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 <tu...@tudorgirba.com> wrote:
> What is the code that you use?
>
> Doru
>
> On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <diego.sanc...@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
>>
>>
>>
>> --
>> View this message in context:
>> http://forum.world.st/ZipFileMember-contentStream-has-leading-characters-tp4784238.html
>> Sent from the Pharo Smalltalk Users mailing list archive at Nabble.com.
>>
>>
>
>
> --
> www.tudorgirba.com
>
> "Every thing has its own flow"
>
--
Saludos
DiegoS
------
Diego Sanchez
[image: http://]about.me/sanchez.diego
<http://about.me/sanchez.diego?promo=email_sig>
