sqlite3 COUNT in iOS (specifically ON the iPhone/iPad) - objective-c

Question

I know that the simulator and the actual iOS hardware are not EXACTLY the same, but I'm starting to pull my hair out over this one. I have this code:
sqlite3 *database;
sqlite3_stmt *statement;
int themeCount;
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK)
{
NSString *updateSQL = [NSString stringWithFormat: #"SELECT COUNT(*) FROM Theme"];
const char *update_stmt = [updateSQL UTF8String];
if(sqlite3_prepare_v2(database, update_stmt, -1, &statement, NULL) == SQLITE_OK){
if(sqlite3_step(statement)==SQLITE_ROW)
{
themeCount = sqlite3_column_int(statement, 0);
}
}
sqlite3_finalize(statement);
sqlite3_close(database);
}
With the simulator, it works perfectly fine. Once I push it to my devices, it fails. I've broken it down and came up with the return code where it fails:
if(sqlite3_prepare_v2(database, update_stmt, -1, &statement, NULL) == SQLITE_OK)
If I change that line to capture the code (ie. int x = sqlite3_prepare_v2(...)) it returns 0 with the simulator, 1 with the device. What am I doing wrong here?!?!
Also, for the record, the CREATE statement for the Theme table is:
#"CREATE TABLE Theme (ThemeId INTEGER PRIMARY KEY, ThemeName TEXT, Available BIT);"
(My first thought is that it was case sensitive)

Answer 1

You are not opening the database you think you are opening. The sqlite3_prepare_v2 is the first statement that needs the schema to be present. I suspect your databasePath is incorrect.
You can be more specific with sqlite3_open_v2 by omitting the SQLITE_OPEN_CREATE flag which is the default with sqlite3_open so you don't notice that a new database is being created by the open call. See SQLite3 docs. With the result of
sqlite3_open_v2([databasePath UTF8String], &database, SQLITE_OPEN_READWRITE, NULL)
you will see that the database does not exist.