从浏览器Android的推出意图 [英] Android Intent launch from browser

问题描述

这个话题之前已经覆盖，但我找不到具体要问什么，我的答案。

This topic has been covered before, but I can't find an answer specific to what I'm asking.

我在哪里：我跟着hackmod第一块在这里建议：<一href=\"http://stackoverflow.com/questions/3469908/make-a-link-in-the-android-browser-start-up-my-app/3472228#3472228\">Make在Android浏览器链接启动我的应用程序？并得到这个在网页的链接工作。

Where I am: I followed hackmod's first piece of advice here: Make a link in the Android browser start up my app? and got this to work with a link in the webpage.

不过，我无法理解的第二个选项（意向的URI）。这里是我得到了什么：

However, I'm having trouble understanding the second option (intent uri's). here's what I've got:

    <activity android:name="com.myapps.tests.Layout2"
        android:label="Auth Complete"
        >
        <intent-filter>
          <action android:name="android.intent.action.VIEW" />
          <category android:name="android.intent.category.DEFAULT" />
          <category android:name="android.intent.category.BROWSABLE" />
          <data android:scheme="http" android:host="mydomain.com"
                android:path="/launch" />
        </intent-filter>
    </activity>

现在，与我可以去mydomain.com/launch，并启动我的活动。这一切运作良好，但我得到的选择器。我要的是它只是启动我的活动没有给予选项。

Now, with that I can go to "mydomain.com/launch" and it launches my activity. this all works well, except that I get the chooser. what I want is for it to just launch my activity without giving options.

从在岗的解释，我引用它看起来像多数民众赞成意图的URI是的，但我无法找到一个简单的例子。我应该在我的我的网页链接看起来像以没有选择器推出呢？

From the explanation in the post I referenced it looks like thats what intent uris are for,but I can't find a straightforward example. what should my link in my webpage look like in order to launch this with no chooser?

我见过几个例子，看起来是这样的：

I've seen a couple of examples that look something like this:

<a href="intent:#Intent;action=com.myapp.android.MY_ACTION;end">

不过，这似乎并不当我尝试它的工作。

However, that doesn't seem to work when I try it.

我的测试设备是Galaxy Tab的2

My test device is a Galaxy Tab 2.

任何帮助将是AP preciated。

any help would be appreciated.

推荐答案

根据您的意图过滤此链接应该工作：

Depending on your intent filter this link should work:

<a href="http://mydomain.com/launch">start my app</a>

但是你要注意的是，Android系统将询问用户是否您的应用程序或其他任何浏览器应启动。

But you should note that the android system will ask the user if your app or any other browser should be started.

如果你想避免这种实现自定义协议型处理器。所以只要你的应用程序将侦听，用户将无法获得意图选择器。

If you want to avoid this implement a custom protcol handler. So just your app will listen for that and the user won't get the intent chooser.

尝试添加该数据的意图：

Try to add this data intent:

<data android:scheme="mycoolapp" android:host="launch" />

随着code本以上链接应该工作：

With the code above this link should work:

<a href="mycoolapp://launch">start my app</a>

这篇关于从浏览器Android的推出意图的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！