Javascript 如何按条件删除对象中的对象

我有这个：

Object {1: Object, 2: Object, 3: Object, 4: Object, 5: Object, 6: Object, 7: Object, 8: Object, 9: Object, 10: Object, 11: Object, 12: Object, 13: Object, 14: Object, 15: Object}1: Object2: Object3: Object1: Object2: Object3: Object4: Object5: Object__proto__: Object4: Object1: Object2: ObjectfromLevel: "1"toLevel: "6"__proto__: Object3: Object4: Object5: ObjectfromLevel: "1"toLevel: "6"__proto__: Object__proto__: Object5: Object6: Object1: Object2: Object3: ObjectfromLevel: "1"toLevel: "6"__proto__: Object4: Object5: Object__proto__: Object7: Object8: Object9: Object10: Object11: Object12: Object1: Object2: ObjectfromLevel: undefinedtoLevel: undefined__proto__: Object3: Object4: Object5: Object__proto__: Object13: Object14: Object15: Object1: Object2: Object3: Object4: ObjectfromLevel: undefinedtoLevel: undefined__proto__: Object5: Object6: Object7: Object8: Object9: Object10: Object11: Object12: Object13: Object14: Object15: Object16: Object17: Object18: Object__proto__: Object__proto__: Object

我需要删除每个具有未定义属性的对象。 如果父对象中没有任何子对象，则需要删除此父对象。 换句话说，我只需要具有属性不等于未定义的对象。

工作JSFIDLE:

如果还要删除空对象，请使用以下检查：

JSON.stringify(obj[key]) === '{}' // true if obj[key] is empty object

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您真的需要在一行中包含所有对象结构吗？难道一个小得多的格式化对象也不能说明你的观点吗？到目前为止你做了什么？请添加一个有效的对象。`var JObj=new object（）；$.each（$（$）ol），函数（n）{if（！JObj.hasOwnProperty（$（this）.attr（$）表）{JObj[$（this.attr（$）表）]=new Object（）；}if（！JObj[$（this.attr（$）hasOwnProperty（$（this）.attr（$）列）{JObj[$（this.attr（$）表）][$（this.attr（$）列）]=new Object（）(“table”）][$（this.attr（“column”）].fromLevel=$（this.find”）.attr（“row”）；JObj[$（this.attr（“table”）][$（this.attr（“column”）].toLevel=$（this.find（'.ui selected:first'）.attr（“row”）；`这是我添加项目的代码。@Viruscom1000欢迎您！

JSON.stringify(obj[key]) === '{}' // true if obj[key] is empty object