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();
}
}
});
}
解决方法
由于您尝试将其实现为“smsto：”,因此“text / plain”类型将帮助您.如果没有帮助,请尝试额外的“sms_body”.
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(),Toast.LENGTH_SHORT)
.show();
}
