Question

使用python zipfile模块我创建了一个zip文件，如下所示：

CREATE TABLE #myTable ( 
    Name VARCHAR(50), 
    Year INT, 
    Hours INT
) 

INSERT INTO #myTable *.... some values*

DECLARE @var INT 
SET @var = 2015

DECLARE @DynamicPivot NVARCHAR(MAX) 

SET @DynamicPivot = '
SELECT * 
FROM #myTable
PIVOT( 
SUM(Hours) FOR Year IN ([' + CAST(@var as nvarchar(10)) + '])
) AS PvtTable'

EXEC sp_executesql @DynamicPivot

DROP TABLE #myTable

现在，我希望将此s = StringIO() zip_file = zipfile.ZipFile(s, "w") zip_file.write('/local/my_files/my_file.txt') s.seek(0)保存在我的文件系统中，路径zip_file为/local/my_files/。一般来说，为了保存noraml文件，我使用了以下流程：

my_file.zip

但是，我认为我无法用这个来保存zipfile。任何人都可以帮我完成这件事。

Answer 1

zip_file = zipfile.ZipFile("/local/my_files/my_file.zip", "w")
zip_file.write('/local/my_files/my_file.txt')
zip_file.close()

ZipFile对象初始化的第一个参数是您要保存zip文件的路径。

Answer 2

如果您需要使用StringIO，请尝试以下代码：

from StringIO import StringIO
import zipfile

s = StringIO()
with zipfile.ZipFile(s, "w", compression=zipfile.ZIP_DEFLATED) as zf:
    zf.write('/local/my_files/my_file.txt')

with open('/local/my_files/my_file.zip', 'wb') as f_out:
    f_out.write(s.getvalue())

或者你可以用更简单的方式做到：

import zipfile

with zipfile.ZipFile("/local/my_files/my_file.zip", "w", compression=zipfile.ZIP_DEFLATED) as zf:
    zf.write("/local/my_files/my_file.txt")

如何在python中将创建的Zip文件保存到文件系统？

2 个答案: