python做工资
好开心啊，在旁边大神的帮助下，终于实现了发工资的python脚本，之前用shell写的老出错，刚才测试，发80个人工资详情，妥妥的，代码如下：
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
from email.header import Header
#from handle_data_for_salary import main
#import handle_data_for_salary
import smtplib
import xlrd
import xdrlib,sys
def open_excel(file='test1.xls'):
try:
data = xlrd.open_workbook(file)
return data
except Exception as e:
print(str(e))
def excel_table_byname(file='test1.xls',colnameindex=0,by_name='Sheet1'):
data = open_excel(file)
table = data.sheet_by_name(by_name)
nrows = table.nrows #行数
colnames = table.row_values(colnameindex) #某一行数据
list = []
for rownum in range(2,nrows):
row = table.row_values(rownum)
if row:
app = {}
for i in range(len(colnames)):
app[colnames[i]] = row[i]
list.append(app)
return list
def get_value():
tables = excel_table_byname()
for row in tables:
global mail
mail = row["邮箱地址"]
global name
name = row["姓名"]
global context
authInfo = {}
authInfo['server'] = 'smtp.163.com'
authInfo['user'] = 'footprints19940807@163.com'
authInfo['password'] = '邮箱密码'
fromAdd = 'footprints19940807@163.com'
subject = 'Email Subject'
context = str(row)
toAdd = [mail, ]
sendEmail(authInfo, fromAdd, toAdd, subject, context)
def sendEmail(authInfo, fromAdd, toAdd, subject, plainText):
strFrom = fromAdd
strTo = '; '.join(toAdd)
server = authInfo.get('server')
smtpPort = 25
sslPort = 465
user = authInfo.get('user')
passwd = authInfo.get('password')
if not (server and user and passwd) :
print('incomplete login info, exit now')
return
# 设定root信息
msgRoot = MIMEMultipart('related')
msgRoot['Subject'] = subject
msgRoot['From'] = '%s<%s>' % (Header('工资详情', 'utf-8'), strFrom)
msgRoot['To'] = strTo
# 邮件正文内容
msgText = MIMEText(plainText, 'plain', 'utf-8')
msgRoot.attach(msgText)
msgAlternative = MIMEMultipart('alternative')
msgRoot.attach(msgAlternative)
try:
#发送邮件
smtp = smtplib.SMTP()
smtp.connect(server, smtpPort)
#ssl加密方式，通信过程加密，邮件数据安全
# smtp = smtplib.SMTP_SSL()
# smtp.connect(server, sslPort)
#设定调试级别，依情况而定
# smtp.set_debuglevel(1)
smtp.login(user, passwd)
smtp.sendmail(strFrom, toAdd, msgRoot.as_string())
smtp.quit()
print("邮件发送给%s成功!"%(name))
except Exception as e:
print("失败：" + str(e))
if __name__ == '__main__' :
get_value()
注意：python环境为3.5，实验环境在windows，使用工具为pycharm，需要用到的模块:pip3 install xlrd，其中的工资表，如下图截图部分所示：后面还有很多列，看自己的需求加不，反正都是一列一列的数据
实验结果如下：
现在只是实现了需求，发的邮件多了估计有问题，还有邮件的格式太丑了，等下回家再改改，加油加油
python做工资
好开心啊，在旁边大神的帮助下，终于实现了发工资的python脚本，之前用shell写的老出错，刚才测试，发80个人工资详情，妥妥的，代码如下：
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
from email.header import Header
#from handle_data_for_salary import main
#import handle_data_for_salary
import smtplib
import xlrd
import xdrlib,sys
def open_excel(file='test1.xls'):
try:
data = xlrd.open_workbook(file)
return data
except Exception as e:
print(str(e))
def excel_table_byname(file='test1.xls',colnameindex=0,by_name='Sheet1'):
data = open_excel(file)
table = data.sheet_by_name(by_name)
nrows = table.nrows #行数
colnames = table.row_values(colnameindex) #某一行数据
list = []
for rownum in range(2,nrows):
row = table.row_values(rownum)
if row:
app = {}
for i in range(len(colnames)):
app[colnames[i]] = row[i]
list.append(app)
return list
def get_value():
tables = excel_table_byname()
for row in tables:
global mail
mail = row["邮箱地址"]
global name
name = row["姓名"]
global context
authInfo = {}
authInfo['server'] = 'smtp.163.com'
authInfo['user'] = 'footprints19940807@163.com'
authInfo['password'] = '邮箱密码'
fromAdd = 'footprints19940807@163.com'
subject = 'Email Subject'
context = str(row)
toAdd = [mail, ]
sendEmail(authInfo, fromAdd, toAdd, subject, context)
def sendEmail(authInfo, fromAdd, toAdd, subject, plainText):
strFrom = fromAdd
strTo = '; '.join(toAdd)
server = authInfo.get('server')
smtpPort = 25
sslPort = 465
user = authInfo.get('user')
passwd = authInfo.get('password')
if not (server and user and passwd) :
print('incomplete login info, exit now')
return
# 设定root信息
msgRoot = MIMEMultipart('related')
msgRoot['Subject'] = subject
msgRoot['From'] = '%s<%s>' % (Header('工资详情', 'utf-8'), strFrom)
msgRoot['To'] = strTo
# 邮件正文内容
msgText = MIMEText(plainText, 'plain', 'utf-8')
msgRoot.attach(msgText)
msgAlternative = MIMEMultipart('alternative')
msgRoot.attach(msgAlternative)
try:
#发送邮件
smtp = smtplib.SMTP()
smtp.connect(server, smtpPort)
#ssl加密方式，通信过程加密，邮件数据安全
# smtp = smtplib.SMTP_SSL()
# smtp.connect(server, sslPort)
#设定调试级别，依情况而定
# smtp.set_debuglevel(1)
smtp.login(user, passwd)
smtp.sendmail(strFrom, toAdd, msgRoot.as_string())
smtp.quit()
print("邮件发送给%s成功!"%(name))
except Exception as e:
print("失败：" + str(e))
if __name__ == '__main__' :
get_value()
注意：python环境为3.5，实验环境在windows，使用工具为pycharm，需要用到的模块:pip3 install xlrd，其中的工资表，如下图截图部分所示：后面还有很多列，看自己的需求加不，反正都是一列一列的数据
实验结果如下：
现在只是实现了需求，发的邮件多了估计有问题，还有邮件的格式太丑了，等下回家再改改，加油加油