Android 在openDatabase之后尝试获取可读性时,出现数据库未打开错误
执行此代码后,在mydb=this.getReadableDatabase()的行中;错误显示“数据库未打开”。原因是什么Android 在openDatabase之后尝试获取可读性时,出现数据库未打开错误,android,sqlite,Android,Sqlite,执行此代码后,在mydb=this.getReadableDatabase()的行中;错误显示“数据库未打开”。原因是什么 private SQLiteDatabase mydb; public Cursor select(String table,String where){ mydb.openDatabase(DATABASE_PATH, null, 1); mydb =this.getReadableDatabase()
private SQLiteDatabase mydb;
public Cursor select(String table,String where){
mydb.openDatabase(DATABASE_PATH, null, 1);
mydb =this.getReadableDatabase();
String sql = "SELECT * FROM "+table+" WHERE "+where;
xLog.info(sql);
Cursor result=null;
try {
result = mydb.rawQuery("SELECT * FROM "+table+" WHERE "+where, null);
} catch (Exception e) {
xLog.error(e.getMessage());
}
return result;
}
这是整个数据库类,当然我简化了它并删除了不必要的数据:
public class DbHelper extends SQLiteOpenHelper{
private SQLiteDatabase mydb;
// private static DbHelper sInstance = null;
private static final String DATABASE_NAME = "shareholders.db";
private static final String DATABASE_PATH = "/data/data/com.example.shareholders/databases/shareholders.db";
public DbHelper(Context context) {
super(context, "shareholders.db", null, 1);
}
@Override
public void onCreate(SQLiteDatabase db) {
try {
sql = "CREATE TABLE IF NOT EXISTS settings (name text,value text)";
db.execSQL(sql);
} catch (Exception e) {
xLog.error(e.getMessage());
}
ContentValues cv = new ContentValues()
cv.clear();
cv.put("name", "picture");
cv.put("value", "");
db.insert("settings", null, cv);
db.close();
}
@Override
public void onUpgrade(SQLiteDatabase arg0, int arg1, int arg2) {
// TODO Auto-generated method stub
}
public Cursor selectAll(String table){
mydb.openDatabase(DATABASE_PATH, null, 1);
mydb =this.getReadableDatabase();
String sql = "SELECT * FROM "+table;
xLog.info(sql);
Cursor result=null;
try {
result = mydb.rawQuery(sql, null);
} catch (Exception e) {
xLog.error(e.getMessage());
}
return result;
}
}
根据你的密码
DbHelper helper = new DbHelper(context);
mydb = helper.getReadableDatabase();
这不是在Android中实现数据库评估的最佳方式 把你的数据库类的代码放在这里,可能你的数据库还没有被创建,这是因为y系统无法打开它。我该怎么处理你的答案呢?你试图在不创建数据库的情况下打开数据库。您需要初始化SQLiteOpenHelper。将这些添加到selectAll()DbHelper=newdbhelper(上下文);mydb=helper.getReadableDatabase();