Hi, A quick look at the source (Zope/lib/python/Zpublisher/HTTPRequest.py) suggests that you should be accessing the filename attribute of the FileUpload instance to get the location of the uploaded file ie. something like: <dtml-var expr=picture1.filename> You can then read the contents of the file and do your insert. Also, the docstring says: File upload objects can be used just like files. So presumably you can access the file data by read() ing from picture1 directly. HTH michael
-----Original Message----- From: zope-bounces@zope.org [mailto:zope-bounces@zope.org] On Behalf Of Quince Gibson Sent: Wednesday, January 21, 2004 4:37 PM To: zope@zope.org Cc: us3media@aol.com Subject: [Zope] FileUpload, MySQL, Zope ZSQL problem
Tried finding an answer to this question but I can't. I'm trying to upload pictures to a MySQL database with the normal conditions:
<input type="file" name="picture1"> enctype="multipart/form-data" method="post"
My ZSQL method has the following DTML: <dtml-var expr=picture1.data>
I've also tried the same thing with dtml-sqlvar using type=string.
I'm always greeted with the following error whenever I feel I'm going to make a breakthrough:
FileUpload instance has no attribute 'data'
removing the ".data" section gives me: <ZPublisher.HTTPRequest.FileUpload instance at 0x932c35c> as the entry for that particular field.
Is it possible?????? Am I trying the impossible??? if not... how can I make it possible to get these pictures into this database. (i'm starting to lose my cool on this computer).
Thanks for any help. If it's not possible I guess I'll have to bloat my Object Database rather than my relational databse.
Quince Baltimore, MD.