Newbie question on re-defining an XML schema in my Django data model...
I need some help in resetting my thinking on creating my Django data model. I have previously created an XML schema definition for my data model, but am now trying to re-create this XML data model in a Django data model. In my XML schema, I had defined "collections" of XML complex types that essentially were lists that I could XQuery. I used the ID & IDREF tags to cross-reference my elements between lists of elements. As XML is a tree structure, my schema defined a top down tree hierarchy. Now I want to redefine this in a Django data model. In my primary scenario, I had a list of "singular" workflows that constituted a "parent" element "catalog" of workflows, i.e. a workflow catalog. Multiple "singular" workflows can be combined into an ORDERED collection of workflows called a "reference" or "composite" workflow. There are many composite/reference workflow instances with varied ordered combinations of singular workflows. In my XML hierarchy, I used the IDREF tags to reference the "singular" workflows from within the "reference" workflow element instance. In my Django object model, I can define a class for a "singular" workflow...easy enough, and I can query the Django database model to select the "singular" workflows. But, what is the correct method of now creating the "composite" workflow that will reference the singular workflows? My pseudo-code for the composite workflow class looks like this... class SingularWorkFlow(models.Model): pass class CompositeWorkflow(models.Model): OrderedListOfSingularWorkflows = list(models.ForeignKey(SingularWorkFlow) I don't think this will work, but would kindly ask for some guidance. Regards, Marc -- 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.
Retrieve object model admin userid & password?
I am working through some authentication examples. When I created my first database, I "recall" entering a username and password, but entering what I recall was the correct input is not authenticating. Is there an easy way to reset or retrieve the username and password that was used to create the object model? -- 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.
Conditional choice fields for model specification...
I would like to create my choice fields in my models according to a
hierarchy of conditions, but I can't think of how to specify this in
Python.
For example...pseudo code...
class mod1(models.Model):
choice1 = (
('a', 'A'),
('b', 'B'),
)
choice_a_if_choice1_is_A = (
('z', 'Z'),
('y', 'Y'),
)
choice_a_if_choice1_is_B = (
('m', 'M')
('n', 'N')
)
field1 = models.CharField(max_length=30, choices=choice1)
field2 = models.CharField(max_length=30,
choices=choice_a_if_choice_is_A)
field2 = models.CharField(max_length=30,
choices=choice_a_if_choice_is_B)
How to I set the choices for field2 depending upon the choice
selected in field1??
)
Kind regards, Marc
--
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Can't figure out this Object Model error...
I am receiving the following error from the Django debugger when trying to access my EDA_AppCatalog object. My class definition is listed below. I have similar defined objects that are working with no problem, but I can't seem to clear this error message up. Thanks, Marc DatabaseError at /admin/bookmarks/eda_appcatalog/ no such column: bookmarks_eda_appcatalog.eda_app_id Request Method: GET Request URL:http://127.0.0.1:8000/admin/bookmarks/eda_appcatalog/ Django Version: 1.3.1 Exception Type: DatabaseError Exception Value: no such column: bookmarks_eda_appcatalog.eda_app_id Exception Location: /usr/local/lib/python2.7/dist-packages/django/db/ backends/sqlite3/base.py in execute, line 234 Python Executable: /usr/bin/python2.7 Python Version: 2.7.2 Python Path: ['/home/jmarcedwards/KeplerDevelopment/KeplerEclipseWorkspace_1/ JME_DjangoTest_3/JME_DjangoTest_3', --- class EDA_AppCatalog(models.Model): eda_app = models.ForeignKey(EDA_App) -- 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: Can't figure out this Object Model error...
