Home > php教程 > PHP开发 > body text

A quick solution to the problem that Angular's lazy loading mechanism cannot roll back after refreshing

高洛峰
Release: 2016-12-27 15:24:56
Original
1178 people have browsed it

I encountered a very strange problem in the project today. I used oclazyload to lazily load the angular module. After refreshing the page, I could not return to the previous page by clicking the back button. It is estimated that the lazy loading mechanism was used to destroy the state inside angular. Association, resulting in the inability to return to the previous state (the $stateChangeStart event of clicking the back button ui-routre will not be triggered). Of course, this is just a guess, and the source code has not been explored in depth due to the event relationship.

Solution to the inability to roll back after the angular lazy loading mechanism is refreshed:

By looking at the source code of angular (ionic), we found that there is an onUrlChange method on the $browser service. When we change the url address from outside angular , the event processing function registered in this method will be called, as shown in the following figure:

A quick solution to the problem that Angulars lazy loading mechanism cannot roll back after refreshing

In this case, you can register a function at the program entrance

//当通过浏览器回退/前进按钮跳转state时,重新加载页面,如果用系统state,则不会进入此方法 
$browser.onUrlChange(function (url) { 
//TODO 解析url中的state,使用懒加载去加载state模块,实现页面刷新
});
Copy after login

Through this function, you can refresh the page again when going back and forward...

The above is a quick solution to the problem that the Angular lazy loading mechanism cannot be rolled back after refreshing. I hope it is helpful to you. If you have any questions, please leave me a message and I will reply to you in time. I would also like to thank you all for your support of the PHP Chinese website!

For more related articles on the quick solution to the problem that Angular’s ​​lazy loading mechanism cannot be rolled back after refreshing, please pay attention to the PHP Chinese website!

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Recommendations
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!