对 console.log 的连续调用会产生不一致的结果

2022-01-21 00:00:00 angularjs javascript console.log angular-ui-router

好吧，我完全被这个吓傻了.(我可能忽略了一些明显的东西，但是......)

Ok, I'm completely dumbfounded by this. (and I might be overlooking something obvious but...)

我连续两次调用 console.log.他们之间没有别的事了

I have 2 consecutive calls to console.log. There isn't anything else between them

console.log($state);
console.log($state.current);

这是生成结果的图像

为什么两者会产生不同的电流"?对象?怎么会这样?

Why do the 2 produce different "current" objects? How can this happen?

这些调用是在 ajax 调用中进行的，同时解决路由依赖关系.如果您需要更多代码或上下文，请告诉我.

Those calls are made inside an ajax call while resolving a route dependencies. If you need more code or context let me know.

在 Chrome 和 Firefox 中确认了同样的问题

Confirmed the same issue in Chrome and Firefox

Ajax 调用和包装函数(没有任何修改)

normaCtrl.publicNorma = ['$http', '$state', '$stateParams', 'baseUrl', function ($http, $state, $stateParams, baseUrl)
{
    var id = $stateParams.id;
    return $http.get(baseUrl + "api/public/norma/" + id).then(
        function (response) {
            console.log($state);
            console.log($state.current);
            console.log($state.current.title);
            return response.data;
        }
    );
}];

可能的相关问题

  • javascript console.log在同一个对象上显示不同的值

    AngularJS:带有异步 chrome.storage.local.get() 的奇怪 console.log 行为

    推荐答案

    好吧，这就是那些偶然发现这个问题的人的答案.

    Well, here's the answer for those that stumble upon this.

    Console.log 在执行的最后状态显示深层可变对象，而不是在调用 console.log 时的状态.

    Console.log shows deep mutable objects at the last state of execution, not at the state when console.log was called.

    基本上，在使用可变深层对象时，Console.log 会存储对所述对象的引用，而不是存储对象克隆.

    Basically, when working with mutable deep objects, Console.log stores the reference to said object instead of storing an object clone.

    由于存储和可视化之间存在时间间隔，当您单击箭头进行进一步检查时，您看到的实际上是对象的当前状态，而不是状态调用 console.log 时的对象.

    Since there is a time gap between storing and visualization, when you click the arrow for further inspection what you're seeing is actually the current state of the object and not the the state of the object when console.log was called.

    始终确保您使用对象快照"的一种方法是调用 Json.stringify 或在可用时使用 console.dir.

    One way to always make sure you're using an "object snapshot" is to call Json.stringify or use console.dir when available.

相关文章