This is the confusing part. The EDA_AppCatalog imports from my bookmarks package with no problem. My class definition for this object has only 1 field, a ForeignKey to an EDA_App. I know that the Object Model will be assigning an "ID" to the object when it stores it in the database. I am not assigning this ID field. I also don't do this with the other object models that I have defined and that seem to be working with no problem. Why the admin interface is complaining about this ID field even before I have even instantiated an initial EDA_App to add to the database is strange. I have another object defined that uses 3 ForeignKeys that works fine in the admin interface. There seems to be something that is not defined properly with this particular object definition. -M On Dec 5, 7:35 pm, Furbee wrote: > Don't mean to sound remedial, but have you verified that the table > bookmarks_eda_appcatalog exists and that it has a column named eda_app_id? > Did you use manage.py syncdb to create the tables? > > Furbee > > > > > > > > On Mon, Dec 5, 2011 at 4:24 PM, Marc Edwards wrote: > > I am receiving the following error from the Django debugger when > > trying to access my EDA_AppCatalog object. My class definition is > > listed below. I have similar defined objects that are working with no > > problem, but I can't seem to clear this error message up. > > > Thanks, Marc > > > DatabaseError at /admin/bookmarks/eda_appcatalog/ > > > no such column: bookmarks_eda_appcatalog.eda_app_id > > > Request Method: GET > > Request URL: http://127.0.0.1:8000/admin/bookmarks/eda_appcatalog/ > > Django Version: 1.3.1 > > Exception Type: DatabaseError > > Exception Value: > > > no such column: bookmarks_eda_appcatalog.eda_app_id > > > Exception Location: /usr/local/lib/python2.7/dist-packages/django/db/ > > backends/sqlite3/base.py in execute, line 234 > > Python Executable: /usr/bin/python2.7 > > Python Version: 2.7.2 > > Python Path: > > > ['/home/jmarcedwards/KeplerDevelopment/KeplerEclipseWorkspace_1/ > > JME_DjangoTest_3/JME_DjangoTest_3', > > > --- > > > > class EDA_AppCatalog(models.Model): > > > eda_app = models.ForeignKey(EDA_App) > > > -- > > 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.
Re: Can't figure out this Object Model error...
Even when I delete the class definition, re-sync the db, and then re- define the object, even setting the primary_key=True on a new CharField, I am still receiving the same error. Can it be that the database file has become corrupted in some manner for this object. I have not modified my SQLite file in any way except through the admin interface. On Dec 5, 7:44 pm, Marc Edwards wrote: > This is the confusing part. The EDA_AppCatalog imports from my > bookmarks package with no problem. > > My class definition for this object has only 1 field, a ForeignKey to > an EDA_App. > > I know that the Object Model will be assigning an "ID" to the object > when it stores it in the database. I am not assigning this ID field. > I also don't do this with the other object models that I have defined > and that seem to be working with no problem. > > Why the admin interface is complaining about this ID field even before > I have even instantiated an initial EDA_App to add to the database is > strange. > > I have another object defined that uses 3 ForeignKeys that works fine > in the admin interface. There seems to be something that is not > defined properly with this particular object definition. > > -M > > On Dec 5, 7:35 pm, Furbee wrote: > > > > > > > > > Don't mean to sound remedial, but have you verified that the table > > bookmarks_eda_appcatalog exists and that it has a column named eda_app_id? > > Did you use manage.py syncdb to create the tables? > > > Furbee > > > On Mon, Dec 5, 2011 at 4:24 PM, Marc Edwards wrote: > > > I am receiving the following error from the Django debugger when > > > trying to access my EDA_AppCatalog object. My class definition is > > > listed below. I have similar defined objects that are working with no > > > problem, but I can't seem to clear this error message up. > > > > Thanks, Marc > > > > DatabaseError at /admin/bookmarks/eda_appcatalog/ > > > > no such column: bookmarks_eda_appcatalog.eda_app_id > > > > Request Method: GET > > > Request URL: http://127.0.0.1:8000/admin/bookmarks/eda_appcatalog/ > > > Django Version: 1.3.1 > > > Exception Type: DatabaseError > > > Exception Value: > > > > no such column: bookmarks_eda_appcatalog.eda_app_id > > > > Exception Location: /usr/local/lib/python2.7/dist-packages/django/db/ > > > backends/sqlite3/base.py in execute, line 234 > > > Python Executable: /usr/bin/python2.7 > > > Python Version: 2.7.2 > > > Python Path: > > > > ['/home/jmarcedwards/KeplerDevelopment/KeplerEclipseWorkspace_1/ > > > JME_DjangoTest_3/JME_DjangoTest_3', > > > > --- > > > > > > class EDA_AppCatalog(models.Model): > > > > eda_app = models.ForeignKey(EDA_App) > > > > -- > > > 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.
