Re: Import error
Hi Emily, On 22 Jun 2012, at 15:46, Emily wrote: > This is the class I created... > > import string > import random > > class Helpers: > > def random_password(): Take random_password outside of the Helpers class and you should be OK. On the import statement you need to refer to a 'top level object' within the imported file. The only top level object you have is the class Helpers - which it doens't look like you really need. Cheers, aid -- 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: ForeignKey problem
On 25 Jun 2012, at 17:41, Soviet wrote: > I'm new to this Django thing and I run into first problem :). > > Let's say I have two models and in each I have field with ForeignKey > relating to field in other model (hope it's clear). Now that I want to > run migrate with South, I'm getting "NameError: name 'ModelName' is > not defined". This is clearly (?) an issue with the fact that one > model is on top of the other one and second Model hasn't been defined > yet. Changing position of the models is, obviously, not helping. How > do I approach this? In the first of the two models, use a string containing the second model's name as the first argument to the ForeignKey method. Regards, aid -- 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: ForeignKey problem
On 25 Jun 2012, at 17:53, Soviet wrote: > Thank you kind sir for your fast response, that worked brilliantly. > Can I be cheeky and ask why does it work? :) Magic! ;-) Although seriously, Django obviously has some code in there to handle just the situation you have come across. Sorry, I don't have a deeper answer than that! Cheers, aid -- 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: Generating random alphanumeric codes
On 26 Jun 2012, at 15:53, Sithembewena Lloyd Dube wrote:
> Would anyone have tips on how to generate random 4-digit alphanumeric codes
> in python? Also, how does one calculate the number of possible combinations?
For upper case, lower case and digits we have,
26 + 26 + 10 = 62
...possibilities for each position. If you have four positions you have,
62 * 62 * 62 * 62 = 14776336
... possible combinations.
You can create a random number in this range by,
import Random from random
def build_population(spans):
for span in spans:
for x in range(ord(span[0]), ord(span[1]) + 1):
yield chr(x)
population = list(build_population( (('A','Z'), ('a', 'z'), ('0', '9'))
))
r = Random()
r.sample(population, 4)
['C', 'l', 'q', 'y']
You may need to look into getting some entropy into the random number
generator...
Cheers
aid
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Django: Object Oriented or Relational Database?
Hi, I'm creating a DB using Django for managing our companies data - customers, services, orders, billing etc. I'm using the Django admin interface to manage the data. I seem to be consistently at odds with myself and would be good to hear the community's point of view on this. I'm defining the DB in Python; using objects and inheritance. As such, when I'm doing this I'm thinking in an object orientated (OO) manner - a service uses a number of resources so I create a Service object with with either foreign key or many-to-many (as appropriate) relationship to the Resource objects. This gives me nice referential integrity and I can be sure that the Service has valid connections to the Resources it requires. It seems, however, that Django expects the models not to be defined in an OO manner. For example; should I wish resources to be listed as Inlines on the Service's page I can only do this if the foreign key is with the Resource - pointing 'up' the stack to the services - as possible to being pointed to by the Service. Django seems to be defining stricture using an OO language; but expecting that structure in a more traditional, relational database, style where entries point up to their owners rather than the owners pointing down to their children. In order to try to fit in with Django Admin I've ended up using both styles. Unsurprisingly this mixture of styles is now causing me problems with circular dependencies. I know I need to jump one way or the other; but would be great to get some comments from experienced Django coders before I decide which way... Any comments/advice welcome... Cheers, aid -- 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: Django: Object Oriented or Relational Database?
On 26 Jan 2011, at 11:55, Daniel Roseman wrote:
>
> This is what's known as the object-relational impedance mismatch [1].
> Unfortunately, as you've noticed, OO concepts don't map completely cleanly
> onto the relational model, and this is an issue with all systems that attempt
> to do it. Django does what it can to smooth over the gaps, but can't always
> do everything.
Thanks for the link; interesting reading.
> However, you should not be getting circular dependencies. Can you provide
> some examples? One place where I often see them is when two models in
> separate applications are related via ForeignKey, and the developer has
> imported each model into the other's models.py. But it isn't necessary to
> import them at all, as you can refer to models via strings:
>
> myfkfield = models.ForeignKey('otherapp.OtherModel')
>
> so no circular dependency is declared. Does that solve your problem? If not,
> some examples will be helpful.
I've used the string form for forward references of models; but I still seem to
get problems when performing the initial migration (using South) of the
database as the table to which the forward reference points does not exist at
the time of creating the initial table.
I've just tried using a basic syncdb and that works fine. I presume I'm seeing
problems as South's migrations work one app at a time; in contrast to syncdb
that seems to install the whole project - only worrying about dependencies at
the end.
Still; I feel circular dependencies are bad; even work arounds are possible.
Is there a recommended path to take with Django?
Kind regards,
aid
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Re: Customising the Admin Save Models in DJango 1.1
Hu,
On 26 Jan 2011, at 07:22, Prof Falken wrote:
> I have two models, the first is a 'change request ticket' and the
> second reflects the status of that ticket. They are linked by a
> ForeignKeyField on the first table.
