, value]) => {
    const nullish = value ?? null;
    return nullish !== null;
  });

  return Object.fromEntries(newEntries);
};

为了进行深度搜索，我使用了下面的代码，也许它对任何研究这个问题的人都有用(它不适用于循环依赖关系):

function removeEmptyValues(obj) {
        for (var propName in obj) {
            if (!obj[propName] || obj[propName].length === 0) {
                delete obj[propName];
            } else if (typeof obj[propName] === 'object') {
                removeEmptyValues(obj[propName]);
            }
        }
        return obj;
    }

下面是一个使用reduce的超级干净的Typescript解决方案:

const removeUndefinedFields = <T>(obj: T): T =>
  Object.keys(obj).reduce(
    (acc, key) =>
      obj[key as keyof T] === undefined
        ? { ...acc }
        : { ...acc, [key]: obj[key as keyof T] },
    {} as T
  )

如果你可以使用Lodash，你可以添加DeepDash递归库，并实现你想要的一些非常简洁的代码:

const prune = obj => _.filterDeep(obj, (v) => !(_.isUndefined(v) || _.isNull(v)));

调用prune(anObjectWithNulls)将返回没有未定义或空值的对象。

如果有人需要欧文(和埃里克)答案的递归版本，这里是:

/**
 * Delete all null (or undefined) properties from an object.
 * Set 'recurse' to true if you also want to delete properties in nested objects.
 */
function delete_null_properties(test, recurse) {
    for (var i in test) {
        if (test[i] === null) {
            delete test[i];
        } else if (recurse && typeof test[i] === 'object') {
            delete_null_properties(test[i], recurse);
        }
    }
}