我正在尝试配置一个简单的SQLiteOpenHelper,在执行. android.database.sqlite时出现错误。表people_table没有名为Minutes的列(代码1):,编译时:INSERT INTO people_table(分钟,小时)VALUES(?,?)我不知道问题出在哪里?
package radiofm.arabel;
import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteOpenHelper;
import android.util.Log;
public class DatabaseHelper extends SQLiteOpenHelper {
private static final String TAG = "DatabaseHelper";
private static final String TABLE_NAME = "people_table";
private static final String COL1 = "ID";
private static final String COL2 = "Hour";
private static final String COL3 = "Minutes";
public DatabaseHelper(Context context) {
super(context, TABLE_NAME, null, 1);
}
@Override
public void onCreate(SQLiteDatabase db) {
String createTable = "CREATE TABLE " + TABLE_NAME + " (" +
COL1 + "ID INTEGER PRIMARY KEY AUTOINCREMENT," +
COL2 + "TEXT," +
COL3 + "TEXT " +
");";
db.execSQL(createTable);
}
@Override
public void onUpgrade(SQLiteDatabase db, int i, int i1) {
db.execSQL("DROP IF TABLE EXISTS " + TABLE_NAME);
onCreate(db);
}
public boolean addData(String item1,String item2) {
SQLiteDatabase db = this.getWritableDatabase();
ContentValues contentValues = new ContentValues();
contentValues.put(COL2, item1);
contentValues.put(COL3, item2);
Log.d(TAG, "addData: Adding " + item1 + " to " + TABLE_NAME);
long result = db.insert(TABLE_NAME, null, contentValues);
//if date as inserted incorrectly it will return -1
if (result == -1) {
return false;
} else {
return true;
}
}
}
2条答案
按热度按时间6g8kf2rb1#
您在创建表时忘记了空格。您的列现在名为
MinutesTEXT
。这本身不是错误,因为您不必在SQLite中指定列类型。
Reference
要解决这个问题,请确保删除当前数据库,并将创建表查询替换为该查询。
pu3pd22g2#
您的onCreate在SQL列定义中似乎缺少空格。另外,您的ID列必须定义为_id。
试试这个。
还有