Zipping Directory or file(s) in python - python-2.7

Question

I'm trying to zip files present in a directory and give a specific name (destination folder) to it. I want to pass the source and destination folders as input to the program.
But when ever I'm passeing the source file path it's giving me and error. I guess I'll face the same problem with the destination file path.
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
The code I've written is as follow:
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)

Answer 1

i have made some changes to your code as follows:
import os
import zipfile
def zip(src, dst):
zf = zipfile.ZipFile(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()
# Changed from input() to raw_input()
SrcPath = raw_input("Enter source directory: ")
print SrcPath # checking source path
# done the same and also added the option of specifying the name of Zipped file.
DestZipFileName = raw_input("Enter destination Zip File Name: ") + ".zip" # i.e. test.zip
DestPathName = raw_input("Enter destination directory: ")
# Here added "\\" to make sure the zipped file will be placed in the specified directory.
# i.e. C:\\Users\\md_sarfaraz\\Desktop\\
# i.e. double \\ to escape the backlash character.
DestPath = DestPathName + "\\" + DestZipFileName
print DestPath # Checking Destination Zip File name & Path
zip(SrcPath, DestPath)
Good Luck!