You may find it best to collapse them into a single object; using the 'choice'
option to allow the user to set the ticket's state.
> The two problems that I have are:
>
> 1) Some of the 'states' should mark the ticket as closed, so when the
> ticket is saved with the state of 'closed', it should be marked as
> 'completed' with the date and time the action was carried out.
You can jump in and change the model's data before the model is saved by
over-riding the 'save' method in your model.
Below is one of my models. My frequency attribute could be quite similar to
what you need for your state field. Below you can see where I'm setting a
number of my attributes (based upon the parent object when the Charge object is
created as an Inline) before this Charge is saved,
class Charge (models.Model):
CHARGE_FREQUENCY_CHOICES = (
(u'NRC', u'Non-Recurring'),
(u'MRC', u'Monthly'),
(u'QRC', u'Quaterly'),
(u'ARC', u'Annual')
)
frequency = models.CharField(max_length=15,
choices=CHARGE_FREQUENCY_CHOICES)
def save(self, *args, **kwargs):
self.description = self.service
self.customer = self.service.customer
self.legacy_charge_id = None
self.prorata = False
self.created_date = datetime.date.today()
self.started_date = self.service.completed_date
self.expiry_date = self.service.expiry_date
self.terminated_date = self.service.terminated_date
super(Charge, self).save(*args, **kwargs)
In order that the user does not set (or need to set) the above fields, set the
fields attribute of the Admin object so only the values you want the user to
control are visible. Here's mine,
class ChargeInlineAdmin(admin.TabularInline):
fields = ('frequency','amount', 'currency', 'vat_chargeable')
model = Charge
extra = 0
Regards,
aid
Adrian Bool
a...@logic.org.uk
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Re: forbid clones
On 26 Jan 2011, at 12:26, Jaroslav Dobrek wrote: >> >> http://docs.djangoproject.com/en/1.2/ref/models/fields/#unique > Example: > > This should be allowed: > > car1: manufacturer = "foo", name = "bar" > car2: manufacturer = "foo", name = "baz" > > This should not be allowed: > > car1: manufacturer = "foo", name = "bar" > car2: manufacturer = "foo", name = "bar" Doesn't look to be directly possible from the Django's API; but I guess you could, class Car(models.Model): manufacturer = models.CharField(max_length=127) name = models.CharField(max_length=127) manufacturer_name = models.CharField(max_length=256, unique=True) def save(self, *args, **kwargs): manufacturer_name = "%s_%s" % (self.manufacturer, self.name) super(Charge, self).save(*args, **kwargs) aid -- 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: forbid clones
Please ignore this dumn-ass p On 26 Jan 2011, at 12:34, Adrian Bool wrote: > > On 26 Jan 2011, at 12:26, Jaroslav Dobrek wrote: > >>> >>> http://docs.djangoproject.com/en/1.2/ref/models/fields/#unique >> Example: >> >> This should be allowed: >> >> car1: manufacturer = "foo", name = "bar" >> car2: manufacturer = "foo", name = "baz" >> >> This should not be allowed: >> >> car1: manufacturer = "foo", name = "bar" >> car2: manufacturer = "foo", name = "bar" > > Doesn't look to be directly possible from the Django's API; but I guess you > could, > > class Car(models.Model): > manufacturer = models.CharField(max_length=127) > name = models.CharField(max_length=127) > manufacturer_name = models.CharField(max_length=256, unique=True) > >def save(self, *args, **kwargs): >manufacturer_name = "%s_%s" % (self.manufacturer, self.name) >super(Charge, self).save(*args, **kwargs) > Urgh. Please ignore this dis-information. Hat off to Chris for knowing what he types. aid -- 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: Django: Object Oriented or Relational Database?
Hi Bruno, On 26 Jan 2011, at 12:55, bruno desthuilliers wrote: > Some will talk about "OO/relational impedance mismatch" (ok, I cheated > - Daniel just did;)), but I don't see it that way as far as I'm > concerned - I'm using a relational model wrapped into an OO > representation, and that's just what I want for most apps. Many thanks for your comments; although it wasn't really what I wanted to hear - as I prefer the OO model over the relational. I now feel that if I decide to use Django - not following a relational style will only cause me difficulties. Kind regards, aid -- 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: problem with compressed egg in mysqldb module
On 26 Jan 2011, at 23:24, Cindy wrote: > > On Jan 26, 2:13 pm, Cindy wrote: >> Hi, all. I'm getting this error, when trying to export my django site >> to a new place: Do you mean to a new server? >> >> TemplateSyntaxError: Caught ImproperlyConfigured while rendering: >> Error loading MySQLdb module: bad local file header in /usr/local/lib/ >> python2.6/site-packages/MySQL_python-1.2.3-py2.6-freebsd-8.1-RELEASE- >> amd64.egg If so; is the destination server running a 64bit OS? Regards, aid -- 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: Customising the Admin Save Models in DJango 1.1
On 27 Jan 2011, at 13:28, Matthew Macdonald-Wallace wrote: >def save(self,*args,**kwargs): >if not self.id: >self.created = datetime.date.today() >if self.Status.ClosesChangeRequest == True: >self.completed = True >super(ChangeHeader, self).save(*args,**kwargs) > > but I assume the issue I'm running in to is that the Save handler > doesn't know the current state of the Status field. > > Is there a better approach which avoids hard-coding? If the user had selected a valid ChangeStatus for the Status attribute on the change form; then I would expect you to be able to query the Status in the way you do above (although my situation was slightly different; so just because I would't expect something doesn't mean that is the way it is ;-)). If you run with the above code, do you get an Exception? aid -- 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: pyodbc: Error: ('IM002', '[IM002] [unixODBC][Driver Manager]Data source name not found, and no default driver specified (0) (SQLDriverConnectW)')
On 30 Jan 2011, at 09:31, Orgil wrote:
> My code is below:
> ---
> import pyodbc
> conn = pyodbc.connect('DRIVER={SQL Server};SERVER=testserver
> \mssql2008;DATABASE=eoffice;UID=erp;PWD=123')
> ---
I'd try replacing 'testserver\mssql2008' with a domain name that your ubuntu
box can ping - or the IP address of the Windows 7 box.
