build application from admin code
Hello, I just started using Django and I am still "marvelling at all the code I didn't have to write". I wonder if this code, the admin interface, could be used for the application itself. I would like to do something like this: Each user of my application gets his very own predefined objects of a certain kind, which he can use, change, delete, and so on. He gets his own urls and views that are but modifications of the automatically generated admin urls and views. Like so: example.org/juan/grammar/verbs/add/ (a modification of example.org/admin/grammar/verbs/add/) or so: example.org/merche/recipes/delete (a modification of example.org/admin/recipes/delete) Juan and Merche are users of example.org, but they can have and administrate their own users. Juan, e.g., might have the users Maria and EmacsGuru. Maria and EmacsGuru can only view and manipulate the pages that Juan allows them to. And Juan can only view and manipulate the pages that the admin of example.org allows him to. Has anyone done something of this kind? Is there example code? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
compiled methods
Hello, is there such a thing as "compiled methods" in Django i.e. methods whose return value is calculated only once and then stored? Or is there a canonical Djangoish way to implement this? An example: Suppose we have phone objects that have a canonical form, such as "alcatel a-341 i" and a paraphrase type p and a method variants() that calculates all variants of the canonical form, given the paraphrase type p, such as: alcatel a-341 i, alcatel a 341 i, alcatel a341 i, alcatel a-341-i, alcatel a 341-i, alcatel a341-i, alcatel a-341i, alcatel a 341i, alcatel a341i, a-341 i, a 341 i, a341 i, a-341-i, a 341-i, a341-i, a-341i, a 341i, a341i Obviously it is very bad for speed to calculate the variants of each phone object every time the object is needed for some action. Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: compiled methods
Thank you very much! The term "memoize" was exactly what I was looking for. Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
override save()
Hello, I have got a model called ContentClassifier which uses regular expressions to classify the content of messages. It has a method compile() that compiles an instance's regular expressions into self.posrx. When the classify method calls compile, i.e. when each ContentClassifier compiles its regular expressions every time it's classify-method is called, then the ContentClassifiers work as expected. (I.e. if, in the code given below snippet 2 instead of snippet 1 is used). If, on the other hand, I want each instance to compile its regular expressions when it is saved (i.e. if I use snippet 1 instead of snippet 2), Django says AttributeError: ... ContentClassifier' object has no attribute 'posrx' Why? # class ContentClassifier(models.Model): def save(self, *args, **kwargs): # snippet 1 self.compile() # snippet 1 super(ContentClassifier, self).save(*args, **kwargs) # snippet 1 def compile(self): ... self.posrx = re.compile(regex_string, re.IGNORECASE) def classify(self, message): # self.compile() # snippet 2 # -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: override save()
Sorry the formatting of the code went wrong. Here it is again: # class ContentClassifier(models.Model): # def save(self, *args, **kwargs): self.compile() super(ContentClassifier, self).save(*args, **kwargs) # def compile(self): ... self.posrx = re.compile(regex_string, re.IGNORECASE) def classify(self, message): # # self.compile() # # -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: override save()
> Since these attributes > are "ordinary" attributes, they aren't saved to the database, so they > aren't loaded from it neither. It seems that this was the problem. Thank you. Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
slight model change makes app unusable
Hello,
I have got a model MobilePhone that has a producer, such as "alcatel",
a canonical name, such as "b-500 s" and a paraphrase type, such as
"a-341 i". Each instance of MobilePhone is supposed to calculate all
its variants (such as alcatel b-500 s, alcatel b 500 s, alcatel b-500-
s, b-500 s, b 500 s, b-500-s, and so on). This worked perfectly as
long as the producer wasn't a field of its own, but part of the
canonical name. Since I introduced producer names (in order to
simplify the paraphrase types and to allow for variants of producer
names (cf. HP, Hewlett Packard and so on), the app became unusable:
Whenever one wants to see the list of all (461) mobile phones in the
admin interface, the development server becomes very slow and
eventually breaks down with a memory error. Below is the class
MobilePhone. What can I do?
##
class MobilePhone(models.Model):
producer = models.ForeignKey(MobilePhoneProducer)
canonical_name = models.CharField(max_length=200)
paraphrase_type = models.CharField(max_length=50, choices=(
('n-127', 'n-127'),
('a-341 i', 'a-341 i'),
('neotouch', 'neotouch'),
))
lexical_variants = models.ManyToManyField(LexicalVariant,
blank=True)
n127_rx = re.compile('^(?P[a-z]+)\-(?P[0-9]+)$')
a341_i_rx = re.compile('^(?P[a-z]+)\-(?P[0-9]+) (?
