如何在特定号码的应用程序中共享文本

Question

使用此代码只打开特殊号码的聊天但文本不是共享.我该怎么做？

public class MainActivity extends AppCompatActivity {
Button Wa;
String id = "+919000000000";
EditText txt;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    txt = (EditText)findViewById(R.id.editText);
    Wa = (Button)findViewById(R.id.btn_whatsapp);
    Wa.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            Uri uri = Uri.parse("smsto:" + id);
            Intent waIntent = new Intent(Intent.ACTION_SENDTO,uri);

            String text = "testing message";
            waIntent.setPackage("com.whatsapp");
            if (waIntent != null) {
                waIntent.putExtra(Intent.EXTRA_TEXT, text);
                startActivity(Intent.createChooser(waIntent, text));
            } else {
                Toast.makeText(getApplicationContext(), "WhatsApp not found", Toast.LENGTH_SHORT)
                        .show();
            }

         }
    });

}

Answer 1

既然你试图实现它 as "smsto:"，"text/plain"as 类型会帮助你。尝试额外的，好像"sms_body"它没有帮助。

Uri uri = Uri.parse("smsto:" + id);
Intent waIntent = new Intent(Intent.ACTION_SENDTO,uri);
String text = "testing message";
waIntent.setPackage("com.whatsapp");
if (waIntent != null) {
    waIntent.setType("text/plain");
    //waIntent.putExtra(Intent.EXTRA_TEXT, text);
    waIntent.putExtra("sms_body", text); 
    startActivity(Intent.createChooser(waIntent, text));
} else {
    Toast.makeText(getApplicationContext(), "WhatsApp not found", Toast.LENGTH_SHORT)
            .show();
}