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我想压缩目录中的文件并给它一个特定的名称(目标文件夹)。我想将源文件夹和目标文件夹作为输入传递给程序。在python中压缩目录或文件
但是,当我走过源文件路径它给我和错误。我想我会面对与目标文件路径相同的问题。
d:\SARFARAZ\Python>python zip.py
Enter source directry:D:\Sarfaraz\Python\Project_Euler
Traceback (most recent call last):
File "zip.py", line 17, in <module>
SrcPath = input("Enter source directry:")
File "<string>", line 1
D:\Sarfaraz\Python\Project_Euler
^
SyntaxError: invalid syntax
我写的代码如下:
import os
import zipfile
def zip(src, dst):
zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
abs_src = os.path.abspath(src)
for dirname, subdirs, files in os.walk(src):
for filename in files:
absname = os.path.abspath(os.path.join(dirname, filename))
arcname = absname[len(abs_src) + 1:]
print 'zipping %s as %s' % (os.path.join(dirname, filename),arcname)
zf.write(absname, arcname)
zf.close()
#zip("D:\\Sarfaraz\\Python\\Project_Euler", "C:\\Users\\md_sarfaraz\\Desktop")
SrcPath = input("Enter source directry:")
SrcPath = ("@'"+ str(SrcPath) +"'")
print SrcPath # checking source path
DestPath = input("Enter destination directry:")
DestPath = ("@'"+str(DestPath) +"'")
print DestPath
zip(SrcPath, DestPath)