Question

我想我已经看过这样一个jquery函数，但找不到它。

var x = {a:1, b:2};
var y = {a:5,     c:3};

var z = $.extend({}, x, y);   // {a: 5, b: 2, c: 3}

// want a similar jquery function which returns {a: 5, b: 2}
// that is, returns only attributes which x has (a,b), but adds no new attributes (c)

Answer 1

在这里编写自己的方法可能会更容易。

浏览Y的所有道具并覆盖X

for (var yItem in Y) {
  if (X.hasOwnProperty(yItem)) { 
    X[yItem] = Y[yItem];
  }
}

＆＃13;

var X = {a:1, b:2};
var Y = {a:5,     c:3};
for (var yItem in Y) {
  if (X.hasOwnProperty(yItem)) { 
    X[yItem] = Y[yItem];
  }
}
alert(JSON.stringify(X));

＆＃13;

Answer 2

只需使用Object.assign()和Object.defineProperty分享另一种解决方案。

var x = {a:1, b:2 };
var y = {a:5,     c:3, d:5};

Object.leftJoin = function(leftObj, rightObj) {
  var acceptKeyList = Object.keys(leftObj);
  var returnObj = {};
  
  for(var key of Object.keys(rightObj)) {
    if(acceptKeyList.indexOf(key) < 0) {
      Object.defineProperty(rightObj, key, {
        enumerable: false
      });
    }
  }
  return Object.assign(returnObj, leftObj, rightObj);
}

console.log(Object.leftJoin(x, y));

类似于$ .extend的函数，但不添加任何新属性

2 个答案: