Question

我一直在努力在我的iPhone应用程序中加入一个sqlite数据库。数据库已创建，我已成功将元素插入数据库。现在，当我尝试从db中进行选择时，我的prepare语句不符合SQLITE_OK的要求。这是我创建数据库的代码：

const char *sql_stmt = "CREATE TABLE IF NOT EXISTS EXPENSES (id integer primary key autoincrement, unix_date integer, date_time text, purpose text, start_mile text, end_mile text, distance text, fees text, party_id integer)";

这是我用来提取数据的代码：

    const char *dbpath = [databasePath UTF8String];
sqlite3_stmt    *statement;
printf("in getExpense\n");
//expenses = [[NSMutableArray alloc] init];

if (sqlite3_open(dbpath, &expenseDB) == SQLITE_OK)
{
    NSString *querySQL = [NSString stringWithFormat: @"SELECT * FROM EXPENSES"];

    const char *query_stmt = [querySQL UTF8String];

    if (sqlite3_prepare_v2(expenseDB, query_stmt, -1, &statement, NULL) == SQLITE_OK)
    {

        printf("right before the while loop\n");
        while (sqlite3_step(statement) == SQLITE_ROW)
        {
            printf("in the while loop\n");
            VehicleExpense *expense = [[VehicleExpense alloc] init];

            NSInteger *newTimeStamp = sqlite3_column_int(statement, 1);

            NSString *newDate = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 2)];

            NSString *newPurpose = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 3)];

            NSString *newStart = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 4)];
            NSString *newEnd = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 5)];
            NSString *newDistance = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 6)];
            NSString *newFees = [[NSString alloc] initWithUTF8String:(const char *) sqlite3_column_text(statement, 7)];
            NSInteger *newPartyId = sqlite3_column_int(statement, 8);

            expense.unix_date = newTimeStamp;
            expense.date = newDate;
            expense.purpose = newPurpose;
            expense.start_mile = newStart;
            expense.end_mile = newEnd;
            expense.distance = newDistance;
            expense.fees = newFees;
            expense.party_id = newPartyId;

            [expenses addObject: expense];
            printf("expense add");

            [newDate release];
            [newPurpose release];
            [newStart release];
            [newEnd release];
            [newDistance release];
            [newFees release];

            [expense release];

            expense_count++;

        }
        sqlite3_finalize(statement);
    }else{
        printf("the prepare statement is not ok\n");
    }
    sqlite3_close(expenseDB);
}else{
    printf("couldn't open the db\n");
}

第二个if语句因某种原因而失败？

有什么想法吗？

Answer 1

printf("the prepare statement is not ok\n");根本没有用处。尝试像

这样的东西

NSLog(@"prepare failed (%s)", sqlite3_errmsg(expenseDB));

sqlite语句不行......为什么？ - 苹果手机

1 个答案: