April 2002 - Zope - Zope lists

[PATCH] to Zope's Management Interface
by Jerome Alet 29 Apr '02

29 Apr '02

Hi the attached patch to apply to zope/lib/python/OFS/dtml/main.dtml allows people who want to use ZShell to type ZShell commands directly from the Zope's Management Interface. It should be transparent for people who don't want to use ZShell at all. to use this patch, you have to : - download ZShell v1.42 from : http://www.librelogiciel.com/software/ This is mandatory to stay in the ZMI when running commands, otherwise, with older ZShell versions you'll enter ZShell user interface after entering a command. - extract it in your home directory - cd to the Zope's home, e.g. /usr/lib/zope - type the following command (as root) : patch -p0 <~myusername/zshell-1.42/main.dtml.patch - restart Zope Enjoy !!! (At least you should) This was tested against Zope 2.5.0-4 (Debian package) but applying the patch manually to other versions if needed is very easy. As always any comment is very welcome. PS : This procedure and the patch are now included in the ZShell 1.42 package. Jerome Alet --- lib/python/OFS/dtml/main.dtml.orig Mon Apr 29 12:18:12 2002 +++ lib/python/OFS/dtml/main.dtml Mon Apr 29 12:38:34 2002 @@ -60,6 +60,15 @@ </table> </dtml-if> +<dtml-if "_.hasattr(this(), 'zshell')"> + <form action="zshell" method="GET"> + <input type="text" name="zshellscript:string" width="60" /> + <input type="hidden" name="zshellredirect:string" value="&dtml-URL0;" /> +   + <input type="submit" name="zshellbutton" value="ZShell!" /> + </form> +</dtml-if> + <form action="&dtml-URL1;/" name="objectItems" method="post"> <dtml-if objectItems> <table width="100%" cellspacing="0" cellpadding="2" border="0">

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RE: [Zope] How to use a ZSQLMethod from a Python Script
by Igor Leturia 29 Apr '02

29 Apr '02

>Igor Leturia writes: > > I want to call a ZSQLMethod from a Python Script. In Zope.org, I > > couldn't find a specific how-to on the subject, but accidentally I found > > some other how-to's in which there are examples that use ZSQLMethods > > from Python scripts. It seemed very easy, so I tried it that way: > > > > res = context.mysqlmethod() > > > > It gives an error message, 'resource not found'. Both the >script and > > the sql method are in the same folder. What's going wrong? Dieter Maurer writes: >When script and method are in the same folder, it may >help to replace "context" by "container". > >If it does not help, the precise error message (Error Type, Error Value >and traceback) is necessary to help... > 'container' doesn't work either. As for the error type and value, I don't know, this is the message I get: Site Error An error was encountered while publishing this resource. Resource not found Sorry, the requested resource does not exist. Check the URL and try again. ------------------------------------------------------------------------ -------- Troubleshooting Suggestions The URL may be incorrect. The parameters passed to this resource may be incorrect. A resource that this resource relies on may be encountering an error. For more detailed information about the error, please refer to the HTML source for this page. If the error persists please contact the site maintainer. Thank you for your patience. I'm sure the URL is correct and the ZSQLMethod doesn't need any parameters. There isn't any name problem, as the ZSQLMethod's name is unique, I have never used it before. I'm completely lost. The only thing I can think of is that I need some import clause. Is that possible? Any ideas will be welcome. Thanks in advance, Igor Leturia

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namespace tree-tags
by Nico Niepraschk 29 Apr '02

29 Apr '02

Hello, how can I change (and combine two) namespaces of a tree tag ? <dtml-with ...> <dtml-tree branches=objectValues> </dtml-tree> </dtml-with> ... doesn't work. Have anybody an idea ? Thanks for your help! Nico Niepraschk

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quick test - can current user execute method?
by A M Thomas 29 Apr '02

29 Apr '02

Hi all, In a product's Python script method, I'd like to be able to do a quick test to see whether the current user can execute a particular management script, without hard-coding into the script what role(s) the script requires. Is there an easy way to do this? Or a hard way? I've read what I could in my two books and looked in the "Zope Cookbook", as well as in the mailing list archives and searched zope.org - it seems like a basic enough piece of functionality, and it's probably staring me in the face, but I can't see it. Thank you very much, Am