Novice question...generation of model instances within a "for" loop...
I'm looping through some JSON code and generating new model instances that I am saving to my SQLiteDB. I have 3 nested "for" loops, like this: for app_catalog in JSON_Object['TSS_WorkFlow_Catalog']: for eda_app in app_catalog['EDA_App']: new_eda_app = EDA_App(isv=eda_app['isv'],product_name=eda_app['product_name'],product_category=eda_app['product_category']) new_eda_app.save() for cmd_str_pair in eda_app['CmdStrings']['name']: new_cmd_str = CmdString(name=cmd_str_pair['Text'], cmd=cmd_str_pair['Value'], eda_app=new_eda_app) new_cmd_str.save() My JSON file has 3 EDA_Apps, each EDA_App has 3, 4, and 4 CmdStrings, respectively. When I loop through the second nested "for" for the CmdStrings, the .save() is writing to the same model instance, and at the end of the first nested "for" loop, I have 3 CmdString instances in my DB, NOT 3+4+4=11 CmdString instances. Since my DB has 3 EDA_Apps in the DB, I think that the code to instance a new model instance to save to the DB seems to be fine. It is the second nested "for" loop that is acting fishy. I don't know if I am doing something wrong relative to the DB, because when I step through the 2nd nested "for" loop, I can clearly see the "new_cmd_str" taking on the next values from the JSON file. This must be related to how I am using the DB. When I try using .save(force_insert=True) on the 2nd nested "for" loop save, I receive a Django DEBUG page message that the "PRIMARY KEY must be unique". Any thoughts or observations on my code style? Kind regards, Marc -- 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: Novice question...generation of model instances within a "for" loop...
OK...adding the .pk=None did the trick, but why didn't I have to do this on the outer loop? On Dec 8, 11:52 am, Brian Schott wrote: > I think if you set: > new_cmd_str.id = None > That will force a new PK. Not sure why the CmdString constructor isn't > clearing this.. > > Brian Schott > bfsch...@gmail.com > > On Dec 8, 2011, at 11:47 AM, Marc Edwards wrote: > > > > > > > > > I'm looping through some JSON code and generating new model instances > > that I am saving to my SQLiteDB. > > > I have 3 nested "for" loops, like this: > > > for app_catalog in JSON_Object['TSS_WorkFlow_Catalog']: > > for eda_app in app_catalog['EDA_App']: > > new_eda_app = > > EDA_App(isv=eda_app['isv'],product_name=eda_app['product_name'],product_cat > > egory=eda_app['product_category']) > > new_eda_app.save() > > for cmd_str_pair in eda_app['CmdStrings']['name']: > > new_cmd_str = CmdString(name=cmd_str_pair['Text'], > > cmd=cmd_str_pair['Value'], eda_app=new_eda_app) > > new_cmd_str.save() > > > My JSON file has 3 EDA_Apps, each EDA_App has 3, 4, and 4 CmdStrings, > > respectively. > > > When I loop through the second nested "for" for the CmdStrings, > > the .save() is writing to the same model instance, and at the end of > > the first nested "for" loop, I have 3 CmdString instances in my DB, > > NOT 3+4+4=11 CmdString instances. > > > Since my DB has 3 EDA_Apps in the DB, I think that the code to > > instance a new model instance to save to the DB seems to be fine. It > > is the second nested "for" loop that is acting fishy. I don't know if > > I am doing something wrong relative to the DB, because when I step > > through the 2nd nested "for" loop, I can clearly see the "new_cmd_str" > > taking on the next values from the JSON file. This must be related to > > how I am using the DB. > > > When I try using .save(force_insert=True) on the 2nd nested "for" loop > > save, I receive a Django DEBUG page message that the "PRIMARY KEY must > > be unique". > > > Any thoughts or observations on my code style? > > > Kind regards, Marc > > > -- > > 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 > > athttp://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.
Re: Novice question...generation of model instances within a "for" loop...
