Re: Script to make Windows XP-readable ZIP file
softwindow [EMAIL PROTECTED] wrote: Carl Banks is right Did he write to check out: http://groups.google.com/support/bin/answer.py?answer=14213 ? Why didn't you do so? -- John MexIT: http://johnbokma.com/mexit/ personal page: http://johnbokma.com/ Experienced programmer available: http://castleamber.com/ Happy Customers: http://castleamber.com/testimonials.html -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
On 2006-05-19, softwindow [EMAIL PROTECTED] wrote: Carl Banks is right That would be valuable information if we know what he was right about. -- Grant Edwards grante Yow! Bo Derek ruined at my life! visi.com -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
pac wrote:
I'm preparing to distribute a Windows XP Python program and some
ancillary files,
and I wanted to put everything in a .ZIP archive. It proved to be
inordinately
difficult and I thought I would post my solution here. Is there a
better one?
Suppose you have a set of files in a directory c:\a\b and some
additional
files in c:\a\b\subdir. Using a Python script, you would
like to make a Windows-readable archive (.zip) that preserves this
directory structure, and where the root directory of the archive is
c:\a\b. In other words, all the files from c:\a\b appear in the
archive
without a path prefix, and all the files in c:\a\b\subdir have a path
prefix of \subdir.
This looks like it should be easy with module zipfile and the handy
function os.walk. Create a zip file, call os.walk, and add the files
to
the archive like so:
import os
import zipfile
z =
zipfile.ZipFile(rc:\a\b\myzip.zip,mode=w,compression=zipfile.ZIP_DEFLATED)
for dirpath,dirs,files in os.walk(rc:\a\b):
for a_file in files:
a_path = os.path.join(dirpath,a_file)
z.write(a_path) # Change, see below
z.close()
This creates an archive that can be read by WinZip or by another Python
script
that uses zipfile. But when you try to view it with the Windows
compressed folder
viewer it will appear empty. If you try to extract the files anyway
(because
you know they are really there), you get a Windows Security Warning and
XP
refuses to decompress the folder - XP is apparently afraid it might be
bird flu
or something.
If you change the line marked #Change to z.write(a_path,file),
explicitly naming
each file, now the compressed folder viewer will show all the files in
the archive.
XP will not treat it like a virus and it will extract the files.
However, the
archive does not contain a subdirectory; all the files are in a single
directory.
Some experimentation suggests that Windows does not like any filename
in the
archive that begins with either a drive designator like c:, or has a
path containing
a leading slash like \a\b\afile.txt. Relative paths like
subdir\afile.txt are
okay, and cause the desired behavior when the archive is extracted,
e.g., a new directory subdir is created and afile.txt is placed in it.
Since the method ZipFile.write needs a valid pathname for each file,
the correct
solution to the original problem entails messing around with the OS's
current
working directory. Position the CWD in the desired base directory of
the archive,
add the files to the archive using their relative pathnames, and put
the CWD back
where it was when you started:
import os
import zipfile
z =
zipfile.ZipFile(rc:\a\b\myzip.zip,mode=w,compression=zipfile.ZIP_DEFLATED)
cwd = os.getcwd()
os.chdir(base_dir)
try:
for dirpath,dirs,files in os.walk(''): # This starts the walk at
the CWD
for a_file in files:
a_path = os.path.join(dirpath,a_file)
z.write(a_path,a_path) # Can the second argument be
omitted?
z.close()
finally:
os.chdir(cwd)
This produces an archive that can be extracted by Windows XP using its
built-in
capability, by WinZip, or by another Python script. Now that I have
the solution it
seems to make sense, but it wasn't at all obvious when I started.
Paul Cornelius
Others have addressed your specific question, I wanted to make a
more general suggestion.
You should really take a look at Inno Installer and py2exe combination
for creating .zip library and Windows distribution. I PROMISE it will
be worth your time on future projects. Rolling your own installer will
take much more time/effort over the long haul.
