使用jquery $ .extend（）时忽略属性？ [英] Ignore a property while using jquery $.extend()?

问题描述

是否有一种有效的方法来克隆对象但是遗漏了指定的属性？理想情况下不重写$ .extend函数？

Is there an efficient way to clone an object yet leave out specified properties? Ideally without rewriting the $.extend function?

var object = {
  "foo": "bar"
  , "bim": Array [1000]
};

// extend the object except for the bim property
var clone = $.extend({}, object, "bim");
// = { "foo":"bar" }

我的目标是保存不要复制我不会使用的东西。

My goal is to save resources by not copying something I'm not going to use.

推荐答案

jQuery.extend 需要无数个参数，因此无法重写它以适合您请求的格式，打破功能。

jQuery.extend takes an infinite number of arguments, so it's not possible to rewrite it to fit your requested format, without breaking functionality.

但是，您可以使用 删除 运营商：

You can, however, easily remove the property after extending, using the delete operator:

var object = {
    "foo": "bar"
  , "bim": "baz"
};

// extend the object except for the bim property
var clone = $.extend({}, object);
delete clone.bim;
// = { "foo":"bar" }

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