我使用游标正确使用SQLite?

问题描述 投票:3回答:1

我在“实验”与Android工作室和SQLite的第一次。我的应用程序的屏幕变成白色,并在打开新的活动应该出现在我的SQL查询值冻结。

public class DatabaseHelper extends SQLiteOpenHelper {

public DatabaseHelper (Context context) {
    super(context,"book.db", null, 1);

}

@Override
public void onCreate(SQLiteDatabase db) {
    db.execSQL("create table books (ID INTEGER NOT NULL PRIMARY KEY, " +
            "BOOK_TITLE TEXT NOT NULL, BOOK_CATEGORY TEXT NOT NULL, " +
            "BOOK_DESCRIPTION TEXT NOT NULL)");
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    db.execSQL("drop table if exists books");
    onCreate(db);
}

public boolean insertData(String book_title_, String book_category_, String book_description_)
{
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues cValues = new ContentValues();
    cValues.put("BOOK_TITLE", book_title_);
    cValues.put("BOOK_CATEGORY", book_category_);
    cValues.put("BOOK_DESCRIPTION", book_description_);
    long result = db.insert("book", null, cValues);
    if (result == -1)
        return false;
    return true;
}

public Cursor getAllData()
{
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor res = db.rawQuery("SELECT * FROM book", null);
    return res;
}
}

我发现,问题出在从另一个类调用getAllData()机能的研究。通过我的逻辑应当一切正常?不过,我觉得这个问题是在我的查询,但据谷歌我所做的一切良好。有什么建议么?

java android sqlite
1个回答
0
投票

试试这个方法。这是非常有效的方法:

public List<Movie> getAllData()
{
    List<Movie> movies_list = new ArrayList<>();
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor res = db.rawQuery("SELECT * FROM movie_collection", null);
    while(res.moveToNext())
    {
        movies_list.add(new Movie(res.getString(1)+"\n", res.getString(2)+"\n", res.getString(3)+"\n", R.drawable.gladiator));
    }
    res.close();
    return movies_list;
}

和在viewAll()方法:

public void viewAll()
{
    movies_list = movie_database.getAllData();
    // do whatever you want
}
© www.soinside.com 2019 - 2024. All rights reserved.