Probably because I was setting the ForeignKey in the new_cmd_str to the new_eda_app that was in the outer loop. That created a new reference to that instance, and Django knew that it would need a new primary key. Without another external reference, it seemed that the same primary key was being used. Would this explanation seem reasonable? On Dec 8, 12:52 pm, Marc Edwards wrote: > OK...adding the .pk=None did the trick, but why didn't I have to do > this on the outer loop? > > On Dec 8, 11:52 am, Brian Schott wrote: > > > > > > > > > I think if you set: > > new_cmd_str.id = None > > That will force a new PK. Not sure why the CmdString constructor isn't > > clearing this.. > > > Brian Schott > > bfsch...@gmail.com > > > On Dec 8, 2011, at 11:47 AM, Marc Edwards wrote: > > > > I'm looping through some JSON code and generating new model instances > > > that I am saving to my SQLiteDB. > > > > I have 3 nested "for" loops, like this: > > > > for app_catalog in JSON_Object['TSS_WorkFlow_Catalog']: > > > for eda_app in app_catalog['EDA_App']: > > > new_eda_app = > > > EDA_App(isv=eda_app['isv'],product_name=eda_app['product_name'],product_cat > > > egory=eda_app['product_category']) > > > new_eda_app.save() > > > for cmd_str_pair in eda_app['CmdStrings']['name']: > > > new_cmd_str = CmdString(name=cmd_str_pair['Text'], > > > cmd=cmd_str_pair['Value'], eda_app=new_eda_app) > > > new_cmd_str.save() > > > > My JSON file has 3 EDA_Apps, each EDA_App has 3, 4, and 4 CmdStrings, > > > respectively. > > > > When I loop through the second nested "for" for the CmdStrings, > > > the .save() is writing to the same model instance, and at the end of > > > the first nested "for" loop, I have 3 CmdString instances in my DB, > > > NOT 3+4+4=11 CmdString instances. > > > > Since my DB has 3 EDA_Apps in the DB, I think that the code to > > > instance a new model instance to save to the DB seems to be fine. It > > > is the second nested "for" loop that is acting fishy. I don't know if > > > I am doing something wrong relative to the DB, because when I step > > > through the 2nd nested "for" loop, I can clearly see the "new_cmd_str" > > > taking on the next values from the JSON file. This must be related to > > > how I am using the DB. > > > > When I try using .save(force_insert=True) on the 2nd nested "for" loop > > > save, I receive a Django DEBUG page message that the "PRIMARY KEY must > > > be unique". > > > > Any thoughts or observations on my code style? > > > > Kind regards, Marc > > > > -- > > > 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 > > > athttp://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.
Re: Novice question...generation of model instances within a "for" loop...
Brian: I'm getting the same behavior with the new_cmd_str.id=None. I'm going to get a sandwich. I'll think about this more over lunch. -M J. Marc Edwards Lead Architect - Semiconductor Design Portals Nimbis Services, Inc. Skype: (919) 747-3775 Cell: (919) 345-1021 Fax: (919) 882-8602 marc.edwa...@nimbisservices.com www.nimbisservices.com On 12/08/2011 11:52 AM, Brian Schott wrote: > I think if you set: > new_cmd_str.id = None > That will force a new PK. Not sure why the CmdString constructor isn't > clearing this.. > > Brian Schott > bfsch...@gmail.com > > > > On Dec 8, 2011, at 11:47 AM, Marc Edwards wrote: > >> I'm looping through some JSON code and generating new model instances >> that I am saving to my SQLiteDB. >> >> I have 3 nested "for" loops, like this: >> >>for app_catalog in JSON_Object['TSS_WorkFlow_Catalog']: >>for eda_app in app_catalog['EDA_App']: >>new_eda_app = >> EDA_App(isv=eda_app['isv'],product_name=eda_app['product_name'],product_category=eda_app['product_category']) >>new_eda_app.save() >>for cmd_str_pair in eda_app['CmdStrings']['name']: >>new_cmd_str = CmdString(name=cmd_str_pair['Text'], >> cmd=cmd_str_pair['Value'], eda_app=new_eda_app) >>new_cmd_str.save() >> >> My JSON file has 3 EDA_Apps, each EDA_App has 3, 4, and 4 CmdStrings, >> respectively. >> >> When I loop through the second nested "for" for the CmdStrings, >> the .save() is writing to the same model instance, and at the end of >> the first nested "for" loop, I have 3 CmdString instances in my DB, >> NOT 3+4+4=11 CmdString instances. >> >> Since my DB has 3 EDA_Apps in the DB, I think that the code to >> instance a new model instance to save to the DB seems to be fine. It >> is the second nested "for" loop that is acting fishy. I don't know if >> I am doing something wrong relative to the DB, because when I step >> through the 2nd nested "for" loop, I can clearly see the "new_cmd_str" >> taking on the next values from the JSON file. This must be related to >> how I am using the DB. >> >> When I try using .save(force_insert=True) on the 2nd nested "for" loop >> save, I receive a Django DEBUG page message that the "PRIMARY KEY must >> be unique". >> >> Any thoughts or observations on my code style? >> >> Kind regards, Marc >> >> -- >> 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. <>
Newbie question on forms using ChoiceField and "choices" field...
