使用 pymssql 写入 CSV 并使用列名作为标题答案

【问题标题】：Use pymssql to write to CSV and use column names as headers使用 pymssql 写入 CSV 并使用列名作为标题
【发布时间】：2017-07-20 17:52:24
【问题描述】：

我有以下脚本用于 i) 连接到数据库，ii) 运行查询，iii) 将查询结果保存为 csv 和 iv) 通过电子邮件发送输出。

一切正常，除了我似乎无法将列名作为文件的标题。我已经浏览了 SO，但找不到适合我的解决方案。我的脚本如下。任何帮助都会很棒：

import pymssql
import csv
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email import encoders
fromaddr='XXXX@gmail.com'
toaddr='XXXX@gmail.com'
msg = MIMEMultipart()
msg['From'] = fromaddr
msg['To'] = toaddr
msg['Subject'] = "This is a test"
body = "This is still a test"
conn = pymssql.connect(server='XXXXXXXXXX.net',
               port=XXXX,
               user='XXXX',
               password='XXXX',
               database='XXXX')
cursor = conn.cursor()
query = 'Select * From Table'
cursor.execute(query)
with open("XXXXXXXX.csv","w") as outfile:
    writer = csv.writer(outfile, quoting=csv.QUOTE_NONNUMERIC)
    for row in cursor:
        writer.writerow(row)
msg.attach(MIMEText(body,'plain'))
filename = "XXXXXX.csv"
attachment = open("XXXXXXXXXXX.csv","rb")
from email.mime.base import MIMEBase
part = MIMEBase('application','octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition',"attachment; filename= %s" % filename)
msg.attach(part)
text = msg.as_string()
server = smtplib.SMTP('smtp.office365.com',587)
server.starttls()
server.login(fromaddr,"XXXXXX")
text = msg.as_string()
server.sendmail(fromaddr,toaddr,text)

【问题讨论】：

标签： python csv export-to-csv pymssql

【解决方案1】：

根据DB-API -- 其中PyMSSQL complies -- 有一个名为.description 的游标属性，您可以在此处使用。

with open("XXXXXXXX.csv","w") as outfile:
    writer = csv.writer(outfile, quoting=csv.QUOTE_NONNUMERIC)
    writer.writerow(col[0] for col in cursor.description)
    for row in cursor:
        writer.writerow(row)

【讨论】：

您也可以用writer.writerows(cursor) 代替for 循环。

【解决方案2】：

使用cursor 对象上的.description 并循环遍历它以获取列的名称，例如：

colNameList = []
for i in range(len(cursor.description)):
    desc = cursor.description[i]
    colNameList.append(desc[0])

    colNames = ','.join(colNameList)
    print colNames

【讨论】：