我创建了一个简单的android活动,充当拨号盘。 它有一个用于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){
}
}
});
}
}
当我运行我的应用时,必须发布
如果我点击通话按钮并授予权限,则在我再次点击之前不会调用该意图
我不知道如何检查是否授予了权限
答案 0 :(得分:4)
使用onRequestPermissionResult,如果用户按允许和 DENY ,它会处理操作,只需在“如果用户按下允许”条件下调用意图:
@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;
}
}
}
希望这有帮助。
答案 1 :(得分:0)
if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_GRANTED)
是检查许可权结果的糟糕逻辑,我不知道Google为什么实施如此糟糕的代码。
特别是当您检查多个权限时,情况一片混乱。假设您要
CAMERA
,ACCESS_FINE_LOCATION
和ACCESS_NETWORK_STATE
。
您需要检查ACCESS_FINE_LOCATION,但用户仅在第一次运行时才授予CAMERA,并且您要检查GrantResults [1],但在第二次运行中,ACCESS_FINE_LOCATION成为索引为0的权限。
您应该使用
int size = permissions.length;
boolean locationPermissionGranted = false;
for (int i = 0; i < size; i++) {
if (permissions[i].equals(Manifest.permission.ACCESS_FINE_LOCATION)
&& grantResults[i] == PackageManager.PERMISSION_GRANTED) {
locationPermissionGranted = true;
}
}
或更简单的一个
if (ContextCompat.checkSelfPermission(this, Manifest.permission.ACCESS_FINE_LOCATION)
== PackageManager.PERMISSION_GRANTED) {
// Do something ...
}
采用onPermissionRequestResult
方法。