[delphizip] Re: error 11002 Zip file could not be opened

  • From: "Edwin" <mindvisualizer@xxxxxxxxx>
  • To: "delphizip@xxxxxxxxxxxxx" <delphizip@xxxxxxxxxxxxx>
  • Date: Thu, 28 Jun 2007 11:52:32 +0800

Hi Peters,
what do you mean by these lines?:
------------------------------------------------------------------------------------------------
but=20
aZip.ZipFileName :=3D ???    // set to name of zip file
------------------------------------------------------------------------------------------------

According to the most recently investigation, it maybe related to unicode, the 
value(a path, I don't know yet and waiting for the user's reply) assigned to 
ZipFileName is a WideString, my solution for now is to use 
Tntsystem.WideStringToStringEx to convert it into string before assigning to 
ZipFileName and see this error exists or not.




Best Regards,
Edwin
http://www.innovationgear.com
MindVisualizer -- Productive Visual Mind Mapping Software 
2007-06-28



Sender£º RPeters
Sent£º 2007-06-28 09:49:16
Recipient£º delphizip@xxxxxxxxxxxxx
CC£º 
Subject£º [delphizip] Re: error 11002 Zip file could not be opened

Not certain if you just left it out of the supplied code or just missed =
it
but=20
aZip.ZipFileName :=3D ???    // set to name of zip file

(The name supplied to ExtractFileToStream is the filename within the zip
file.)

The most likely reason for that error number in that function is because =
it
cannot open ZipFileName.
If that file exists then make certain it is not being used by something =
else
and that it contains files.

- Russell Peters
> -----Original Message-----
> From: delphizip-bounce@xxxxxxxxxxxxx [mailto:delphizip-
> bounce@xxxxxxxxxxxxx] On Behalf Of Edwin
> Sent: Wednesday, 27 June 2007 11:43 PM
> To: delphizip@xxxxxxxxxxxxx
> Subject: [delphizip] Re: error 11002 Zip file could not be opened
>=20
> Sorry for not providing detailed information Roger.
> It's a single zip and when this error occured the program is reading
> the zip file, and the code looks like:
>=20
>     aZip.ZipStream.Clear;
>     aZip.ExtractFileToStream(con_XmlFileName); //the file
> con_XmlFileName exists in the zip file
>     if aZip.ZipStream.Size  < 1 then
>       raise Exception.CreateFmt(err_ErrorReadMindMap, [aZip.ErrCode]);
>=20
>=20
>=20
>=20
> Best Regards,
> Edwin
> http://www.innovationgear.com
> MindVisualizer -- Productive Visual Mind Mapping Software
> 2007-06-27
>=20
>=20
>=20
> Sender=A3=BA Roger Aelbrecht
> Sent=A3=BA 2007-06-27 21:25:37
> Recipient=A3=BA delphizip@xxxxxxxxxxxxx
> CC=A3=BA
> Subject=A3=BA [delphizip] Re: error 11002 Zip file could not be opened
>=20
> Edwin wrote:
>  > Dear all,
>  >
>  > my program got a error message "Zip file could not be opened" and =
the
> error code  is11002.
>  >
>  > Please adivse what reason might cause this error? Thanks.
>  >
>  >
>  > --------------
>  > Best Regards,
>  > Edwin
>  > http://www.innovationgear.com
>  > MindVisualizer -- Productive Visual Mind Mapping Software
>  > 2007-06-27
>  >
>=20
> With this little information it is very difficult to answer your
> question.
> That error 11002 =3D DS_FileOpen is used on 9 diferent places int he
> code.
> So you need to give more info:
> - is it a single zip or a multi part zip, if it is a multi part zip
> then
> there are various reasons why the last part could not be found/opened.
> - what kind of operation are you performing, most probably you are
> setting the ZipFileName and therefore List and OpenEOC is trying to
> fond
> the EOC and fails because the file could not be opened for reading
>=20
> --
> Roger Aelbrecht
>=20
>=20
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>=20
>=20
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>=20
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>=20
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