Regards,
aid
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Re: XML parsing
Hi,
Remove any carriage returns and extra spaces from your binary.xml file, like so,
http://schemas.xmlsoap.org/soap/envelope/";>http://flypp.infy.com/sms/v2010r1_0";>12345343661t1p1p2DICT
test
Seems to parse ok then.
aid
On 31 Jan 2011, at 09:18, sami nathan wrote:
> HI
>I am trying to parse my response by the following method
> elif request.method == 'POST':
>reqp=request.raw_post_data
>response = HttpResponse(mimetype='text/xml')
>response.write(reqp)
>print response
>xmldoc = minidom.parse('binary.xml')
>reflist = xmldoc.getElementsByTagName('message')
>print reflist
>return response
> But its giving me error of "TO MANY VALUE TO UNPACK"
> --
> my request looks like this
>
> xmlns:soapenv="http://sc
> hemas.xmlsoap.org/soap/envelope/"> xmlns:ns1="http://flypp.infy.com/sms/v2010r1_0";>1234534366<
> applicationID>1t1p
> 1p2DICT
> test ns1:receiveShortMessageRequest>
>
>
> Now my ultimate aim is to send error response to client sending me the above
> request without DICT in message
>
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Adrian Bool
a...@logic.org.uk
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Re: working django with existing database
Arief, Just do a search and replace in your text file along the lines of, search for, max_digits=0, decimal_places=-127 replace, max_digits=10, decimal_places=2 And Django should be happy. aid On 31 Jan 2011, at 10:29, arief nur andono wrote: > oh please... > > i need only to query the table, have no access to modified the table > > why there is no way query the database and then make the output as object?? > > 2011/1/31 Ian > On Jan 31, 12:20 am, arief nur andono > wrote: > > class TempJmlGangguanPyl(models.Model): > > singkatpyl = models.CharField(max_length=4, blank=True) > > singkatgrd = models.CharField(max_length=4, blank=True) > > jumlah_gangguan = models.DecimalField(null=True, max_digits=0, > > decimal_places=-127, blank=True) > > lama_gangguan = models.DecimalField(null=True, max_digits=0, > > decimal_places=-127, blank=True) > > class Meta: > > db_table = u'temp_jml_gangguan_pyl' > > It appears that your numeric columns are all decimal-precision > floating-point (i.e. "FLOAT" or "NUMBER" with no precision or scale > specified), which is a bit of a corner case. The basic problem is > that Django core has no field for this. DecimalField can only handle > fixed-point, and FloatField uses binary precision. If these columns > are actually used to store fixed-point data, you could manually update > the generated models (or preferably the original schema) with the > actual precision and scale of the data. If they are truly floating- > point, then you would probably need to write a custom DecimalField > class. I wouldn't recommend changing them to FloatField in any case, > because that could result in a loss of precision. > > A similar problem can be encountered with "NUMBER(p, s)" columns where > s < 0 or s > p. Both these scenarios are perfectly valid in Oracle, > but are (somewhat arbitrarily) disallowed by the DecimalField > validation. > > I'll make a note of this. At the very least, there should be a more > descriptive error message for when this comes up. > > Ian > > -- > 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. > > > > -- > 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. Adrian Bool a...@logic.org.uk -- 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: DRY and static attribute for multiple classes.
On 1 Feb 2011, at 21:05, Marc Aymerich wrote: > Hi all, > I want to provide an encapsulated static attribute called _registry > for several classes. > > I try to use inheritance in order to make it DRY: all classes inherit > from a BaseClass that implements the _registry encapsulation. But with > inheritance it doesn't work how I want, because a single instance of > the _registry is shared between all of the inherited classes, and I > want to have an independent _registry for every class. > > How can I do that without coping all the code in every class? Create your base class with the _registry attribute like so, class MyBaseClass(models.Model) _registry = models.IntegerField() class Meta: abstract = True With 'abstract = True'; when you inherit from MyBaseClass the child classes will have their own _registry columns in the DB (rather than sharing a common table). Cheers, aid -- 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.