-Larry Bates
--
http://mail.python.org/mailman/listinfo/python-list
Script to make Windows XP-readable ZIP file
I'm preparing to distribute a Windows XP Python program and some
ancillary files,
and I wanted to put everything in a .ZIP archive. It proved to be
inordinately
difficult and I thought I would post my solution here. Is there a
better one?
Suppose you have a set of files in a directory c:\a\b and some
additional
files in c:\a\b\subdir. Using a Python script, you would
like to make a Windows-readable archive (.zip) that preserves this
directory structure, and where the root directory of the archive is
c:\a\b. In other words, all the files from c:\a\b appear in the
archive
without a path prefix, and all the files in c:\a\b\subdir have a path
prefix of \subdir.
This looks like it should be easy with module zipfile and the handy
function os.walk. Create a zip file, call os.walk, and add the files
to
the archive like so:
import os
import zipfile
z =
zipfile.ZipFile(rc:\a\b\myzip.zip,mode=w,compression=zipfile.ZIP_DEFLATED)
for dirpath,dirs,files in os.walk(rc:\a\b):
for a_file in files:
a_path = os.path.join(dirpath,a_file)
z.write(a_path) # Change, see below
z.close()
This creates an archive that can be read by WinZip or by another Python
script
that uses zipfile. But when you try to view it with the Windows
compressed folder
viewer it will appear empty. If you try to extract the files anyway
(because
you know they are really there), you get a Windows Security Warning and
XP
refuses to decompress the folder - XP is apparently afraid it might be
bird flu
or something.
If you change the line marked #Change to z.write(a_path,file),
explicitly naming
each file, now the compressed folder viewer will show all the files in
the archive.
XP will not treat it like a virus and it will extract the files.
However, the
archive does not contain a subdirectory; all the files are in a single
directory.
Some experimentation suggests that Windows does not like any filename
in the
archive that begins with either a drive designator like c:, or has a
path containing
a leading slash like \a\b\afile.txt. Relative paths like
subdir\afile.txt are
okay, and cause the desired behavior when the archive is extracted,
e.g., a new directory subdir is created and afile.txt is placed in it.
Since the method ZipFile.write needs a valid pathname for each file,
the correct
solution to the original problem entails messing around with the OS's
current
working directory. Position the CWD in the desired base directory of
the archive,
add the files to the archive using their relative pathnames, and put
the CWD back
where it was when you started:
import os
import zipfile
z =
zipfile.ZipFile(rc:\a\b\myzip.zip,mode=w,compression=zipfile.ZIP_DEFLATED)
cwd = os.getcwd()
os.chdir(base_dir)
try:
for dirpath,dirs,files in os.walk(''): # This starts the walk at
the CWD
for a_file in files:
a_path = os.path.join(dirpath,a_file)
z.write(a_path,a_path) # Can the second argument be
omitted?
z.close()
finally:
os.chdir(cwd)
This produces an archive that can be extracted by Windows XP using its
built-in
capability, by WinZip, or by another Python script. Now that I have
the solution it
seems to make sense, but it wasn't at all obvious when I started.
