Question

我正在编写一个简单的应用程序，要求获得呼叫权限，用户可以在EditText中键入要拨打的电话号码，然后按“拨打电话”。

当我拨打一个真实的号码时，它最终会转到手机但是它拨打一些长号码，例如2637643787767877.无论是否发生这种情况我使用模拟器或我自己的galaxy s6。

我当然在我的清单文件中有这个：

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

这是我的代码。

package cornez.com.phonecall;

import android.Manifest;
import android.content.Intent;
import android.content.pm.PackageManager;
import android.net.Uri;
import android.support.annotation.NonNull;
import android.support.v4.app.ActivityCompat;
import android.support.v4.content.ContextCompat;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.view.WindowManager;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends AppCompatActivity {

    EditText phoneNumToCall;
    Button makeCallButton;
    int permissionCheck;
    private static final int PERMISSION_REQUEST_CALL_PHONE = 100;
    private static final String LOG_TAG = "MainActivity";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        phoneNumToCall = (EditText) findViewById(R.id.editText);

        makeCallButton = (Button) findViewById(R.id.button);

        //onclicklistener for BUTTON

        permissionCheck = ContextCompat.checkSelfPermission(this,
                Manifest.permission.CALL_PHONE);

        makeCallButton.setOnClickListener(new View.OnClickListener() {
            public void onClick(View v) {
                // your handler code here


                if (permissionCheck == PackageManager.PERMISSION_GRANTED) {
                    makeCall();
                }

                else {
                    // request the permission
                    ActivityCompat.requestPermissions(MainActivity.this,
                            new String[]{Manifest.permission.CALL_PHONE},
                            PERMISSION_REQUEST_CALL_PHONE);
                }

            } });





        //makeCallButton.setOnClickListener();

        phoneNumToCall.requestFocus();
        getWindow().setSoftInputMode(WindowManager.LayoutParams.
                SOFT_INPUT_STATE_VISIBLE);
    }

    public void onRequestPermissionsResult(int requestCode,
                                           @NonNull String permissions[],
                                           @NonNull int[] grantResults)
    {
        if (requestCode == PERMISSION_REQUEST_CALL_PHONE)
        {
            if (grantResults.length > 0 &&
                    grantResults[0] == PackageManager.PERMISSION_GRANTED)
            {
                makeCall();
            }
            else
            {
                // notify user that permission was denied
                Toast.makeText(this, "The permission was denied", Toast.LENGTH_SHORT).show();
            }
        }
    }


    private void makeCall()
    {
          try
          {
              Uri uri = Uri.parse("tel:" + phoneNumToCall);
              Intent intent = new Intent(Intent.ACTION_CALL, uri);
              startActivity(intent);
          }
          catch (SecurityException ex)
          {
              String errorMsg = "No permission to make phone call.";
              Log.e(LOG_TAG, errorMsg, ex);
          }
     }




    }

问题是，这有什么问题？

Answer 1

您没有从EditText解析字符串你只需声明EditText并调用它。

phoneNumToCall = (EditText) findViewById(R.id.editText);

试试这个

用此

替换makeCall()方法

private void makeCall()
    {
          try
          {
              Uri uri = Uri.parse("tel:" + phoneNumToCall.getText().toString());
              Intent intent = new Intent(Intent.ACTION_CALL, uri);
              startActivity(intent);
          }
          catch (SecurityException ex)
          {
              String errorMsg = "No permission to make phone call.";
              Log.e(LOG_TAG, errorMsg, ex);
          }
     }

Answer 2

String uri = "tel:" + callNumber;
            String deviceName = android.os.Build.MODEL;
            Intent intent;
            if ("Nexus 5".equalsIgnoreCase(deviceName) || "XT1092".equalsIgnoreCase(deviceName) || "XT1572".equalsIgnoreCase(deviceName) || "XT1575".equalsIgnoreCase(deviceName) ) {
                intent = new Intent(Intent.ACTION_DIAL);
            } else {
                intent = new Intent(Intent.ACTION_CALL);
            }
            intent.setData(Uri.parse(uri));
            startActivity(intent)

试试这段代码。

PhoneCall - 拨打奇怪的长号而不是输入的号码

2 个答案: