On Sun, Nov 21, 2010 at 8:42 PM, Andy Dustman <farcep...@gmail.com> wrote:
>
>
> You never actually set sql anywhere, so you'll always get a NameError
> instead of IOError. It would probably be better to not catch the
> exception at all in this function.
> --
> Question the answers
> _______________________________________________
>
Thank you, Andy!
I missed that. That's why open source code is a good thing. More eyes
looking finds problems.
This time I actually entered the code into a file and executed it. A
working version follows:
<code>
from __future__ import print_function
import sqlite3
import sys
def createdb(db):
try:
con = sqlite3.connect(db)
cur = con.cursor()
sql = '''
CREATE TABLE t1
(
kid INTEGER PRIMARY KEY,
c1 TEXT,
c2 TEXT
)'''
cur.execute(sql)
sql = '''
CREATE TABLE t2
(
kid INTEGER PRIMARY KEY,
c1 TEXT,
c2 TEXT
)'''
cur.execute(sql)
sql = '''
CREATE TABLE t3
(
kid INTEGER PRIMARY KEY,
c1 TEXT,
c2 TEXT
)'''
cur.execute(sql)
con.commit()
except sqlite3.Error:
print("ERROR: createdb did not commit.", file=sys.stderr)
print("tried this sql:", file=sys.stderr)
print(sql, file=sys.stderr)
raise
finally:
cur.close()
con.close()
if __name__ == '__main__':
createdb('myDbName')
</code>
This script will run once without errors.
If you run it a second time, you get:
<console output>
ERROR: createdb did not commit.
tried this sql:
CREATE TABLE t1
(
kid INTEGER PRIMARY KEY,
c1 TEXT,
c2 TEXT
)
Traceback (most recent call last):
File "x.py", line 44, in <module>
createdb('myDbName')
File "x.py", line 14, in createdb
cur.execute(sql)
sqlite3.OperationalError: table t1 already exists
</console output>
Which is exactly what you should expect. Note that using the naked "raise"
statement results in a higher quality traceback with error text pinpointing
the problem.
I used print as a function, so this example should work in python versions
from 2.6 to 3.2.
--
Vernon
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