Python SQL update query: string or integer address expected instead of instance instance

Question

I use Python 2.7 and pypyodbc in order to run SQL queries but whenever I run the update query using python,

cursor.execute("UPDATE tbl_User SET gender = ? WHERE id = 1", ['male'])

I get the error:

TypeError: string or integer address expected instead of instance instance

The same query works if I run it directly on SQL Server.

Are you able to get it to work without passing the parameter and hard coding it into the UPDATE statement? — kayla210
– kayla210, Commented Jun 3, 2016 at 13:17
you mean something like that: cursor.execute("UPDATE tbl_User SET gender = 'male' WHERE id = 1")? — Shani Gamrian
– Shani Gamrian, Commented Jun 3, 2016 at 13:18
Yes, just to ensure the code is executable through pypyodbc. :) — kayla210
– kayla210, Commented Jun 3, 2016 at 13:23
I get the error :/ but as I said, it does work if I run it directly on SQL Server — Shani Gamrian
– Shani Gamrian, Commented Jun 3, 2016 at 13:27

Shani Gamrian · Accepted Answer · 2016-06-03 13:48:22Z

3

The problem was I didn't put a semicolon at the end of the query. The solution is:

cursor.execute("UPDATE tbl_User SET gender = ? WHERE id = 1;", ['male'])

answered Jun 3, 2016 at 13:48

Shani Gamrian

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2 Comments

You may want to amend your answer and tag it to include what database you're using, since this would seem to be database specific.

Ah, that makes sense. SQL Server's really adamant about those semi-colons, but it'd been a while since I've used it, so that didn't pop out to me.

Ma0 · Accepted Answer · 2016-06-03 13:39:19Z

0

How about that:

gender = 'male'
cursor.execute("UPDATE tbl_User SET gender = {:s} WHERE id = 1".format(gender))

Also make sure that the python version of sqlite and the one used by the SQL Server are the same. For the python one type:

import sqlite3
print(sqlite3.sqlite_version)

answered Jun 3, 2016 at 13:20

Ma0

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I get the error: pypyodbc.ProgrammingError: (u'42S22', u"[42S22] [Microsoft][ODBC SQL Server Driver][SQL Server]Invalid column name 'male'.")