java 如何从android中的URL启动android应用程序?
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How to launch android application from URL in android?
提问by Jayeshkumar Sojitra
Can anybody please tell me how to open android application from URL in android.
谁能告诉我如何从android中的URL打开android应用程序。
I am using following method to open application in android but it doesn't works for me.
我正在使用以下方法在 android 中打开应用程序,但它对我不起作用。
<data android:scheme="application"/>
I am passing source of URL "application://[RandomText]"
我正在传递 URL“application://[RandomText]”的来源
Is it correct way to open application?
打开应用程序的方法正确吗?
回答by Ashish Jaiswal
Try this:
试试这个:
<category android:name="android.intent.category.DEFAULT" />
<category android:name="android.intent.category.BROWSABLE" />
<data android:host="www.put-your-website.com" />
<data android:scheme="https" />
<data android:scheme="http" />
<data android:pathPattern=".*" />
</intent-filter>
回答by pshirishreddy
Read about BroadcastReceiver. An application must register itself to listen to intents sent using sendBroadcast(). After this the user is presented with a pop up to launch that application. (Similar to send functionality where you can choose to send using different applications)
阅读有关BroadcastReceiver 的信息。应用程序必须注册自己以侦听使用 sendBroadcast() 发送的意图。在此之后,用户将看到一个弹出窗口以启动该应用程序。(类似于发送功能,您可以选择使用不同的应用程序发送)