Issues using iter_rows

[zakp...@gmail.com]

When trying to use iter_rows I get an error saying that "Worksheet object has no attribute iter_rows". I have the latest version of both openpyxl (1.5.8) and python (2.7.3) using office 2003 on windows 7. I was using the snippet of code below when the error cropped up:

from openpyxl.reader.excel import load_workbook

from openpyxl.workbook import Workbook

import time, sys

***other code***

col_name = 'A'
start_row = 1
end_row = 99

range_expr = "{col}{start_row}:{col}{end_row}".format(
    col=col_name, start_row=start_row, end_row=end_row)

for (time_cell,) in ws.iter_rows(range_string=range_expr):

    print time_cell.value.hour 

There isn't anything extra I have to import for this to work is there? 

[Eric Gazoni]

Don't forget to load_workbook with iterators support :

workbook = load_workbook('/path/to/file', use_iterators=True)

Cheers,
Eric

Le 21/06/12 16:07, zakp...@gmail.com a écrit :

[Zak]

Ah, something so simple, thanks.

Another question, while I have an active thread, how would I go about getting values from a column? I know using iter_rows like this will pass through a column until the end of the file. My code as following is supposed to pass through column C and pull out integer values until there is no more data: 

   address = []

   col_name = 'C'

   start_row = 4

   end_row = ws.get_highest_row()+1

   range_expr = "{col}{start_row}:{col}{end_row}".format(col=col_name, start_row=start_row, end_row=end_row)

   for cell in ws.iter_rows(range_string=range_expr):

      if str(cell.value).isdigit:

         print 'it lives!!'

      else:

         print 'it doesnt!!'

   return address

But crashes at the if statement. saying that 'tuple' object has no attribute 'value'. What makes no sense to me is that the documentation uses cell.value throughout. I have a feeling im missing something terribly simple again.

[Eric Gazoni]

iter_rows returns a tuple of cell, one per row.

if you have a range that is 1-col wide, then you can do

for row in ws.iter_rows(...):
    cell = row[0]
    if str(cell.value)......

Cheers,
Eric

Le 21/06/12 16:35, Zak a écrit :

[Zak]

Hm that doesnt seem to work, it gives "IndexError: tuple index out of range" at the line:

cell = row[0] 

[Eric Gazoni]

that means your iter_rows is not retrieving anything, try to print "row" and see if that looks like what you expect.

Le 21/06/12 16:48, Zak a �crit�:

Hm that doesnt seem to work, it gives "IndexError: tuple index out of range" at the line:

cell = row[0]�

[Zak]

Hmm, printing the row (even after manually changing where the program is going in the spreadsheet) always shows an empty set of parens -> ()

[Zak]

is there any reason that iter_rows wouldnt be returning anything??

[Zak]

found a fix, all i needed was:

range_expr = 'C:C'

for row in ws.iter_rows(row_offset=4):

            '''code'''