ZipFileMember>>contentStream has leading characters

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.

What is the code that you use?

Doru

On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <[hidden email]> 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"

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 <[hidden email]> wrote:

What is the code that you use?

Doru

On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <[hidden email]> 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 about.me/sanchez.diego

| zipper file |
	zipper := ZipArchive new readFrom: 'tp1.zip'.

        "file contains test.st"
	file := zipper members at: 1.
	
	"file contentStream returns the stream with file's information uncompressed"

On 13 October 2014 08:22, Diego Sánchez <[hidden email]> wrote:

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 <[hidden email]> wrote:

What is the code that you use?

Doru

On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <[hidden email]> 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 about.me/sanchez.diego

--

Saludos

DiegoS

------

Diego Sanchez about.me/sanchez.diego

You are likely referring to the issue reported here:

https://pharo.fogbugz.com/f/cases/12826/Zip-FileSystem-does-not-work

This should work now in the latest Pharo 4.0. If you are using Pharo 3.0, loading the slice manually from the Pharo40Inbox should fix the issue as well.

What version are you using?

Cheers,

Doru

On Mon, Oct 13, 2014 at 1:23 PM, Diego Sánchez <[hidden email]> wrote:

| zipper file |
	zipper := ZipArchive new readFrom: 'tp1.zip'.

        "file contains test.st"
	file := zipper members at: 1.
	
	"file contentStream returns the stream with file's information uncompressed"

On 13 October 2014 08:22, Diego Sánchez <[hidden email]> wrote:

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 <[hidden email]> wrote:

What is the code that you use?

Doru

On Mon, Oct 13, 2014 at 1:31 AM, DiegoSanchez <[hidden email]> 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 about.me/sanchez.diego

--

Saludos

DiegoS

------

Diego Sanchez about.me/sanchez.diego

--

www.tudorgirba.com

"Every thing has its own flow"

---

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NAML

On 13/10/14 01:31, DiegoSanchez wrote:
> Hi everybody!  This is my first post

Welcome :)
In which image version are you?

> 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.
>
>

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NAML