P[a-z])$')
neotouch_rx = re.compile('^(?P[a-z]+)$')
def save(self, *args, **kwargs):
self.canonical_name =
whitespace_normalize(self.canonical_name)
super(MobilePhone, self).save(*args, **kwargs)
def __unicode__(self):
return self.canonical_name
def variants(self):
self.save()
paraphrases = self.paraphrase()
for paraphrase in paraphrases:
paraphrases.append(self.producer.name + ' ' + paraphrase)
for variant in self.lexical_variants.all():
paraphrases.append(variant)
paraphrases.append(self.producer.name + ' ' + variant)
if paraphrases:
return ', '.join(paraphrases)
else:
return '(The paraphrase type doesn
\'t seem to match.)'
variants.allow_tags = True
def the_lexical_variants(self):
return ', '.join(self.lexical_variants.all())
the_lexical_variants.short_description = "Lexical variants"
def paraphrase(self):
if self.paraphrase_type == 'n-127':
test = self.n127_rx.search(self.canonical_name)
if test:
chars = test.group('chars')
numb = test.group('numb')
return [
chars + '-' + numb,
chars + numb,
chars + ' ' + numb,
]
elif self.paraphrase_type == 'a-341 i':
test = self.a341_i_rx.search(self.canonical_name)
if test:
chars1 = test.group('chars1')
chars2 = test.group('chars2')
numb = test.group('numb')
return [
chars1 + '-' + numb + ' ' + chars2,
chars1 + ' ' + numb + ' ' + chars2,
chars1 + numb + ' ' + chars2,
chars1 + '-' + numb + '-' + chars2,
chars1 + ' ' + numb + '-' + chars2,
chars1 + numb + '-' + chars2,
chars1 + '-' + numb + chars2,
chars1 + ' ' + numb + chars2,
chars1 + numb + chars2,
]
elif self.paraphrase_type == 'neotouch':
test = self.neotouch_rx.search(self.canonical_name)
if test:
gmodel = test.group('model')
return [
gmodel,
]
return []
###
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use variable that is bound at an object's creation time at the object's creation time
Hello, my model Product has an attribute spelling_variants and a property variants: class Product(models.Model): canonical_name = models.CharField(max_length=200) spelling_variants = models.ManyToManyField(String, related_name="product spellings", blank=True) ... @property def variants(self): variants = [] for v in self.spelling_variants: variants += paraphrase(v) return variants The code works, if product instances are first saved without a vaue for spelling_variants and later modified so as to have some spelling variants. When I add spelling variants at the time when I create a product in the admin interface, I get this error: Caught TypeError while rendering: 'NoneType' object is not iterable I think the problem in these cases is that when Python tries to execute self.variants() it does not yet have a value for self.spelling_variants. How can I make sure that self.spelling_variants has a value when self.variants() is executed? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
blank ModelMultipleChoiceField, sort a ManyToManyField that is allowed to be blank
Hi, why can't a ModelMultipleChoiceField not have the property "blank=True"? I need the ModelMultipleChoiceField to sort the values of a models.ManyToManyField in a form class. Sorting works as expected, but only if the field is obligatory. (Else, I get the error message "TypeError: __init__() got an unexpected keyword argument 'blank'.) Is there another way to sort a models.ManyToManyField that has the property "blank=True"? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: blank ModelMultipleChoiceField, sort a ManyToManyField that is allowed to be blank
> No form fields take that argument - that's for model fields. > > Form fields take the argument "required", which defaults to True. Thank you! That is exactly the information I needed. By the way, is there a special reason why "blank" of model fields translates to "required" of form fields? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: blank ModelMultipleChoiceField, sort a ManyToManyField that is allowed to be blank
Ah, now there is a new problem.
Before, my ManyToManyField was unsorted, but it was possible to add
new items by clicking on the "+" button. Now my items are sorted
thanks to a ModelMultipleChoiceFormField (with "order_by('...')" and
"required=False"), but the "+" button is gone.
Is there a way to get the "+" button back?
