python脚本实现xls(xlsx)转成csv

WBOY
Release: 2016-06-10 15:05:25
Original
2006 people have browsed it

# xls_csv

把xls,xlsx格式的文档转换成csv格式

# 使用
python xls2csv.py

# -*- coding: utf-8 -*-
import xlrd
import xlwt
import sys
from datetime import date,datetime
 
def read_excel(filename):
 
  workbook = xlrd.open_workbook(filename)
  # print sheet2.name,sheet2.nrows,sheet2.ncols
  sheet2 = workbook.sheet_by_index(0)
  
  for row in xrange(0, sheet2.nrows):
    rows = sheet2.row_values(row)
    def _tostr(cell):
      if type(u'') == type(cell): 
        return "\"%s\"" % cell.encode('utf8')
      else:
        return "\"%s\"" % str(cell) 
  
    print ','.join([_tostr(cell) for cell in rows ])
  
if __name__ == '__main__':
  filename = sys.argv[1]
  read_excel(filename)
Copy after login

再给大家分享一则代码

xlsx文件解析处理:openpyxl库 csv文件格式生成:csv

python#coding: utf-8
# 依赖openpyxl库:http://openpyxl.readthedocs.org/en/latest/

from openpyxl import Workbook
from openpyxl.compat import range
from openpyxl.cell import get_column_letter
from openpyxl import load_workbook
import csv
import os
import sys

def xlsx2csv(filename):
try:
 xlsx_file_reader = load_workbook(filename=filename)
 for sheet in xlsx_file_reader.get_sheet_names():
 # 每个sheet输出到一个csv文件中,文件名用xlsx文件名和sheet名用'_'连接
 csv_filename = '{xlsx}_{sheet}.csv'.format(
 xlsx=os.path.splitext(filename.replace(' ', '_'))[0],
 sheet=sheet.replace(' ', '_'))

 csv_file = file(csv_filename, 'wb')
 csv_file_writer = csv.writer(csv_file)

 sheet_ranges = xlsx_file_reader[sheet]
 for row in sheet_ranges.rows:
 row_container = []
 for cell in row:
 if type(cell.value) == unicode:
row_container.append(cell.value.encode('utf-8'))
else:
row_container.append(str(cell.value))
csv_file_writer.writerow(row_container)
csv_file.close()

 except Exception as e:
print(e)

if __name__ == '__main__':
 if len(sys.argv) != 2:
 print('usage: xlsx2csv <xlsx file name>')
else:
xlsx2csv(sys.argv[1])
sys.exit(0)
Copy after login

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template