我创建了一个简单的 Android 活动,充当拨号盘。
它有一个电话号码的编辑文本和一个呼叫按钮
这是代码:(android 6.0 marshmallow)
public class Main2Activity extends AppCompatActivity {
EditText num;
Button call;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main2);
num = (EditText) findViewById(R.id.num);
call = (Button) findViewById(R.id.call);
call.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
try {
// request permission if not granted
if (ActivityCompat.checkSelfPermission(Main2Activity.this, Manifest.permission.CALL_PHONE) != PackageManager.PERMISSION_GRANTED) {
ActivityCompat.requestPermissions(Main2Activity.this, new String[]{Manifest.permission.CALL_PHONE}, 123);
// i suppose that the user has granted the permission
Intent in = new Intent(Intent.ACTION_CALL, Uri.parse("tel:" + num.getText().toString()));
startActivity(in);
// if the permission is granted then ok
} else {
Intent in = new Intent(Intent.ACTION_CALL, Uri.parse("tel:" + num.getText().toString()));
startActivity(in);
}
}
// catch the exception if I try to make a call and the permission is not granted
catch (Exception e){
}
}
});
}
}
当我运行我的应用程序时,我遇到了这些问题
使用 onRequestPermissionResult,它处理用户按下的操作ALLOW and DENY,只需在“如果用户按允许”的条件下调用意图即可:
@Override
public void onRequestPermissionsResult(int requestCode, String permissions[], int[] grantResults) {
super.onRequestPermissionsResult(requestCode, permissions, grantResults);
switch (requestCode) {
case 123: {
if (grantResults.length > 0
&& grantResults[0] == PackageManager.PERMISSION_GRANTED) {
//If user presses allow
Toast.makeText(Main2Activity.this, "Permission granted!", Toast.LENGTH_SHORT).show();
Intent in = new Intent(Intent.ACTION_CALL, Uri.parse("tel:" + num.getText().toString()));
startActivity(in);
} else {
//If user presses deny
Toast.makeText(Main2Activity.this, "Permission denied", Toast.LENGTH_SHORT).show();
}
break;
}
}
}
希望这可以帮助。
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)