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compiled regex as attribute
Hello, I have got objects with very large regular expressions: class Product(models.Model): # ... canonical_name = models.CharField(max_length=200) spelling_variants = models.CharField(max_length=1, blank=True) lexical_variants = models.CharField(max_length=1, blank=True) excluded = models.CharField(max_length=1, blank=True) permutations = models.CharField(max_length=1000, blank=True) When a product has values for all of the above attributes, it can calculate all its possible names. Example: the product 'google android 1.1' has several hundred possible names including these: google android-1.1, oha android-1.1, android-1.1, google android-1,1, oha android-1,1, android-1,1, google android-1-1, oha android-1-1, android-1-1, google android-11, oha android-11, android-11, google android-1 1, oha android-1 1, android-1 1, google android1.1, oha android1.1, android1.1, google android1,1, oha android1,1, android1,1, google android1-1, oha android1-1, android1-1, google android11, oha android11, android11, google android1 1, oha android1 1, android1 1, google android 1.1, oha android 1.1, android 1.1, google android 1,1, oha android 1,1, There are products with many more names. Consider the name of the mobile phone alcatatel one-touch-800-one-touch-chrome. Currently every product calculates all its possible names and a regular expression that includes the disjunction of all the possible names and compiles this regular expression at the time when the product is used. This is very inefficient. I would like to compile and store the regular expression only once: at the time when the new product is saved. However, there doesn't seem to be a models.RegexField. There is a forms.RegexField, but I don't understand how I could use it in the product model. Any tips how to store the compiled regex? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: compiled regex as attribute
> http://pypi.python.org/pypi/django-picklefield/0.1 Thanks for the tip. This works, but it has one great disadvantage: The regexes can't be stored in fixtures. They are stored in the database directly. So it seems I would have to insert all my several thousand products manually again. And if I change some models and need to do a "manage.py syncdb" all the work will be lost ... Or is there a way of loading fixtures that includes that save() is called for each loaded object? -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: compiled regex as attribute
> Since the pickled value is a string, it should work in fixtures. There is no point in storing the regex strings in a pickle field. I already have the regex strings in ordinary django fields. What I want to store is *compiled* regular expressions in order to be able to use them without having to compile them first. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
reimport module every n seconds in main process
Hello, in a Django application I compile certain data into a Python module for efficiency reasons. Each time an admin changes or adds objects, this module also changes. Therefore, I want my main process (the one that is started with "manage.py runserver" or the like) to reimport the data module every n (e.g. 10) seconds. How could this be done? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
'NoneType' object has no attribute 'set_cookie'
Hello,
I didn't use a certain Django project for a few weeks. Now I wanted to
continue developing the project and, as it usually happens with
software that has been left alone for a while, it didn't work
anymore. I am not aware of having changed anything. The system
administrator may, of course, have made some changes to the
system. When I try to add a new object, regardless of which type, in
the admin interface, I get the error messages given below. (I had to
shorten the error message for Google Groups to allow me to post it.)
Any ideas what I could do?
Santiago
##
AttributeError at /admin/monitoring/message/add/
'NoneType' object has no attribute 'set_cookie'
Request Method: GET
Request URL:http://localhost:8000/admin/monitoring/message/add/
Django Version: 1.3
Exception Type: AttributeError
Exception Value:
'NoneType' object has no attribute 'set_cookie'
Exception Location: /usr/local/lib/python2.6/site-packages/django/
middleware/csrf.py in process_response, line 239
Python Executable: /usr/bin/python
Python Version: 2.6.2
Python Path:[...]
Server time:Thu, 31 Mar 2011 10:10:30 +0200
Traceback Switch to copy-and-paste view
[...]
response
None
result
None
kwargs
{}
view_func
* /usr/local/lib/python2.6/site-packages/django/middleware/csrf.py
in process_response
232.
if request.META.get("CSRF_COOKIE") is None:
233.
return response
234.
235.
if not request.META.get("CSRF_COOKIE_USED",
False):
236.
return response
237.
238.
# Set the CSRF cookie even if it's already set, so
we renew the expiry timer.
239.
response.set_cookie(settings.CSRF_COOKIE_NAME,
...
240.
request.META["CSRF_COOKIE"], max_age = 60
* 60 * 24 * 7 * 52,
241.
domain=settings.CSRF_COOKIE_DOMAIN)
242.
# Content varies with the CSRF cookie, so set the
Vary header.
243.
patch_vary_headers(response, ('Cookie',))
244.
response.csrf_processing_done = True
245.
return response
▶ Local vars
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django-admin.py startproject mysite
Hello, the Django (1.3) tutorial says: >From the command line, cd into a directory where you’d like to store your code, then run the command django-admin.py startproject mysite. This will create a mysite directory in your current directory. When I run django-admin.py startproject mysite, I get this error: Unknown command: 'startproject' Type 'django-admin.py help' for usage. Using django-admin.py help, I get these subcommands: Available subcommands: changepassword cleanup collectstatic compilemessages createcachetable createsuperuser dbshell diffsettings dumpdata findstatic flush inspectdb loaddata makemessages reset runfcgi runserver shell sql sqlall sqlclear sqlcustom sqlflush sqlindexes sqlinitialdata sqlreset sqlsequencereset startapp syncdb test testserver validate Does the Django tutorial still match reality? Santiago -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
