小编Arj*_*Rao的帖子

应用关闭点击Google标志按钮

我试图将谷歌登录集成到我的应用程序中.当我点击按钮时,应用程序关闭(不会崩溃).我在这里犯的错是什么?我尽力遵循谷歌开发者网站告诉我的内容.我提前为长代码道歉.

我的Java代码:

      import static com.google.android.gms.common.SignInButton.COLOR_DARK;

    public class User_Name extends AppCompatActivity implements
            GoogleApiClient.OnConnectionFailedListener{
        private PrefManager2 prefManager;
        SignInButton signInButton;
        String email, name;
        LoginButton loginButton;
        GoogleApiClient mGoogleApiClient;

        private static final int RC_SIGN_IN = 9001;
        SQLiteDatabase db;
        private GoogleSignInOptions googleSignInOptions;


        private static final String EMAIL_PATTERN = "^[a-zA-Z0-9#_~!$&'()*+,;=:.\"(),:;<>@\\[\\]\\\\]+@[a-zA-Z0-9-]+(\\.[a-zA-Z0-9-]+)*$";
        private Pattern pattern = Pattern.compile(EMAIL_PATTERN);
        private Matcher matcher; //validate email

        Button normalLogin;

        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);


            prefManager = new PrefManager2(this);
            if (!prefManager.isFirstTimeLaunch()) {
                launchHomeScreen(); //custom method
                finish();
            }


            setContentView(R.layout.activity_user__name);




            googleSignInOptions = new GoogleSignInOptions.Builder(GoogleSignInOptions.DEFAULT_SIGN_IN).requestEmail().requestId().build();
            mGoogleApiClient …
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android googlesigninapi

3
推荐指数
1
解决办法
542
查看次数

无法解析符号 getReadableDatabase();

我正在编写可能是最简单的程序来创建一个游标,所以我可以演示一个查询。我遇到了错误“无法解析符号 getReadableDatabase()”

PS:我已经尝试使缓存无效并重新启动。

    package com.example.arjunrao.databasedemo;

import android.app.Activity;
import android.app.AlertDialog;
import android.content.DialogInterface;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteException;
import android.database.sqlite.SQLiteOpenHelper;
import android.os.Bundle;
import android.view.View;
import android.database.Cursor;
import android.widget.EditText;

public class MainActivity extends Activity {

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

    }

    public void onclick(View view){
        EditText editText = (EditText) findViewById(R.id.editText);
         String drink_name = (String)editText.getText().toString();

        try{
            SQLiteOpenHelper database = new Database_class(this);
            SQLiteDatabase db = new database.getReadableDatabase();
            db.query("DRINK",new String[] {"NAME","DESCRIPTION","FAVORITE"});
        }catch(SQLiteException ex){
           final AlertDialog.Builder alertDialog = new AlertDialog.Builder(MainActivity.this);
            alertDialog.setTitle("SQLITE EXCEPTION GENERATED");
            alertDialog.setPositiveButton("OK", new …
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android android-sqlite

1
推荐指数
1
解决办法
2458
查看次数

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android ×2

android-sqlite ×1

googlesigninapi ×1