Question

我正在尝试使用SQLite数据库实现ListView，但我一直收到错误＆＃39;无法解析符号TABLE_STUDENTEN＆＃39;。＆＃39; TABLE_STUDENTEN＆＃39;是数据库中表的名称。这是我的DBHandler代码：

 public class DBHandler extends SQLiteOpenHelper {

private static final int DATABASE_VERSION = 1;
private static final String DATABASE_NAME = "StudentDB";
public static final String TABLE_STUDENTEN = "Studenten";
public static String DB_PATH;
public static final String COLUMN_ID = "_id";
public static final String COLUMN_NAAM = "_naam";
public static final String COLUMN_STUDENTNUMMER = "_studentnummer";
public static final String COLUMN_KLAS = "_klas";
public static final String COLUMN_CIJFER = "_cijfer";
public static final String COLUMN_OPMERKINGEN = "_opmerkingen";
private Context currentContext;


public DBHandler(Context context){
 super(context, DATABASE_NAME, null, DATABASE_VERSION);
   Context currentContext = context;

    DB_PATH = "/data/data/" + context.getPackageName()+ "/databases";
    CreateDatabase();

}

@Override
public void onCreate(SQLiteDatabase db) {

}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    }

private void CreateDatabase() {

    boolean dbExists = checkDbExists();
    if (dbExists) {
        // niks doen
    } else {


        SQLiteDatabase DB = currentContext.openOrCreateDatabase(DATABASE_NAME, 0, null);
        DB.execSQL("CREATE TABLE IF NOT EXISTS" + TABLE_STUDENTEN +
                "(ID INT, naam VARCHAR, studentnummer VARCHAR, klas VARCHAR, cijfer INT, opmerkingen VARCHAR)" );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Abrahamse, Peter', 's1078846', 'INF1D'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Bremen, Karel', 's1087311', 'INF1C'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Dam', Lisa 's1074243', 'INF1G'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Evertsen, Frederik', 's1087599', 'INF1B'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Frederiksen, Willem', 's1071475', 'INF1A'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Greppel, Maria', 's1087412', 'INF1E'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Joker, Dorine', 's1077851', 'INF1F'); " );
        DB.execSQL("INSERT INTO" + TABLE_STUDENTEN + "Values ('Knopper, Jan', 's1071288', 'INF1H'); " );
    }

}
private  boolean checkDbExists(){
    SQLiteDatabase checkDB = null;

    try {
       String myPath = DB_PATH + DATABASE_NAME;
        checkDB = SQLiteDatabase.openDatabase(myPath, null, SQLiteDatabase.OPEN_READONLY);

    } catch (SQLiteException e) {
        // database bestaat nog niet
    }
    if (checkDB != null); {
        checkDB.close(); 
    }

    return checkDB != null ? true : false; 
}


}

这是该课程的代码：

package com.ipmedt4.challengeweek_v2;

import android.app.ListActivity;
import android.content.Context;
import android.content.Intent;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteException;
import android.support.v7.app.ActionBarActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.ArrayAdapter;
import android.widget.Button;

import java.util.ArrayList;


public class OverzichtStudenten extends ListActivity {
private ArrayList<String> allestudentenresultaat = new ArrayList<String>();
private SQLiteDatabase StudentDB;

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

    displayResultatenLijst();
}

private void displayResultatenLijst(){
setListAdapter(new ArrayAdapter<String>(this, android.R.layout.activity_list_item, allestudentenresultaat));
    getListView().setTextFilterEnabled(true);

 }

private void openAndQueryDatabase(){
    try {

        DBHandler dbHandler = new DBHandler(this.getApplicationContext());
        StudentDB = dbHandler.getWritableDatabase();
        Cursor c = StudentDB.rawQuery("SELECT naam, studentnummer, klas FROM" + TABLE_STUDENTEN, null);
        if (c != null && c.moveToFirst()) {
            do {
                String naam = c.getString(c.getColumnIndex("naam" ));
                String studentnummer = c.getString(c.getColumnIndex("studentnummer" ));
                String klas = c.getString(c.getColumnIndex("klas" ));
            }
            while (c.moveToNext());
        }

    }
        catch (SQLiteException se){
            Log.e(getClass().getSimpleName(), "Kon database niet maken of openen");

        }
    finally {
        if (StudentDB != null)
            StudentDB.execSQL("DELETE FROM" + StudentDB);
            StudentDB.close();

    }
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.menu_overzicht_studenten, menu);
    return true;
}

@Override
public boolean onOptionsItemSelected(MenuItem item) {
    // Handle action bar item clicks here. The action bar will
    // automatically handle clicks on the Home/Up button, so long
    // as you specify a parent activity in AndroidManifest.xml.
    int id = item.getItemId();

    //noinspection SimplifiableIfStatement
    if (id == R.id.action_settings) {
        return true;
    }

    return super.onOptionsItemSelected(item);
}
}

抱歉这些奇怪的名字，我是荷兰人。类DBHandler不包含任何错误。我无法解决的唯一错误是在行＆＃39; Cursor c = StudentDB.rawQuery（＆＃34; SELECT naam，studentnummer，klas FROM＆＃34; + TABLE_STUDENTEN，null）;＆＃39;该错误无法解析符号TABLE_STUDENTEN＆＃39;

希望你能帮助我。谢谢！

Answer 1

那是因为您尝试使用类的public static变量而不使用类名。

使用此...

Cursor c = StudentDB.rawQuery("SELECT naam, studentnummer, klas FROM" + DBHandler.TABLE_STUDENTEN, null);

Answer 2

您已在DBHandler类中将表名定义为静态常量，并且您尝试访问的类OverzichtStudenten中无法访问此变量。

所以不要使用

Cursor c = StudentDB.rawQuery("SELECT naam, studentnummer, klas FROM" + TABLE_STUDENTEN, null);

将其用作：

Cursor c = StudentDB.rawQuery("SELECT naam, studentnummer, klas FROM" + DBHandler.TABLE_STUDENTEN, null);'

无法解析符号Tablename

2 个答案: