Python3 SQL Update not committing

Question

I have the following Python3 code:

def updateDB(databaseLoc, post_id, new_class):
    conn = sqlite3.connect(databaseLoc)
    curs = conn.cursor()
    vals = [new_class, post_id, ]
    curs.execute('UPDATE POST SET CLASSIFICATION=? WHERE POSTID=?', vals)
    conn.commit()
    conn.close()

which is called occasionally with

if js[0]['data']['children'][0]['data']['ups'] > 2000:
    print(post_id, "made it")
    updateDB('retrieved_data.db',1,post_id)

However, even after the update and the commit, when I go into the database, nothing has changed (classification is still set to the initial value 0). I have tested the SQL string manually and it works fine.

I have comparable code for insertion that works fine:

def pushToDB(databaseLoc, posts):
    conn = sqlite3.connect(databaseLoc)
    curs = conn.cursor()
    for p in posts:
        vals = [p.id, p.title, p.s, p.time, p.day, "", p.first, p.second, p.image, 0, ]
        try:
            curs.execute('INSERT INTO POST VALUES(?,?,?,?,?,?,?,?,?,?)',vals)
        except:
            pass
    conn.commit()
    conn.close()

Is there an additional step that needs to be taken for a Python3 update? I have looked through the documentation and it doesn't specify anything additional, but then it only mentions update once and just glosses over it. I have also found very little on update examples in python

Ian Henry · Accepted Answer · 2013-12-01 03:49:51Z

1

The signature for the function is:

def updateDB(databaseLoc, post_id, new_class):

But you appear to be calling it with:

updateDB(databaseLoc, new_class, post_id)

Resulting in trying to update the post where POSTID = 1 every time, which is probably not what you wanted.

answered Dec 1, 2013 at 3:49

Ian Henry

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