我创建了一个简单的数据库用来插入数据库列表,但是当我想把数据发送到数据库时,有下面的错误:
08-28 15:30:43.015: ERROR/Database(822): Failure 1 (no such column: Sharma) on 0x276e50 when preparing 'INSERT INTO friends Values(Sharma,SaiGeetha,18);""
相关的代码:
sampleDB = this.openOrCreateDatabase(SAMPLE_DB_NAME, MODE_PRIVATE, null);
ArrayList<String>FirstName = new ArrayList<String>();
ArrayList<String>LastName = new ArrayList<String>();
ArrayList<Integer >Age = new ArrayList<Integer>();
FirstName.add("SaiGeetha");
FirstName.add("Vivek");
FirstName.add("Rahul");
LastName.add("Sharma");
LastName.add("Lilani");
LastName.add("Lami");
Age.add(18);
Age.add(20);
Age.add(23);
sampleDB.execSQL("CREATE TABLE IF NOT EXISTS " +
SAMPLE_TABLE_NAME +
" (LastName VARCHAR, FirstName VARCHAR," +
" Age INT(3));");
for(int i=0;i<3;i++)
{
sampleDB.execSQL("INSERT INTO " +
enter code here
SAMPLE_TABLE_NAME +
" Values("+LastName.get(i)+","+FirstName.get(i)+","+Age.get(i)+");"+"\"\"");
}
/*sampleDB.execSQL("INSERT INTO " +
SAMPLE_TABLE_NAME +
" Values ('Makam','Sai Geetha','India',25);");
sampleDB.execSQL("INSERT INTO " +
SAMPLE_TABLE_NAME +
" Values ('Chittur','Raman','India',25);");
sampleDB.execSQL("INSERT INTO " +
SAMPLE_TABLE_NAME +
" Values ('Solutions','Collabera','India',20);");*/
Cursor c = sampleDB.rawQuery("SELECT FirstName, Age FROM " +
SAMPLE_TABLE_NAME +
" where Age > 10 LIMIT 5", null);
if (c != null ) {
if (c.moveToFirst()) {
do {
String firstName = c.getString(c.getColumnIndex("FirstName"));
int age = c.getInt(c.getColumnIndex("Age"));
Log.e("---LIST FROM DATABASE--","---VALUE---"+firstName);
Log.e("---LIST FROM DATABASE--","---VALUE---"+age );
}while (c.moveToNext());
}
}
sampleDB.close();
} catch (SQLiteException se )
{
Log.e(getClass().getSimpleName(), "Could not create or Open the database");
} finally {
if (sampleDB != null)
sampleDB.execSQL("DELETE FROM " + SAMPLE_TABLE_NAME);
sampleDB.close();
}
}
哪里的问题呢?