Question

当我想在包含一组字符的json对象中排除某些属性时，我该怎么做？

var obj1 = {name: "James", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""}
var obj2 = {name: "Maria", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""}

所以现在我要删除包含字符串foo

的所有属性

var result = _.isEqual(
  _.omit(obj1, ['\*foo\*']),
  _.omit(obj2, ['\*foo\*'])
);

像这样......

我有办法做到这一点吗？

Answer 1

在香草JS中，

可能不是最好的方法。但你可以这样做

过滤具有console.log(/"(\/ab\/)"/.test('"/ad/"')); // false console.log(/"(\/[^\/]*)"/.test('"/ad/"')) // false console.log(/"(\/ab\/)|(\/[^\/]*)"/.test('"/ad/"')); //true等
根据过滤后的密钥
创建最终对象

foo

Answer 2

您可以尝试跟随（它操纵同一个对象）

＆＃13;

function removeProps (obj, prop) {
  Object.keys(obj).forEach((key) => {
    if(key.indexOf(prop) !== -1) delete obj[key];
  });
}
var obj1 = {name: "James", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""};

removeProps(obj1, "foo");
console.log(obj1);

＆＃13;

如果您希望保留对象

＆＃13;

function removeProps (obj, prop) {
  obj = JSON.parse(JSON.stringify(obj));
  Object.keys(obj).forEach((key) => {
    if(key.indexOf(prop) !== -1) delete obj[key];
  });
  return obj;
}
var obj1 = {name: "James", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""};

var obj1Updated = removeProps(obj1, "foo");
console.log(obj1);
console.log(obj1Updated);

＆＃13;

Answer 3

如你所说，你需要使用正则表达式的解决方案，所以你去：

var obj1 = {name: "James", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""}
var obj2 = {name: "Maria", age: 17, creation: "13-02-2016", deletion: "13-04-2016", foo_x:"", foo_y:"", foo_z:""};

Object.prototype.filterRegex = function(regex) {
    // check if regex is passed
    let filtered = Object.keys(this).filter((key) => !regex.test(key));
    return filtered.reduce((obj, key) => {
        obj[key] = this[key];
        return obj;
    }, {});
};

obj1.filterRegex(/foo/);
obj2.filterRegex(/foo/);

Answer 4

您可以将省略的函数传递给第二个参数：

const result = _.omit(obj1, (value, key) => key.includes('foo'))

在比较期间通过通配符或正则表达式省略多个键

4 个答案: