Thanks a lot for replying... I still a newbie at this ...
in order for shapelib to read shapefile, you need a set of three files(.shp,
.shx, .dbf).
so, if I have to download this to a temp location into my computer filesystem,
could you show me how this is done?
----- Original Message ----
From: Andreas Jung <[EMAIL PROTECTED]>
To: Allen Huang <[EMAIL PROTECTED]>; Zope <zope@zope.org>
Sent: Sunday, December 24, 2006 2:47:52 PM
Subject: Re: Fw: [Zope] problem with the shapelib module.. help
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- --On 23. Dezember 2006 22:27:29 -0800 Allen Huang <[EMAIL PROTECTED]> wrote:
>
> 2. but when I save the shapefiles (taiwan1.shp, taiwan1.shx, taiwan1.dbf)
> into zope and call it using dtml it return the same error message.
> import shapelib, dbflib
> def readshp(filename):
> shp = shapelib.ShapeFile(filename)
> return shp.info()
>
> and in zope
> <dtml-var expr="readshp(taiwan1.shp)">
> or
> <dtml-var expr="readshp('http://localhost/pytest/taiwan1.shp')";>
No idea what shapelib is doing but reading a file from the locale
filesystem as it seems to work from an external method is *different* from
a accessing
content that is stored within the ZODB. I really wonder why you think it
would work the same way? The hierarchical object storage of Zope looks
similar to a filesystem but it is not a filesystem and the Python APIs for
accessing filesystems don't apply. We have no idea what the ShapeFile
constructors expects as data. If it expects a filename of a file within the
filesystem then you must obtain the content of your data stored within the
ZODB, create a temporary file and call the constructor as you did within
your external method. Possibly the constructor accepts the data directly
as *string*..in this case this would make things a bit easier. But it is up
to *you* to check the Shapelib API and take appropriate action.
- -aj
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