Android的按钮,点击网页视图链接退出(或删除的sessionId) [英] Android button as WebView click link for Logout (or delete sessionId)

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问题描述

在我的WebView的Andr​​oid,我希望用户注销按钮点击或任何其他方式。
我不想点击该按钮后显示出任何网页。该应用程序后,应自动退出(System.exit(0),将做到这一点)

In my WebView android, I want the user to logout on button click or any other way. I don't want to show any webpage after clicking that button. After that app should auto-exit(System.exit(0) will do that)

点击我要发是:

http://192.100.1.1:10000/app?service=direct/1/UserSummary/$UserBorder.logoutLink

什么是这样做的最佳方式?

What's the best way of doing this?

推荐答案

这背后的问题的主要原因是该注销链接无关的显示。所以,我想要的东西,如点击注销。在主要活动简单的Andr​​oid按键并没有帮助这一点。

The main reason behind this question was this logout link has nothing to display. So I wanted something like click to logout. Simple Android button in Main Activity didn't help for this.

所以,我创建新的空白活动,用我所用相同的布局,但不是显示空白的布局,我加了 ProcessDialog ,并使其非可取消的如下:

So, I created new blank activity, used same layout I was using, but instead of showing blank layout, I added ProcessDialog and made it non-cancel-able as follows:

    final ProgressDialog pd = new ProgressDialog(Logout.this);
    pd.setCancelable(false);
    pd.setCanceledOnTouchOutside(false);
    pd.setMessage("Please wait...");
    pd.show();

而在公共无效onPageFinished(的WebView视图,字符串URL)方法,我驳回了这一ProcessDialog和解雇我的应用程序的登录页面的意图。也去了应用程序的重写公共布尔shouldOverrideUrlLoading(的WebView视图,字符串URL)来prevent它。

And in public void onPageFinished(WebView view, String url) method, I dismissed this ProcessDialog and fired an intent for login page of my app. Also overridden the public boolean shouldOverrideUrlLoading(WebView view, String url) to prevent it from going to out of app.

在此之后,只需 webView.loadUrl(URL);

这样,我固定我的问题。
(如果你知道更好,更简单的做这件事的方式,请发表。)

This way I fixed my problem. (If you know better and simpler way of doing it, please post.)

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