November 2001 - Zope-Dev

Hexadecimal-sequence of a jpeg-image from Python
by Juan Javier Carrera Obrero 19 Nov '01

19 Nov '01

Hello, I am trying to generate a RTF (Rich Text Format) file with images from Python 2.1. Firstly with MS-Word 97, I have saved a file with RTF format which contained a jpeg-image. I have edited this RTF file with a text editor, and I have seen as MS-Word converts the jpeg-image in a hexadecimal sequence. I would like to do the same process for converting a binary jpeg-image in a hexadecimal-sequence from Python. For it, I have maked the next process (using Python 2.1 and PIL library, in Windows): import Image im = Image.open("photo.jpg") file = open("filename.txt", "wb") file.write ( im.tostring ( "hex") ) file.close() With this process, I obtain a hexadecimal-sequence of the jpeg-image (in "filename.txt"). However, I am trying to work with this sequence but I can not do anything with it. I think this sequence is no valid, because I have replaced from a text editor the hexadecimal-sequence of MS-Word to the hexadecimal-sequence obtained by me, however when I open this file, MS-Word do not show the jpeg-image. Also, the hexadecimal-sequence generates by MS-Word is not equal to the hexadecimal-sequence generates by the previous process. Anybody could help me about this ? Thank you very much.

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<dtml-mime> and MIME-Version
by Julián Muñoz Domínguez 18 Nov '01

18 Nov '01

(I report this here because the bug collector is down) Having some problems sending attachements with <dtml-sendmail>, I have been reported that the bug is: <dtml-mime> doesn't generate "MIME-Version: 1.0" header. Such a header is required by MIME in order to recognize the message as MIME message and not an old RFC822 message. Apparently, many mail readers are not as strict and recognize a MIME message even if the "Mime-Version" header is missing. "pine" seems to be strict.... "dtml-mime" should generate the "MIME-Version" header. -- __o _ \<_ (_)/(_) Saludos de Julián EA4ACL -.-

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Where to put time.sleep(0.01) to get around solaris threading issue?
by Tony McDonald 16 Nov '01

