Android:通过Intent.ACTION_SEND共享(某些东西),然后自动返回到我的应用 [英] Android: Sharing (something) via Intent.ACTION_SEND, then automatically return to my app
问题描述
我想分享我的应用程序中的某些内容。
共享(例如,发送消息)后,我希望我的应用程序再次处于活动状态,而发送的应用程序消失。
使用以下代码,我希望调用 onActivityResult
,但从未调用过。
I would like to share something from my application.
Once it is shared (e.g - message sent), I want my application to be active again and the sending app to disappear.
Using the code below, I expected onActivityResult
to be called, but it is never called.
发送电子邮件后,我的应用程序再次出现,但是发送SMS(消息传递)后,消息传递应用程序仍然保留。 (从未调用 onActivityResult
)
After sending email, my application appears again, but after sending SMS ('Messaging') the messaging app remains. (onActivityResult
is never called)
谢谢:-)
Intent sharingIntent = new Intent(android.content.Intent.ACTION_SEND);
sharingIntent.setType("text/plain");
String shareBody = "This is a test";
sharingIntent.putExtra(android.content.Intent.EXTRA_SUBJECT, "You have to see this!");
sharingIntent.putExtra(android.content.Intent.EXTRA_TEXT, shareBody);
startActivityForResult(Intent.createChooser(sharingIntent, "Share via"),1);
getFragmentManager().popBackStack();
推荐答案
只有在用户按下后退按钮时才会调用按钮。因为您正在启动新的意图,所以您将控制权交给另一个应用程序。
因此,通常情况下,如果用户按下后退按钮(我将以Android用户的身份进行操作),则该用户将返回到您的应用程序。据我所知,这是我更喜欢在这种情况下使用的方式,也是在Android中实现此目的的唯一方法。
This will only be called if the user presses the back button. Because you are starting a new intent, so you give control to another application. So normally, if the user presses the back button (which I would do as an Android user) the user will go back to your app. This the way I would prefer to use for this case, and as far as I know it is also the only way to do this in Android.
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