popup-chrome扩展中的背景页面 [英] Background Page in popup- chrome extension

问题描述

我在弹出窗口中嵌入了一个动态网页。目前它的工作和每次弹出被加载的网页再次加载，因此我失去了我在弹出的网页上做的工作。虽然它很好，但我希望这个网页仍然在后台加载，我只是点击弹出窗口显示它。要做到这一点，我从我的弹出页面（脚本+ html）完整的代码复制到background.html。现在我应该如何在弹出窗口中完全访问页面并直接显示（我也想从后台页面显示html）

谢谢

<解决方案

Popups与背景页面处于相同的过程（扩展过程），一个页面可以获取另一个页面的DOM窗口。弹出窗口通过调用 chrome.extension.getBackgroundPage（）获取后台页面。因此，每次打开弹出窗口时，只需读取并写入背景页面上的某个变量，例如 chrome.extension.getBackgroundPage（）。enteredData =value; 。

或者，您可以使用 HTML5 localStorage 即使在浏览器关闭后也能存储变量;例如 localStorage ['enteredData'] =价值。

I am embedding a dynamic webpage in a popup. Currently its working and every time popup is loaded the webpage is loaded again, thus me losing the work i did on the webpage in popup. Though its fine, but i want that webpage remain loaded in background and i just show it in popup on click. to do this i copied complete code from my pop up page(script+html) to background.html. Now how should i access the page completely in popup and show directly(i want to show html also-from background page)

Thanks

解决方案

Popups live in the same process (the extension process) as the background page, and one page can get the DOM Window of the other. A popup gets the background page by calling chrome.extension.getBackgroundPage(). So every time you open the popup, just read and write to some variable on the background page, for example chrome.extension.getBackgroundPage().enteredData = "value";.

Alternately, you can use HTML5 localStorage to store variables even after the browser is shut down; e.g. localStorage['enteredData'] = "value".

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