CODEWITHSUNDEEP
pythonzopedtml
Nishant
Nishant

Reputation: 21914

How to render a Zope DTML from the REPL

How do you render a Zope DTML from the REPL say without the Folder Object or anything like that for a test case ?

I have a file called /tmp/tmp.dtml with the following content:

<dtml-var test>

I could get this far : 

from AccessControl import ImplC as impl
from App.special_dtml import DTMLFile

#globals has test variable define if that helps?
dtml_page = DTMLFile('/tmp/tmp',globals())
dtml_page() give "None\n"

How do I inject the right namespace into DTMLFile object or during invokation ?

Upvotes: 2

Views: 120

Answers (1)

Nishant
Nishant

Reputation: 21914

This seems to work . I wanted to mock it without Folder though . 

class Test(Folder):
    test = "World"
    dtml = DTMLFile('/tmp/tmp')
dtmlobj = Test()
dtmlobj.dtml()

Class inherited from Folder seems to be important here for not-so-clear reasons ?

I am not getting the significance of why the Test.test where Test is inherited from Folder to render . Is this some type of a Zope convention ?

I tried

dtml = DTMLFile('/tmp/tmp') 
setattr(dtml,'test','foo') 
dtml.render()

I though it was about "self" having the right variables but apparently not .

Upvotes: 0

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