Could zipfile module process the zip data in memory?

[Diez B. Roggisch]

人言落日是天涯，望极天涯不见家 schrieb:
> On Apr 29, 7:37 pm, "Daniel Nogradi" <nogr... at gmail.com> wrote:
>>> I made a C/S network program, the client receive the zip file from the
>>> server, and read the data into a variable. how could I process the
>>> zipfile directly without saving it into file.
>>> In the document of the zipfile module, I note that it mentions the
>>> file-like object? what does it mean?
>>> class ZipFile( file[, mode[, compression[, allowZip64]]])
>>>          Open a ZIP file, where file can be either a path to a file (a
>>> string) or a file-like object.
>> Yes it is possible to process the content of the zipfile without
>> saving every file:
>>
>> [untested]
>>
>>         from zipfile import ZipFile
>>         from StringIO import StringIO
>>
>>         zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' )
>>         for name in zipp.namelist( ):
>>                 content = zipp.read( name )
>>                 s = StringIO( )
>>                 s.write( content )
>>                 # now the file 'name' is in 's' (in memory)
>>                 # you can process it further
>>                 # ............
>>                 s.close( )
>>         zipp.close( )
>>
>> HTH,
>> Daniel
> Thanks!
> Maybe my poor english makes you confusion:-). The client receive the
> zip file data from the server, and keep these data as a variable, not
> as a file in harddisk. such as "zipFileData", but the first argument
> of the "ZipFile" is filename. I would like to let the ZipFile() open
> the file from "zipFileData" directly but not file in harddisk
> 
> zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' )
>                               ^ I don't have this file, all its data
> is in a variable.

You can use cStringIO for that as well. Read the module docs for it.

Diez