主叫chrome.tabs.query后，结果是不可用 [英] After calling chrome.tabs.query, the results are not available

问题描述

我创建（学习）谷歌浏览器的扩展。

I'm creating (learning) an extension for Google Chrome.

要调试某些code，我插入的console.log（），如下：

To debug some code, I inserted console.log(), as follows:

var fourmTabs = new Array();
chrome.tabs.query({}, function (tabs) {
    for (var i = 0; i < tabs.length; i++) {
        fourmTabs[i] = tabs[i];
    }
});
for (var i = 0; i < fourmTabs.length; i++) {
    if (fourmTabs[i] != null)
        window.console.log(fourmTabs[i].url);
    else {
        window.console.log("??" + i);
    }
}

这很简单，code：让所有的选项卡信息到我自己的数组，并打印一些事情。

It's very simple code: get all tabs info into an array of my own, and print some things.

要检查code是否正常工作，因为它应该，我运行code。在这里，问题来了：

To check whether the code works as it should, I run the code. Here comes the problem:

• 当我使用断点（通过开发者工具）时，code运行正常。


• 无断点，没有打印。

任何想法，为什么？

推荐答案

您的问题可以简化为：

/*1.*/ var fourmTabs = [];
/*2.*/ chrome.tabs.query({}, function(tabs) {
/*3.*/     fourmTabs[0] = tabs[0];
/*4.*/ });
/*5.*/ console.log(fourmTabs[0]);

您想到的是， fourmTabs 数组（由3号线）到达5号线时更新。结果
这就是错误，因为 chrome.tabs.query 方法的同步

You expect that the fourmTabs array is updated (by line 3) when line 5 is reached.
That is wrong, because the chrome.tabs.query method is asynchronous.

在试图让你明白异步方面的意义，我将展示具有相同的结构，你的code一个code段和的故事。

In an attempt to make you understand the significance of the asynchronous aspect, I show a code snippet with the same structure as your code and a story.

/*1.*/ var rope = null;
/*2.*/ requestRope(function(receivedRope) {
/*3.*/     rope = receivedRope;
/*4.*/ });
/*5.*/ grab(rope);

• 在1号线，绳子的presence公布。


• 在2-4行，一个的回调函数的被创建，它应该由 requestRope 函数被调用。


• 在第5行，你要抓住通过绳索抢功能。


• At line 1, the presence of a rope is announced.
• At lines 2-4, a callback function is created, which ought to be called by the requestRope function.
• At line 5, you're going to grab the rope via the grab function.

在 requestRope 实施同步后，没有任何问题：结果
＆EMSP;你：你好，我想绳子请的抛绳的^{调用回调函数结果}当你得到了一个。
＆EMSP;她：当然。的抛出绳索的结果
＆EMSP;您：跳转，并抓住绳的 - 您设法得到在另一边，存活的

When requestRope is implemented synchronously, there's no problem:
  You: "Hi, I want a rope. Please throw the rope^{"call the callback function"} when you've got one."
  She: "Sure." throws rope
  You: Jumps and grabs rope - You manage to get at the other side, alive.

在 requestRope 实施异步，如果你把它当作你同步可能有问题：结果
＆EMSP;你：请向我扔一根绳子。结果
＆EMSP;她：好，让我们来看看......结果
＆EMSP;您：跳转，并试图抓住绳子的因为没有绳子，你倒下和死亡结果
＆EMSP;她：抛出绳索的<子>太晚了，当然

When requestRope is implemented asynchronously, you may have a problem if you treat it as synchronous:
  You: "Please throw a rope at me."
  She: "Sure. Let's have a look..."
  You: Jumps and attempts to grab rope Because there's no rope, you fall and die.
  She: Throws rope _{Too late, of course.}

现在你已经看到了一个异步和同步实现功能之间的区别，让我们来解决您原来的问题：

Now you've seen the difference between an asynchronously and synchronously implemented function, let's solve your original question:

var fourmTabs = new Array();
chrome.tabs.query({}, function (tabs) {
    for (var i = 0; i < tabs.length; i++) {
        fourmTabs[i] = tabs[i];
    }
    // Moved code inside the callback handler
    for (var i = 0; i < fourmTabs.length; i++) {
        if (fourmTabs[i] != null)
           window.console.log(fourmTabs[i].url);
        else {
            window.console.log("??" + i);
        }
    }
});
// <moved code inside callback function of chrome.tabs.query>

使用断点，您的code ++工程，因为到达到code的第二部分的时候，回调已被调用。

With breakpoints, your code works, because by the time that the second part of the code is reached, the callback has already been called.

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