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total number of batches (pages)
by Trevor Toenjes 29 Apr '02

29 Apr '02

Is there a built-in to determine how many batches(pages) are returned from a query? or do I have to ... batch_size_max=sequence-step-size batches_total="(_.len(results)-1)/batch_size_max+1" Thanks, Trevor

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checking permissions
by A M Thomas 29 Apr '02

29 Apr '02

Hi there, I'm trying to make my own management view links, and can't find the correct way to test for whether a method can be used by the current user based on just the method name. Here's a section of my python code: =================================== from AccessControl import getSecurityManager checkPermission = getSecurityManager().checkPermission manage_options=context.filtered_manage_options() optionslist = [] for mopt in manage_options: print "<br>checking", context.url() + '/' + mopt['action'] if checkPermission(mopt['action'], context): optionslist.append([mopt['action'], mopt['label']]) =================================== Basically, when I'm logged in as "manager", every method in manage_options checks out. However, when I log in as another user - which user does in fact have access to the appropriate methods - the checkPermission test fails. I'm pretty sure I should be supplying a permission name as the first argument to checkPermission, rather than the method name, but is there a way to test based just on the method name? I can get a handy list of methods using the filtered_manage_options method, but not permissions, as far as I can tell (and no, they're not already tested, as far as I can tell - my test user gets a tab to manage security settings, but can't get a page by clicking on it). Thanks in advance for any help - I appreciate it very much :) - Am

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File upload problem?
by Roger Ineichen 29 Apr '02

29 Apr '02

Hello together, I've a problem with upload a picture and the method cookId(id='', title='', file=file). This method doesn't work for me? The method cookId() doesn't give a id or a title back. Why? Whats wrong? Form -------------------------- <form action="addFile" method="post" enctype="multipart/form-data"> <input type="file" name="file"> <input type="submit" value="upload"> </form> Python (not pythonScript) -------------------------- def addFile(self, file): "This method adds a file" # create the file if(file): id, title = cookId(id='', title='', file=file) self.manage_addProduct['P01_DownloadFolder'].manage_addFile(id=id, title='', file=file) Second Way I tryed (the manage_addFile() call the cookId by standard. -------------------------- def addFile(self, file): "This method adds a file" # create the file self.manage_addProduct['P01_DownloadFolder'].manage_addFile(id='', title='', file=file) cookId is a method from Image.py -------------------------------- def cookId(id, title, file): """ it gets the id from a file """ if not id and hasattr(file,'filename'): filename=file.filename title=title or filename id=filename[max(string.rfind(filename, '/'), string.rfind(filename, '\\'), string.rfind(filename, ':'), )+1:] return id, title The error: ---------- The "id" is not defined cookId does not set the id, How can id get the id from the file in a different way. Is the Form enctype="multipart/form-data" wrong? What should i use? Thanks to all and have a nice week. Roger Ineichen ___________________________ END OF MESSAGE

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checking view permissions
by Tom Rockwell 29 Apr '02

29 Apr '02

Hi, I have a simple file library type page, and am tweeking it a bit so that anonymous users don't see all the file upload options etc. I'd like to hide the "add file" link from anonymous viewers. On the dtml method that shows a file upload form, I changed the "View" permission to "Authenticated" users. On the file library page, I wrapped a <dtml-if> around the code that makes the link to the upload page: <dtml-if "AUTHENTICATED_USER.has_permission('View', upload)"> <a href="upload">Upload</a> </dtml-if> However, for unauthenticated users, Zope gives the error that "upload" doesn't exist. So apparently, if a user can't View an object, they also can't call .has_permission('View' object). Am I missing something? To work around, I just used a <dtml-if upload>, dtml-if has nice behaviour when objects aren't found. I'm using Zope 2.5. Cheers, Tom Rockwell

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python modules problem
by Cedric Bellegarde 29 Apr '02

29 Apr '02

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????

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why file.update_data permission is python only?
by Tom Rockwell 29 Apr '02

29 Apr '02

Hi, Now that the builtin File object can be edited via the web, I'm wondering why the update_data method for Files has permission "Python Only". What is gained by restricted this method, when delete and file upload methods are accessible? I'm using Python 2.5. Thanks, Tom

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