Paul Cornelius
--
http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
i am in win2000 z.write(a_path,a_path) may change to z.write(a_path) but the dirpath is not in zipfile who can tell me? -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
pac wrote: Suppose you have a set of files in a directory c:\a\b and some additional files in c:\a\b\subdir. Using a Python script, you would like to make a Windows-readable archive (.zip) that preserves this directory structure, and where the root directory of the archive is c:\a\b. In other words, all the files from c:\a\b appear in the archive without a path prefix, and all the files in c:\a\b\subdir have a path prefix of \subdir. This looks like it should be easy with module zipfile and the handy function os.walk. Create a zip file, call os.walk, and add the files to the archive like so: import os import zipfile z = zipfile.ZipFile(rc:\a\b\myzip.zip,mode=w,compression=zipfile.ZIP_DEFLATED) for dirpath,dirs,files in os.walk(rc:\a\b): for a_file in files: a_path = os.path.join(dirpath,a_file) z.write(a_path) # Change, see below z.close() (Aside: be careful not to use tabs when posting. I suspect the f-bot will be here to tell you that the above code doesn't work.) Some experimentation suggests that Windows does not like any filename in the archive that begins with either a drive designator like c:, or has a path containing a leading slash like \a\b\afile.txt. Relative paths like subdir\afile.txt are okay, and cause the desired behavior when the archive is extracted, e.g., a new directory subdir is created and afile.txt is placed in it. Since the method ZipFile.write needs a valid pathname for each file, the correct solution to the original problem entails messing around with the OS's current working directory. ZipFile.write takes an optional second argument for the archive filename. You could have done something like this (untested): for dirpath,dirs,files in os.walk(rc:\a\b): for a_file in files: a_path = os.path.join(dirpath,a_file) z_path = a_path[7:] # or whatever z.write(a_path,z_path) z.close() And maybe use a little helper function instead of the string slice to make it more robust (it violates DRY, and I'm not happy to assume the dirpath returned by os.walk has exactly the same prefix as the argument). Position the CWD in the desired base directory of the archive, add the files to the archive using their relative pathnames, and put the CWD back where it was when you started: This may be the best way anyways, unless you have some reason to not change the current directory. Carl Banks -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
pac [EMAIL PROTECTED] wrote: I'm preparing to distribute a Windows XP Python program and some ancillary files, and I wanted to put everything in a .ZIP archive. It proved to be inordinately difficult and I thought I would post my solution here. Is there a better one? heresy maybe, but I use Ant to do such things (with my Perl projects): http://ant.apache.org/ -- John MexIT: http://johnbokma.com/mexit/ personal page: http://johnbokma.com/ Experienced programmer available: http://castleamber.com/ Happy Customers: http://castleamber.com/testimonials.html -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
my code can work, like below: import os import zipfile z = zipfile.ZipFile(rc:\text.zip,mode=w,compression=zipfile.ZIP_DEFLATED) cwd = os.getcwd() try: for dirpath,dirs,files in os.walk(cwd): for file in files: z_path = os.path.join(dirpath,file) z.write(z_path) z.close() finally: if z: z.close() that is true but the archive include the absolute path . can you give me a way to build it with relative path. -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
aha we want to do it with python don't use ant -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
softwindow [EMAIL PROTECTED] wrote: aha we want to do it with python don't use ant :-D I want to do a lot with Perl, but sometimes it's better to use the right tool for the job. And I think that Ant is better at for what I use it compared to a home brew tool I could make. Be careful with Not Invented Here. -- John MexIT: http://johnbokma.com/mexit/ personal page: http://johnbokma.com/ Experienced programmer available: http://castleamber.com/ Happy Customers: http://castleamber.com/testimonials.html -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
Carl Banks is right -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
aha now it's right like this: import os import zipfile z = zipfile.ZipFile(rc:\text.zip,mode=w,compression=zipfile.ZIP_DEFLATED) cwd = os.getcwd() try: for dirpath,dirs,files in os.walk(cwd): for file in files: z_path = os.path.join(dirpath,file) start = cwd.rfind(os.sep)+1 z.write(z_path,z_path[start:]) z.close() finally: if z: z.close() -- http://mail.python.org/mailman/listinfo/python-list
Re: Script to make Windows XP-readable ZIP file
import os import zipfile z = zipfile.ZipFile(rc:\text.zip,mode=w,compression=zipfile.ZIP_DEFLATED) cwd = os.getcwd() try: for dirpath,dirs,files in os.walk(cwd): for file in files: z_path = os.path.join(dirpath,file) start = cwd.rfind(os.sep)+1 z.write(z_path,z_path[start:]) z.close() finally: if z: z.close() * can work -- http://mail.python.org/mailman/listinfo/python-list