16 Nov '01

(this is a bit long - delete it if not interested in Solaris, at least it's not HTML mail! :) Hi, I'm still trying to get our solaris box to work with more than one thread. Briefly, Put the following into a pythonscript; ## Script (Python) "sleeper" ##bind container=container ##bind context=context ##bind namespace= ##bind script=script ##bind subpath=traverse_subpath ##parameters= ##title= ## ## Warnings: ## Prints, but never reads 'printed' variable. ## # Example code: # Import a standard function, and get the HTML request and response objects. from Products.PythonScripts.standard import html_quote from time import time e= time() + float(15) while time() < e: pass print 'active sleep finished' return 'slept' Click 'test'. Try another request on that server. It will not run until the pythonscript has finished. Replace 'sleeper' with a longish running request that does real work and you've suddenly got a single-threaded server. Ok, now try this python script on a non-solaris box (ie one that allows more than a single thread to run, I'm using MacOS-X; (this script is from /usr/local/lib/python2.1/test/test_thread.py) # Very rudimentary test of thread module # Create a bunch of threads, let each do some work, wait until all are done #from test_support import verbose verbose = 1 import random import thread import time import sys mutex = thread.allocate_lock() rmutex = thread.allocate_lock() # for calls to random running = 0 done = thread.allocate_lock() done.acquire() numtasks = 10 def task(ident): global running rmutex.acquire() delay = random.random() * numtasks rmutex.release() if verbose: print 'task', ident, 'will run for', round(delay, 1), 'sec' time.sleep(delay) if verbose: print 'task', ident, 'done' mutex.acquire() running = running - 1 if running == 0: done.release() mutex.release() next_ident = 0 def newtask(): global next_ident, running mutex.acquire() next_ident = next_ident + 1 if verbose: print 'creating task', next_ident thread.start_new_thread(task, (next_ident,)) running = running + 1 mutex.release() for i in range(numtasks): newtask() print 'waiting for all tasks to complete' done.acquire() print 'all tasks done' sys.exit() % [MacOS-X] python2.1 test_thread.py reating task 1 task 1 will run for 7.8 sec creating task 2 task 2 will run for 5.6 sec creating task 3 task 3 will run for 6.3 sec creating task 4 task 4 will run for 4.2 sec creating task 5 task 5 will run for 9.8 sec creating task 6 task 6 will run for 2.8 sec creating task 7 task 7 will run for 0.2 sec creating task 8 task 8 will run for 3.6 sec creating task 9 task 9 will run for 5.1 sec creating task 10 task 10 will run for 3.0 sec waiting for all tasks to complete task 7 done task 6 done task 10 done task 8 done task 4 done task 9 done task 2 done task 3 done task 1 done task 5 done all tasks done Do the same on solaris and you get this; creating task 1 creating task 2 creating task 3 creating task 4 creating task 5 creating task 6 creating task 7 creating task 8 creating task 9 creating task 10 waiting for all tasks to complete task 1 will run for 3.3 sec task 2 will run for 3.5 sec task 3 will run for 3.7 sec task 4 will run for 5.8 sec task 5 will run for 3.3 sec task 6 will run for 0.0 sec task 7 will run for 3.9 sec task 8 will run for 6.6 sec task 9 will run for 1.4 sec task 10 will run for 7.9 sec task 6 done task 9 done task 1 done task 5 done task 2 done task 3 done task 7 done task 4 done task 8 done task 10 done all tasks done Now add the line time.sleep(0.01) before the thread_start.new_thread(task, (next_ident, )) line in the newtask method above. if verbose: print 'creating task', next_ident time.sleep(0.01) thread.start_new_thread(task, (next_ident,)) And you get this on Solaris... creating task 1 task 1 will run for 6.2 sec creating task 2 creating task 3 task 2 will run for 5.9 sec creating task 4 task 3 will run for 3.6 sec creating task 5 task 4 will run for 1.8 sec creating task 6 task 5 will run for 7.3 sec creating task 7 task 6 will run for 3.9 sec creating task 8 task 7 will run for 2.4 sec creating task 9 task 8 will run for 7.5 sec creating task 10 task 9 will run for 8.7 sec waiting for all tasks to complete task 10 will run for 1.7 sec task 10 done task 4 done task 7 done task 3 done task 6 done task 2 done task 1 done task 5 done task 8 done task 9 done all tasks done Ie pretty similar. With this in mind, I've added time.sleep(0.01) lines to the following files in a Zope 2.4.1 source tree. 304 [12:08] % find . -name "*.py" -print | xargs grep -i 'start_new' ./ZServer/PubCore/ZRendezvous.py: thread.start_new_thread(ZServerPublisher, ./ZServer/medusa/contrib/bobo_handler.py: thread.start_new_thread(self._continue_request,(sin,request)) ./ZServer/medusa/monitor_client_win32.py: thread.start_new_thread (reader, (l, s, p)) ./ZServer/medusa/thread/select_trigger.py: thread.start_new_thread (thread_function, (tf, self.count, n)) ./ZServer/medusa/thread/thread_channel.py: thread.start_new_thread ( ./lib/python/ZODB/bpthread.py: start_new_thread=apply And pointed a zope Data.fs that has the test methods at this new distribution. No joy. That is, testing the 'sleeper' script still blocks other requests until it is finished, when those requests are then serviced. Can anyone see what I'm doing wrong here? Tone. -- Dr Tony McDonald, Assistant Director, FMCC, http://www.fmcc.org.uk/ The Medical School, Newcastle University Tel: +44 191 243 6140 A Zope list for UK HE/FE http://www.fmcc.org.uk/mailman/listinfo/zope

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New Engineering Utility
by c27800＠home.ro 16 Nov '01

16 Nov '01

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Error code 4?
by Alex Kelly 15 Nov '01

