I'm using zope and I want in a python script load an image in current folder: -First i've found the method with the os.open fonction,i've created a directory in /usr/lib/zope/lib/python named PythonScriptModules with inside a file named __init__.py: from Products.PythonScripts.Utility import allow_module allow_module(os) I use the container.manage_addProduct['OFSP'].manage_addImage(id+'_img',file,'') function but,when my python script call file = os.open(filename).read(),mozilla tell me my password and loop on authentification -Is there an other method to create an image object without using the os.open function????
Cedric Bellegarde writes:
I'm using zope and I want in a python script load an image in current folder:
-First i've found the method with the os.open fonction,i've created a directory in /usr/lib/zope/lib/python named PythonScriptModules with inside a file named __init__.py: from Products.PythonScripts.Utility import allow_module allow_module(os) I use the container.manage_addProduct['OFSP'].manage_addImage(id+'_img',file,'') function but,when my python script call file = os.open(filename).read(),mozilla tell me my password and loop on authentification "os.open" does not return a file object (but an integer file descriptor). The result does not have a "read" method.
Don't know why you get the "Unauthorized" (usually you are told, what you are not authorized to access, when you refuse to log in. What are you told?)
-Is there an other method to create an image object without using the os.open function???? Use an External Method and "open".
Dieter
Thank you it works.But I have a new problem: <FORM ACTION="AliacomPresse_add" METHOD="post"> <INPUT TYPE="file" NAME="img"> </FORM> when i access img in AlicomPresse_add(python script),he hasn't the value '/home/cedric/my_img.png' but the value 'my_img.png'.It makes that with mozilla and galeon but at home on my computer it works perfectly.Configuration problem???? Dieter Maurer wrote:
Cedric Bellegarde writes:
I'm using zope and I want in a python script load an image in current folder:
-First i've found the method with the os.open fonction,i've created a directory in /usr/lib/zope/lib/python named PythonScriptModules with inside a file named __init__.py: from Products.PythonScripts.Utility import allow_module allow_module(os) I use the container.manage_addProduct['OFSP'].manage_addImage(id+'_img',file,'') function but,when my python script call file = os.open(filename).read(),mozilla tell me my password and loop on authentification "os.open" does not return a file object (but an integer file descriptor). The result does not have a "read" method.
Don't know why you get the "Unauthorized" (usually you are told, what you are not authorized to access, when you refuse to log in. What are you told?)
-Is there an other method to create an image object without using the os.open function???? Use an External Method and "open".
Dieter
Cedric Bellegarde writes:
<FORM ACTION="AliacomPresse_add" METHOD="post"> <INPUT TYPE="file" NAME="img"> </FORM> You should use 'enctype="multipart/form-data"' as an additional "form" attribute.
<http://www.dieter.handshake.de/pyprojects/zope/book/chap3.html> can tell you why ("Web publishing section").
when i access img in AlicomPresse_add(python script),he hasn't the value '/home/cedric/my_img.png' but the value 'my_img.png'.It makes that with mozilla and galeon but at home on my computer it works perfectly.Configuration problem???? Not sure, I understand your problem:
It is fine that "img.filename" is "my_img.png" and not (as in IE) "/home/cedric/my_img.png" as the latter reveals both unnecessary and potentially sensitive information to the server. If you mean: The "img" variable gets bound to the string "my_img.png" rather than the file uploaded from "/home/...", then the missing "enctype" attribute is the cause. Dieter
participants (2)
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Cedric Bellegarde -
Dieter Maurer