Question

Below is my RUN_SQL function:

def RUN_SQL_SAFE(sql, input_tuple=(), get_update_id=False, debug = False):
    conn = GET_MYSQL_CONNECTION()
    cursor = conn.cursor(cursorclass=MySQLdb.cursors.DictCursor)
    cursor.execute(sql, input_tuple)
    conn.commit()
    if get_update_id:
        res = cursor.lastrowid
    cursor.close()
    conn.close()
    if get_update_id:
        return res

I run my code using “RUN_SQL_SAFE(sql, tuple, True)”, here sql is a insert sql and the table is empty but return res with 3. I wonder to know why it doesn’t return 1??
Thanks

Answer 1

Maybe the table has been in use before an therefore has an AUTO_INCREMENT value > 1…

BTW: I don’t think that this repeated if get_update_id: is very elegant. Instead, you could do

def run_sql_safe(sql, input_tuple=(), get_update_id=False, debug = False):
    from contextlib import closing
    conn = get_mysql_connection()
    with closing(conn):
        cursor = conn.cursor(cursorclass=MySQLdb.cursors.DictCursor)
        with closing(cursor):
            cursor.execute(sql, input_tuple)
            conn.commit()
            if get_update_id:
                return cursor.lastrowid

If you could tell your connection to have MySQLdb.cursors.DictCursor as its “basic cursor class”, you could even do

def run_sql_safe(sql, input_tuple=(), get_update_id=False, debug = False):
    from contextlib import closing
    conn = get_mysql_connection(cursorclass=MySQLdb.cursors.DictCursor)
    with closing(conn):
        with conn as cursor:
            cursor.execute(sql, input_tuple)
            if get_update_id:
                return cursor.lastrowid

which does this commit automatically.