问题描述
在没有附加组件的Safari(实际上是大多数其他浏览器)中，console.log将在最后执行状态而不是在调用console.log的状态下显示对象.
In Safari with no add-ons (and actually most other browsers), console.log will show the object at the last state of execution, not at the state when console.log was called.
我必须克隆对象只是为了通过console.log将其输出，以获取该行的对象状态.
I have to clone the object just to output it via console.log to get the state of the object at that line.
示例:
var test = {a: true}
console.log(test); // {a: false}
test.a = false;
console.log(test); // {a: false}
推荐答案
我认为您正在寻找console.dir().
console.log()不会执行您想要的操作，因为它会打印对该对象的引用，并且在您将其弹出时将其更改. console.dir在调用对象时在对象中打印属性的目录.
console.log() doesn't do what you want because it prints a reference to the object, and by the time you pop it open, it's changed. console.dir prints a directory of the properties in the object at the time you call it.
下面的JSON想法是一个好主意；您甚至可以继续解析JSON字符串，并获得可浏览的对象，如.dir()会给您的内容:
The JSON idea below is a good one; you could even go on to parse the JSON string and get a browsable object like what .dir() would give you:
console.log(JSON.parse(JSON.stringify(obj)));
这篇关于如何使console.log显示对象的当前状态?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！