安卓权限请求代码问题

Question

如何请求权限？我尝试过查看文档，但是常量int请求代码MY_PERMISSIONS_REQUEST_CALL_PHONE似乎不能正常工作，还有什么需要注意的反向兼容性吗？

ActivityCompat.requestPermissions(getApplicationContext(),
                            new String[]{Manifest.permission.READ_CONTACTS},
                            MY_PERMISSIONS_REQUEST_CALL_PHONE);

如何声明MY_PERMISSIONS_REQUEST_CALL_PHONE常量int？

- ueen

---

也许这对你有用 http://stackoverflow.com/a/36787464/3436179 - Alexander

请检查此参考链接：http://www.tutorialspoint.com/android/android_phone_calls.htm。 - Damini Mehra

Answer 1

public void makeCall(String s)
{
    Intent intent = new Intent(Intent.ACTION_CALL);
    intent.setData(Uri.parse("tel:" + s));
    if (ActivityCompat.checkSelfPermission(this, Manifest.permission.CALL_PHONE) != PackageManager.PERMISSION_GRANTED){

        requestForCallPermission();

    } else {
        startActivity(intent);

    }
}
public void requestForCallPermission()
{

    if (ActivityCompat.shouldShowRequestPermissionRationale(this,Manifest.permission.CALL_PHONE))
    {
    }
    else {

        ActivityCompat.requestPermissions(this,new String[]{Manifest.permission.CALL_PHONE},PERMISSION_REQUEST_CODE);
    }
}

@Override
public void onRequestPermissionsResult(int requestCode, String permissions[], int[] grantResults)
{
    switch (requestCode) {
        case PERMISSION_REQUEST_CODE:
            if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_GRANTED) {
                makeCall("100");
            }
            break;
    }
}

//现在调用方法makeCall("你想要拨打的电话号码"); makeCall("100"); 更多详情请查看链接

- Karan Shaw

Answer 2

对于较低版本，您只需要在清单中声明权限， 但对于Marshmallow，您需要在要执行代码的位置中给出权限。

在此处，您想发起通话。因此，请将下面提供的代码插入/包含在编写呼叫的代码块中。

   public void makeCall(){
       Intent intent = new Intent(Intent.ACTION_CALL);
       intent.setData(Uri.parse("tel:" + "123456"));
       int result = ContextCompat.checkSelfPermission(context, Manifest.permission.CALL_PHONE);
       if (result == PackageManager.PERMISSION_GRANTED){
           startActivity(intent);
       } else {
           requestForCallPermission(); 
       }
  }

  private void requestPermission(){
    if(ActivityCompat.shouldShowRequestPermissionRationale(activity,Manifest.permission.CALL_PHONE)){} 
    else {
           ActivityCompat.requestPermissions(activity,new String[]{Manifest.permission.CALL_PHONE},PERMISSION_REQUEST_CODE);
       }
  }

  @Override
  public void onRequestPermissionsResult(int requestCode, String permissions[], int[] grantResults) {
    switch (requestCode) {
        case PERMISSION_REQUEST_CODE:
            if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_GRANTED) { 
                makeCall(); 
            } 
            break;
      }
  }

- Vickyexpert

---

是的，我理解了，但是如何声明PERMISSION_REQUEST_CODE？ - ueen

抱歉，我忘记告诉你了，你需要像下面这样在顶部声明它 private static final int PERMISSION_REQUEST_CODE = 1; - Vickyexpert

同时，在requestPermission()方法的else部分，将权限更改为CALL_PHONE而不是ACCESS_FINE_LOCATION。 - Vickyexpert

Answer 3

您还需要在AndroidManifest.xml中指定要使用的权限，例如：

<uses-permission android:name="android.permission.READ_CONTACTS"></uses-permission>

- nandu codes

Answer 4

            if (ActivityCompat.checkSelfPermission(this, Manifest.permission.CALL_PHONE) != PackageManager.PERMISSION_GRANTED) {
                // TODO: Consider calling
                //    ActivityCompat#requestPermissions
                // here to request the missing permissions, and then overriding
                //   public void onRequestPermissionsResult(int requestCode, String[] permissions,
                //                                          int[] grantResults)
                // to handle the case where the user grants the permission. See the documentation
                // for ActivityCompat#requestPermissions for more details.
                return;
            }
Intent intent = new Intent(Intent.ACTION_CALL);
            intent.setData(Uri.parse("tel:" + "123456"));
            startActivity(intent);

试着这样做。

- Sharad Chauhan

---

谢谢，它开始工作了，我猜我把startActivity放在if之前 :) - ueen

是的，那就是唯一的问题。继续学习吧。 :) - Sharad Chauhan

Answer 5

尝试下面的代码，希望能帮到您。 首先，它会要求您允许弹出窗口，然后才会拨打号码。

if (ContextCompat.checkSelfPermission(HomePanelActivity.this, Manifest.permission.CALL_PHONE) != PackageManager.PERMISSION_GRANTED) {
            if (ActivityCompat.shouldShowRequestPermissionRationale(HomePanelActivity.this,
                    Manifest.permission.CALL_PHONE)) {
                ActivityCompat.requestPermissions(HomePanelActivity.this, new String[]{Manifest.permission.CALL_PHONE}, REQUEST_PERMISSION);
            }
        } else {
            Intent callIntent = new Intent(Intent.ACTION_CALL);
            callIntent.setData(Uri.parse("tel:" + phoneNumber));
            if (ActivityCompat.checkSelfPermission(HomePanelActivity.this, Manifest.permission.CALL_PHONE) == PackageManager.PERMISSION_GRANTED) {
                startActivity(callIntent);
            }
        }

@Override
    public void onRequestPermissionsResult(int requestCode, String permissions[], int[] grantResults) {
        switch (requestCode) {
            case 10:
                if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_GRANTED) {
                    Intent callIntent = new Intent(Intent.ACTION_CALL);
                    callIntent.setData(Uri.parse("tel:" + phoneNumberToCall));
                    if (ActivityCompat.checkSelfPermission(HomePanelActivity.this, Manifest.permission.CALL_PHONE) == PackageManager.PERMISSION_GRANTED) {
                        startActivity(callIntent);
                    }
                } else {
                    Snackbar.make(drawerLayout, "You Deny permission", Snackbar.LENGTH_SHORT).show();
                return;
            }
        }
    };

- Drup Desai

---

我需要在启动时获取权限，并在按钮点击时进行调用。 而且我需要这个权限，因此拒绝是不可选的... - ueen

这需要API23，我的应用的最低SDK版本是15。 - ueen

是的，您需要编译SDK版本到23。最低15也可以工作。 - Drup Desai