javascript - 请问大家 js中有没有什么简单的方法比较两个对象是否相等

Question

如题 就像下面的两个
objA={ a:'123', b:'456' };

objB={ a:'123', b:'000' };

很明显不等 应该返回false 求解答

Answer 1

其实这个事情还真的不太简单的，可以看一下 underscore.js 的是如何实现的：
https://github.com/jashkenas/underscore/blob/master/underscore.js#L111...
事实上你可以看到它的代码非常复杂。

当然用起来是非常简单的，文档在这里：
http://underscorejs.org/#isEqual

所以对于楼主的问题，答案就是没有简单的办法比较，除非用别人写好的，underscore.js 有一万多 star，值得信赖。

Answer 2

两个对象都JSON.stringify 然后进行字符串比较

Answer 3

对underscore中的相等判断我写过一篇文章，你可以参考一下：https://github.com/classicemi/blog/issues/7

Answer 4

用 objA == objB 啊，或者 objA === objB 也行，都返回false;

Answer 5

传送门： https://github.com/wh1100717/localDB/blob/develop/src/core/utils.coffe...

<code>coffee</code><code>isEqual = (a, b) -> eq(a, b, [], [])
eq = (a, b, aStack, bStack) ->
        return a isnt 0 or 1 / a is 1 / b if a is b
        return false if a is null and b is undefined
        return false if a is undefined and b is null
        className = toString.call(a)
        return false if className isnt toString.call(b)
        switch className
            when "[object RegExp]" then return "" + a is "" + b
            when "[object String]" then return "" + a is "" + b
            when "[object Number]"
                return +b isnt +b if +a isnt +a
                return if +a is 0 then 1 / +a is 1 / b else +a is +b
            when "[object Date]" then return +a is +b
            when "[object Boolean]" then return +a is +b
        areArrays = className is "[object Array]"
        if not areArrays
            return false if typeof a isnt "object" or typeof b isnt "object"
            aCtor = a.constructor
            bCtor = b.constructor
            return false if (aCtor isnt bCtor) and not (Utils.isFunction(aCtor) and aCtor instanceof aCtor and Utils.isFunction(bCtor) and bCtor instanceof bCtor) and ("constructor" of a and "constructor" of b)
        length = aStack.length
        while length--
            return bStack[length] is b if aStack[length] is a
        aStack.push(a)
        bStack.push(b)
        if areArrays
            size = a.length
            result = size is b.length
            if result
                while size--
                    break if not (result = eq(a[size], b[size], aStack, bStack))
        else
            keys = Utils.keys(a)
            size = keys.length
            result = Utils.keys(b).length is size
            if result
                while size--
                    key = keys[size]
                    break if not (result = Utils.has(b, key) and eq(a[key], b[key], aStack, bStack))
        aStack.pop()
        bStack.pop()
        return result
</code>

Answer 6

简单的办法就是 JSON.stringify(obj_a) === JSON.stringify(obj_b);
但是注意 1） obj_a(b)中不能有环。
2） 性能比较差