相关疑难解决方法(0)

假设我想定义一个URI,例如:

myapp://path/to/what/i/want?d=This%20is%20a%20test

必须由我自己的应用程序或服务处理.请注意,该方案"myapp",而不是"http",或"ftp".这正是我的意图:为Android OS全局定义自己的URI架构.这可能吗？

这有点类似于某些程序已经在做的事情,例如Windows系统,例如Skype(skype://)或任何torrent下载程序(torrent://).

我正在使用Firebase构建一个android instagram克隆应用程序.我在我的应用程序中启用了社交媒体共享按钮,通过Facebook,电子邮件,WhatsApp等分享故事的内容,但不知道如何去做.

看看我尝试过的东西:

    public class InstacloneApp extends AppCompatActivity {

    private RelativeLayout relativeLayout;

    private ImageView postCoverImg, userPhotoUrl;
    private TextView post_Title, post_Descpn, post_Author, postDate;

    private Button commentsBtn;
    private FloatingActionButton shareFAB;

    private String post_details = null;
    private FirebaseAuth mAuth;
    private DatabaseReference postRef;

    private Context mCtx = this;
    private String uid_post = null;

    private ScrollView scrollView;
    private Toolbar toolbar;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_insta_clone_app);

        relativeLayout = (RelativeLayout) findViewById(R.id.activity_blog_posts_view);
        scrollView = (ScrollView) findViewById(R.id.scrollView);
        toolbar = (Toolbar) findViewById(R.id.toolbar);

        toolbar.setTitle("");
        setSupportActionBar(toolbar);

        getSupportActionBar().setDisplayHomeAsUpEnabled(true);

        post_details = …