python to parse excel file csv format

[Jay Jesus Amorin]

This is what i have done and its giving me error.

#!/usr/bin/env python

import csv, sys, os
filename = (sys.argv[1])
reader = csv.reader(open(filename, "rb"), delimiter=',',
quoting=csv.QUOTE_NONE)

try:
    #for row in reader:
    for fmodes,fname in reader:
        os.chmod(fname,fmodes)

except csv.Error, e:
    sys.exit('file %s, line %d: %s' % (filename, reader.line_num, e))


testserver:~> ./parsecsv.py chown.csv
Traceback (most recent call last):
  File "./promote2prod.py", line 10, in ?
    os.chmod(fname,fmodes)
TypeError: an integer is required


Please help I'm a complete newbie to python.

Many thanks.


Jay


On Thu, Dec 4, 2008 at 4:32 AM, Tino Wildenhain <tino at wildenhain.de> wrote:

> MRAB wrote:
>
>> Jay Jesus Amorin wrote:
>>
>>> This is how i do it, but it runs with error. Kindly help
>>>
>>>
>>> #!/usr/bin/env python
>>>
>>> import csv, sys, os
>>> filename = (sys.argv[1])
>>> reader = csv.reader(open(filename, "rb"), delimiter=',',
>>> quoting=csv.QUOTE_NONE)
>>>
>>> try:
>>>        for row in reader:
>>>                os.popen("chown row[0] row[1]")
>>>
>> This should be:
>>
>>    os.popen("chown %s %s" % (row[0], row[1]))
>>
>> or:
>>
>>    os.popen("chown %s %s" % tuple(row))
>>
>
> No, it should really be os.popen(("chown",row[0],row[1]))
> or better yet,
>
> for fmodes,fname in reader:
>    os.popen(("chown",fmodes,fname))
>
> or even plus better:
>
> for fmodes,fname in reader:
>    os.chmod(fname,fmodes)
>
> (Both my examples avoid problems with unquoted filenames)
>
> Regards
> Tino
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>
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