列表内的单个逗号分隔的元组

Question

Using pyodbc I'm working on dealing with executing stored procedures with outputs.

import pyodbc
conn = pyodbc.connect('DRIVER=' + driver + ';PORT=1433;SERVER=' + server +
                      ';PORT=1443;DATABASE=' + database + ';UID=' + username + ';PWD=' + password)
cmd = conn.cursor()
sql = """\
Too long to post
"""
params = ('EOXH39020220', 'EOXH39020245', 'EOXH3E360011')
for item in params:
    cmd.execute(sql, item)
    rst = cmd.fetchall()
    print(rst)
conn.close

The output of a particular stored procedure I am testing with produces the following:

[(False, 1, 3, 2, 967, 6, 'ABC-DE-FGHI', 'LOREM', 'IPSUM', 'CLASS PRODUCT', 'F/P', 'LABELDESCRIPTION', 'N/A', 'DESCRIPTION', 'ASL66', 'ASL10', '3FE50712BA', 'ABC-DE-FGHI', 'PO', 'ITEM#', 'OEM#', 'F/P', '1')]
[(False, 1, 4, 1, 967, 7, 'ABC-DE-FGHI', 'LOREM', 'IPSUM', 'CLASS PRODUCT', 'F/P', 'LABELDESCRIPTION', 'N/A', 'DESCRIPTION', 'ASL66', 'ASL10', '3FE50712BA', 'ABC-DE-FGHI', 'PO', 'ITEM#', 'OEM#', 'F/P', '1')]
[(False, 1, 4, 2, 967, 8, 'ABC-DE-FGHI', 'LOREM', 'IPSUM', 'CLASS PRODUCT', 'F/P', 'LABELDESCRIPTION', 'N/A', 'DESCRIPTION', 'ASL66', 'ASL10', '3FE50712BA', 'ABC-DE-FGHI', 'PO', 'ITEM#', 'OEM#', 'F/P', '1')]

I can't use .split on it since it's not really a string. How can I split up the contents of a tuple inside of a list? What I am trying to do is break the results back out into variables to use. I'm not getting the results I expect to see. Ex:

testlist = [(False, 1, 3, 2, 967, 6, 'ABC-DE-FGHI', 'LOREM', 'IPSUM', 'CLASS 1 PRODUCT', 'F/P', 'DESCRIPTION', 'N/A', 'DESCRIPTION', 'ASL66', 'ASL10', '3FE50712BA', 'ABC-DE-FGHI', 'PO', 'ITEM#', 'OEM#', 'F/P', '1')]
for i in testlist[0]:
    print(testlist[0][i])

results:

False 1 2 3

As suggested by roganjosh, I completely missed what I needed in my loop. first 5 items ex:

testlist = [(False, 1, 3, 2, 967, 6, 'ABC-DE-FGHI', 'LOREM', 'IPSUM', 'CLASS PRODUCT', 'F/P', 'LABELDESCRIPTION', 'N/A', 'DESCRIPTION', 'ASL66', 'ASL10', '3FE50712BA', 'ABC-DE-FGHI', 'PO', 'ITEM#', 'OEM#', 'F/P', '1')]

newlist = []

for x in testlist[0]:
    newlist.append(x)

print(newlist[:5])

[False, 1, 3, 2, 967]

Answer 1

The question is a bit confused but we established the issue in comments. Your title is mostly accurate in what's being returned from fetchall() : a list containing a single tuple, not a dictionary as you go on to state.

fetchall() is generally used when you expect multiple matches, where each row would be a tuple in the list (so you might consider using fetchone() as stated by @Barmar in the comments if you're always getting one result).

In the comments, you state that you tried the following but it didn't work:

for i in testlist[0]: 
    print(testlist[0][i])

It won't - testlist[0] gives you access to the tuple, but then you try to access that as though it were a dictionary with i as the key. Instead, i is the name that is being assigned to each item in that tuple.

The solution is as simple as doing:

for i in testlist[0]: 
    print(i)