15 Nov '01

I have been working on fixing our "Error code: 11" problems, however we also have about the same amount of "Error Code: 4" entries in our debug.log. Does anyone know what that is, and what I can do to clear that one up too? For the record we are running: Solaris 8 on a sparc Python 2.1.1 (compiled from source) Zope 2.4.2 (compiled from source) Thanks, Alex Kelly

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What to do when one mail recipient fails out of several?
by Lennart Regebro 15 Nov '01

15 Nov '01

There is currently a poposal to make MailHost transactional, so you can be sure that you sent the mail. However, when you have multiple recipients, that doesn't solve the problem, because then you actually get "half done" transactions. Smtplib.SMTP.sendmail is nice enough to send back a list of which mails failed, but Zope can't really do much with that list. If it returns an error, the user will resend the mail to all recipients, if it ignores it (as today) the user will think all mails went through. The problem really lies in the binary mode of error/success that web will have in this case. There is no partial success, either you get up an error page or you don't. My thought about how to solve this is to make an addition to smtplib.SMTP.sendmail(), an new parameter "stopOnRcptErr" or something, that defaults to "False", which gives todays functionality. MailHost could then call smtplib.SMTP.sendmail with stopOnRcptErr=True, which would then raise a SMTPSenderRefused error if any of the recipients fail. My questions: 1. What do you think? Is this reasonable? 2. Does anybody know where to send suggestions for changed Python stuff? python-help doesn't seem right...

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Zope Version (As reported by ZMI)
by Alex Kelly 15 Nov '01

15 Nov '01

I noticed that after I compiled zope from the source, the zope version that is reported in the Control Panel says: (Zope 2.4.3 (source release, python 2.1, linux2), python 2.1.1, sunos5) I have done all the compiling (Zope+Python) on one Solaris box, so why is "linux2" reported in that string? Thanks, Alex Kelly

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from Globals import Persistent
by Benjamin Buffereau 15 Nov '01

15 Nov '01

Hello, I'm using the binary version of Zope 2.4.3 for Windows. I'm trying to run the PollProduct example of the Zope Developper's Guide, and it seems to me that there is a big problem with the import of the name "Persistent". Here is what I get trying to import "Persistent" from the Python interpreter: >>> from Globals import Persistent Traceback (most recent call last): File "<stdin>", line 1, in ? File "C:\PROGRA~1\ZopeDev\lib\python\Globals.py", line 90, in ? import Acquisition, ComputedAttribute, App.PersistentExtra, os File "C:\PROGRA~1\ZopeDev\lib\python\App\PersistentExtra.py", line 87, in ? from Persistence import Persistent ImportError: cannot import name Persistent Is this an identified and fixed bug, or should I replace "from Globals import Persistent" with something else?? Cheers, Benjamin.

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Binary Image File in Hexadecimal Format
by Juan Javier Carrera Obrero 15 Nov '01

15 Nov '01

Hello, I am generating a RTF file from Python 2.1 with Zope 2.4, using the "Rich Text Format (RTF) Version 1.6 Specification" (http://msdn.microsoft.com/library/specs/rtfspec.htm) But I can not include a "jpg" image in the RTF file. For including the "jpg" image file, firstly, I have obtainted the hexadecimal representation of the image with Image.encode("hex") from PIL library: Import Image img = Image.open("photo.jpg") file = open("filename","wb") file.write(img.encode("hex")) file.close() However, this hexadecimal representation is no good, because when, for example, I save the same image with Microsoft Word in RTF format, these representations no match. Anybody can help me ? Thank you very much.

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anyone got any code for dealing with batches and LazyFilter?
by Subjectivize 15 Nov '01

15 Nov '01

I'm using a ZCatalog/LazyFilter inside a dtml-in method. Unfortunatly when in comes to batching etc the result is that the list is not longer lazy. Everything gets evaluated. Anyone come accross a similar problem and found a "nice" way of dealing with it? What is needed is something that uses the start index of the underlying list rather the filtered one but then I'm not sure how this can be done while still using all the existing dtml-in machinery (ot the TAL stuff for that matter). Any I making any sense here? Just wondering if anyone had come across the same problem.

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