doitFromStream
	| zipper file |
	zipper := ZipArchive new readFrom: '/Users/diego/tp1-95195.zip'.
	file := zipper members at: 2.
	
	"file contentStream returns the stream with file's information uncompressed"

On 13 October 2014 08:15, Tudor Girba <tudor@tudorgirba.com> wrote:
What is the code that you use?

Doru

On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <diego.sanches@gmail.com> wrote:
Hi everybody!� This is my first post and I'm posting this question after
going around this forum without finding answer for my question.� I couldn't
wrap my mind around it.

Here is the issue (in fact it is not a issue, classes work fine).

I want to uncompress a zip file and file in the content of one of the file
included in zip file.� For instance:
The file 'tp1.zip' contains just one file inside it: 'test.st'.

Once I uncompressed it, using code bellow:



 I found that file stream has a set of leading characters which cannot be
 interpreted by *CodeInterpreter*.

Here you will fine these leading characters:

 /Hexa characters/
*ef� bb� bf *

/Ascii characters/
*ef� bb� bf *

 Indeed, this leading characters are written by the fileOut process and it's
ok (those are magic number which depend on the SO) but I don't know how to
skip it for CodeInterpreter to be able fileIn this file.

My first solution was skipping this characters through the use of
 /Stream>>next: 4/. But I'm looking for a much portable solution.

Somebody went through this problem?

Thanks for help in advance.
DiegoS



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--

"Every thing has its own flow"



--
Saludos
DiegoS

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