Question

我在obj中获得了很多对象。就像这个

Object {MAILING ADDRESS: "P O BOX 59", APN: "066-102-11-1"}
Object {MAILING ADDRESS: "", APN: ""}
Object {MAILING ADDRESS: "P O BOX 3", APN: "066-105-11-1"}
Object {MAILING ADDRESS: "", APN: ""}

当我对它的Null值应用删除函数时，我得到了这个

Object {MAILING ADDRESS: "P O BOX 59", APN: "066-102-11-1"}
Object {}
Object {MAILING ADDRESS: "P O BOX 3", APN: "066-105-11-1"}
Object {}

我想完全删除对象{}并且只想获得2个具有值的对象。我删除此值的代码是

＆＃13;

(function filter(obj_field) {
  $.each(obj_field, function(key, value) {
    if (value === "" || value === null) {
      delete obj_field.key;
    } else if (Object.prototype.toString.call(value) === '[object Object]') {
      filter(value);
    } else if ($.isArray(value)) {
      $.each(value, function(k, v) {
        filter(v);
      });
    }
  });
})(obj_field);

＆＃13;

Answer 1

 (function filter(obj_field) {
   $.each(obj_field, function(key, value) {
     if(isEmpty(value)) {
      delete obj_field.key;
     } else if (Object.prototype.toString.call(value) === '[object Object]') 
     {
        filter(value);
     } else if ($.isArray(value)) {
          $.each(value, function(k, v) {
           filter(v);
      });
        }
       });
   })(obj_field);

   function isEmpty(obj) {
      for(var key in obj) {
         if(obj.hasOwnProperty(key))
             return false;
      }
     return true;
     }

Answer 2

使用delete运算符仅删除引用，它不会删除对象本身。

var myObj = {}
myObj.x = 30;

console.log(myObj.x); // '30'

delete myObj; // It will not delete the object
console.log(myObj.x); // '30'

delete myObj.x; // it will delete the property of the object 
console.log(myObj.x); // 'undefined'

Answer 3

让我们整理一下：

mainObj={
 Object1: {"MAILING ADDRESS": "P O BOX 59", APN: "066-102-11-1"},
 Object2: {"MAILING ADDRESS": "", APN: ""},
 Object3: {"MAILING ADDRESS": "P O BOX 3", APN: "066-105-11-1"},
 Object4: {"MAILING ADDRESS": "", APN: ""} 
};

for(key in mainObj){
  if(!Object.values(mainObj[key]).some(el=>el)){
     delete mainObj[key];
  }
}

这将创建：

mainObj={
  Object1,Object3
};

http://jsbin.com/qazepupupi/edit?console

注意：这个ES6，如果此代码在旧的客户端浏览器上运行，您可以使用下面的代码：

for(key in mainObj){
  if(!Object.keys(mainObj[key]).some(function(inlinekey){ return mainObj[key][inlinekey]})){
     delete mainObj[key];
  }
}

完全删除JavaScript对象

3 个答案: