[Tutor] copy zip file from source folder to destination and unzipall files within destination

[Alan Gauld]

"MARCUS NG" <markersng at gmail.com> wrote

> Currently nothing works due to my limited understanding.

That's a pretty vague description of the problem.
What exactly happens?
Does the file get copied?
Do you get error messages? If so what?

> I am wondering if my syntax is wrong or am I missing anything?

If the syntax is wrong Python should tell you.
Does it?

The more information you give us the better able we
are to diagnose the problem.

> the code is as such. also if there is a better code, I am definitely open 
> to
> it. If it is good, can you also explain steps?

Without any more info I'll pass along a couple of observations.

> ###########
>
> import shutil, zipfile
> shutil.copyfile('C:\Users\blueman\Desktop\myTest1.0.zip',
> 'C:\Users\blueman\Desktop\allFiles')

The first thing is the use of DOS style path names.
You need to do one of the following:

1) Use raw strings - prefix the string with an r
 r'C:\Users\blueman\Desktop\allFiles')

2) Escape the backslashes
 'C:\\Users\\blueman\\Desktop\\allFiles')

3) Use Unix style slashes:
 'C:/Users/blueman/Desktop/allFiles')

Otherwise Python sees the \ as an escape character
and tries to interpret \U, \b \D and \a as special characters
like \n etc

> zipfilepath='C:\Users\blueman\Desktop\allFiles\myTest1.0.zip'
> extractiondir='C:\Users\blueman\Desktop\allFiles\test'

Same here

> def extract(zipfilepath, extractiondir):
>    zip = zipfile(zipfilepath)
>    zip.extractall(path=extractiondir)
>    print 'it works'

While puuting things in functions is usuually a good idea, in this case
you only call it once so I wouldn't bother - at least not until I had the
program working

> extract()

You don't pas any arguments but you defined the functin to
take 2 - Python should tell you about this.

HTH,

-- 
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/