python shutil: only copy part of the file -

the code read xls file directory, convert csv file , copy directory.

filepath = os.path.join('.', 'attachments') filepaths = [f f in os.listdir(filepath) if os.path.isfile(os.path.join(filepath, f)) , f.endswith('.xls')]  f in filepaths:     wb = xlrd.open_workbook(os.path.join(filepath, f))     sheet = wb.sheet_by_index(0)     filename = f + '.csv'     fp = open(os.path.join(filepath, filename), 'wb')     wr = csv.writer(fp, quoting=csv.quote_all)     rownum in xrange(sheet.nrows):        wr.writerow(sheet.row_values(rownum))     fp.close      shutil.copy(os.path.join('.', 'attachments', filename), new_directory) 

the result is: xls file converted csv file, in new_directory, copied file contains part of csv file.

for example, original csv file has 30 rows, in copied file, there 17 rows. idea of why happen?

here's problem:

fp.close 

you need call close method, not reference method. so:

fp.close() 

however, make life easier if use with statements instead of trying figure out explicitly close everything:

with open(os.path.join(filepath, filename), 'wb') fp:     wr = csv.writer(fp, quoting=csv.quote_all)     rownum in xrange(sheet.nrows):         wr.writerow(sheet.row_values(rownum))