Question

我使用地震与下面的Android代码。

当用户摇动消息框时多次出现。我将如何修改代码以使其在摇动之后停止检测＆＃39;并等待消息框被解除，然后再次检测到震动？

import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.util.Log;
import android.app.AlertDialog;
import android.content.DialogInterface;
import com.squareup.seismic.ShakeDetector;
import com.squareup.seismic.ShakeDetector.Listener;

import android.hardware.SensorManager;


public class MainActivity extends AppCompatActivity implements Listener {





    public void clickFunction(View view) {


        new AlertDialog.Builder(this)
                .setTitle("Not Signed In")
                .setMessage("You need to be signed in to use this feature.")
                .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int which) {
                        // continue with delete
                    }
                })
                .setNegativeButton(android.R.string.no, new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int which) {
                        // do nothing
                    }
                })
                .setIcon(android.R.drawable.ic_dialog_alert)
                .show();
    }


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

        Log.i("Hello", "created!");

        SensorManager sensorManager = (SensorManager) getSystemService(SENSOR_SERVICE);
        ShakeDetector sd = new ShakeDetector(this);
        sd.start(sensorManager);




    }



    public void hearShake() {







            Log.i("wooo", "Shaken!");





        new AlertDialog.Builder(this)
                    .setTitle("Not Signed In")
                    .setMessage("SHAKEN!.")
                    .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                        public void onClick(DialogInterface dialog, int which) {
                            // continue with delete
                        }
                    })
                    .setNegativeButton(android.R.string.no, new DialogInterface.OnClickListener() {
                        public void onClick(DialogInterface dialog, int which) {
                            // do nothing
                        }
                    })
                    .setIcon(android.R.drawable.ic_dialog_alert)
                    .show();

        }




}

Answer 1

而不是new AlertDialog.Builder...show();我建议分配一个AlertDialog变量（例如mDialog），而只需添加hearShake()

if(mDialog.isShowing()) return;

更新：让sd成为全局变量

ShakeDetector sd = new ShakeDetector(this);
//...
protected void onCreate(...){
    //...
    sd = new ShakeDetector(this);
    sd.start(sensorManager);
}

public void hearShake(){
    //...
    sd.stop();
    //...
}

防止多重震动Android

1 个答案: