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Home/ Questions/Q 3596206
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T19:57:29+00:00 2026-05-18T19:57:29+00:00

I am trying to change some data in an sqlite3 file and I my

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I am trying to change some data in an sqlite3 file and I my non-existing knowledge in python and google-fu made me end up with this code:

#!/usr/bin/python
# Filename : hello.py

from sqlite3 import *

conn = connect('database')

c = conn.cursor()

c.execute('select * from table limit 2')

for row in c:
    newname = row[1]
    newname = newname[:-3]+"hello"
    newdata = "UPDATE table SET name = '" + newname + "', originalPath = '' WHERE id = '" + str(row[0]) + "'"
    print row
    c.execute(newdata)
    conn.commit()
c.close()

It works like a charm on the first row but for some reason it only runs the loop one time (only the first row in the table gets modified). When I remove “c.execute(newdata)” it loops through the first two rows in the table, as it should. How do I make it work?

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  1. Editorial Team
    Editorial Team
    2026-05-18T19:57:29+00:00Added an answer on May 18, 2026 at 7:57 pm

    It’s doing that because as soon as you do c.execute(newdata) the cursor is no longer pointing at the original result set anymore. I would do it this way:

    #!/usr/bin/python
    # Filename : hello.py
    
    from sqlite3 import *
    
    conn = connect('database')
    
    c = conn.cursor()
    
    c.execute('select * from table limit 2')
    result = c.fetchall()
    
    for row in result:
        newname = row[1]
        newname = newname[:-3]+"hello"
        newdata = "UPDATE table SET name = '" + newname + "', originalPath = '' WHERE id = '" + str(row[0]) + "'"
        print row
        c.execute(newdata)
    conn.commit()    
    c.close()
    conn.close()
    
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