8.9 Generating a Dictionary Mapping from Field Names to Column Numbers

Credit: Tom Jenkins

8.9.1 Problem

You want to access data fetched from a DB API cursor object, but you want to access the columns by field name, not by number.

8.9.2 Solution

Accessing columns within a set of database-fetched rows by column index is neither readable nor robust if columns are ever reordered. This recipe exploits the description attribute of Python DB API's cursor objects to build a dictionary that maps column names to index values, so you can use cursor_row[field_dict[fieldname]] to get the value of a named column:

def fields(cursor):     """ Given a DB API 2.0 cursor object that has been executed, returns     a dictionary that maps each field name to a column index; 0 and up. """     results = {}     column = 0     for d in cursor.description:         results[d[0]] = column         column = column + 1     return results

8.9.3 Discussion

When you get a set of rows from a call to:

cursor.fetch{one, many, all}

it is often helpful to be able to access a specific column in a row by the field name and not by the column number. This recipe shows a function that takes a DB API 2.0 cursor object and returns a dictionary with column numbers keyed to field names.

Here's a usage example (assuming you put this recipe's code in a module that you call dbutils.py somewhere on your sys.path):

>>> c = conn.cursor(  ) >>> c.execute('''select * from country_region_goal where crg_region_code is null''') >>> import pprint >>> pp = pprint.pprint >>> pp(c.description) (('CRG_ID', 4, None, None, 10, 0, 0), ('CRG_PROGRAM_ID', 4, None, None, 10, 0, 1), ('CRG_FISCAL_YEAR', 12, None, None, 4, 0, 1), ('CRG_REGION_CODE', 12, None, None, 3, 0, 1), ('CRG_COUNTRY_CODE', 12, None, None, 2, 0, 1), ('CRG_GOAL_CODE', 12, None, None, 2, 0, 1), ('CRG_FUNDING_AMOUNT', 8, None, None, 15, 0, 1)) >>> import dbutils >>> field_dict = dbutils.fields(c) >>> pp(field_dict) {'CRG_COUNTRY_CODE': 4, 'CRG_FISCAL_YEAR': 2, 'CRG_FUNDING_AMOUNT': 6, 'CRG_GOAL_CODE': 5, 'CRG_ID': 0, 'CRG_PROGRAM_ID': 1, 'CRG_REGION_CODE': 3} >>> row = c.fetchone(  ) >>> pp(row) (45, 3, '2000', None, 'HR', '26', 48509.0) >>> ctry_code = row[field_dict['CRG_COUNTRY_CODE']] >>> print ctry_code HR >>> fund = row[field_dict['CRG_FUNDING_AMOUNT']] >>> print fund 48509.0

8.9.4 See Also

Recipe 8.10 for a slicker and more elaborate approach to the same task.