convert output from sqlite from tuple to list in Python

Question

I am new to programming and am writing a simple python script that will gather some information from a PC for forensic purposes. Some of the information I want is the Chrome history. Basically I need to read the data from the history sqlite database, perform a calculation to convert the time stamp to a human readable format, and then write the data to a CSV file. I have the following code:

connection = sqlite3.connect('c:\history')### need correct path
connection.text_factory = str
cur = connection.cursor()
output_file = open('chrome_history2.csv', 'wb')
csv_writer = csv.writer(output_file,)
headers = ('URL', 'Title', 'Visit Count', 'Last Visit')
csv_writer.writerow(headers)
epoch = datetime(1601, 1, 1)
for row in (cur.execute('select url, title, visit_count, last_visit_time from urls limit 10')): #selects data
        list(row) #convert to list - does not work
        url_time = epoch + timedelta(microseconds=row[3]) #calculates time
        row[3] = url_time #changes value in row to readable time
        csv_writer.writerow(row)
connection.close()

This code returns the error:

TypeError: 'tuple' object does not support item assignment

I understand that I can't manipulate the data in a tuple but I am explicitly converting each row to a list. Can anyone explain a) why list (row) does not work, and b) a better way to do it?

Thanks in advance!

list does not change a value in place, it returns a new list. — 301_Moved_Permanently
– 301_Moved_Permanently, Commented Oct 1, 2015 at 9:15
list(row) returns a new list, which you throw away. row = list(row) would work. — Open AI - Opting Out
– Open AI - Opting Out, Commented Oct 1, 2015 at 9:16

shevron · Accepted Answer · 2015-10-01 09:20:09Z

2

I think that:

        list(row) #convert to list - does not work

Should be:

        row = list(row) #convert to list

In addition, you may want to look into using a csv.DictReader object instead of the default reader, as it may be more convenient (and is mutable, unlike the default tuple). See https://docs.python.org/2/library/csv.html#csv.DictReader for detauls.

answered Oct 1, 2015 at 9:20

shevron

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1 Comment

David White Over a year ago

Yep - that fixed it. Thank you.

ROBOT4321 · Accepted Answer · 2021-03-27 07:28:55Z

0

Another work around would be like this-->

    conn = sqlite3.connect('library.db')
    conn.row_factory = lambda cursor, row: row[0]
    c = conn.cursor()

    id_sql = 'SELECT emp_id FROM teachtable WHERE teach_name = "%s"' % (name1)
    id_no = c.execute(id_sql).fetchall()
    conn.commit()
    conn.close()
    # This list output can be used to iterate through it in a for loop
    
    # print the first element from the list
    idno = id_no[0]
    print(id_no)

Here's a reference link to understand how "row_factory" object is used to return the query output as a list!

edited Mar 27, 2021 at 7:28

answered Mar 27, 2021 at 7:22

ROBOT4321

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