OK...I have the following model and form. *class CmdString(models.Model): name= models.CharField(max_length=50) cmd = models.CharField(max_length=200) eda_app = models.OneToOneField(EDA_App, primary_key=True) def __unicode__(self): return self.cmd * *class EDA_AppSelectForm(forms.Form): app_cmds = forms.ChoiceField(choices=CmdString.objects.all())* When I execute in a Python shell the command *CmdString.objects.all() *I have the following output from my database: *[, , , , , , , , , , , , , , , , , , , , '...(remaining elements truncated)...']* This is a Python list with each element of the list a dictionary object. This list is iterable, i.e. *for cmd in CmdString.objects.all(): ... print cmd* will return: *vcs -a -b -f $FILE1 vhdlan -c -d -f $FILE2 virsim -c -d -f $FILE2 virt1 -a -b -f $FILE1 virt2 -c -d -f $FILE2 virt3 -c -d -f $FILE2 virt4 -c -d -f $FILE2 quartz1 -c -d -f $FILE2 quartz2 -c -d -f $FILE2 quartz3 -a -b -f $FILE1 quartz4 -c -d -f $FILE2 vcs -a -b -f $FILE1 vhdlan -c -d -f $FILE2 virsim -c -d -f $FILE2 virt1 -a -b -f $FILE1 virt2 -c -d -f $FILE2 virt3 -c -d -f $FILE2 virt4 -c -d -f $FILE2 quartz1 -c -d -f $FILE2 quartz2 -c -d -f $FILE2 quartz3 -a -b -f $FILE1 quartz4 -c -d -f $FILE2* Perfect...I'm happy with this output. Now, I wanted to take a look at the output of the form to be rendered as HTML. I'm looking for the "select" tag with the options to be output from the "choices" kwarg. When I to take a look at an instance of this form with "as_p", I get the following error: *>>> samp2 = EDA_AppSelectForm() >>> type(samp2) * *>>> samp2.as_p() >>> Traceback (most recent call last): File "", line 1, in File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 235, in as_p errors_on_separate_row = True) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 180, in _html_output 'field': unicode(bf), File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 408, in __unicode__ return self.as_widget() File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 439, in as_widget return widget.render(name, self.value(), attrs=attrs) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/widgets.py", line 516, in render options = self.render_options(choices, [value]) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/widgets.py", line 533, in render_options for option_value, option_label in chain(self.choices, choices): TypeError: 'CmdString' object is not iterable >>> print samp2.as_p() >>> Traceback (most recent call last): File "", line 1, in File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 235, in as_p errors_on_separate_row = True) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 180, in _html_output 'field': unicode(bf), File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 408, in __unicode__ return self.as_widget() File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/forms.py", line 439, in as_widget return widget.render(name, self.value(), attrs=attrs) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/widgets.py", line 516, in render options = self.render_options(choices, [value]) File "/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/forms/widgets.py", line 533, in render_options for option_value, option_label in chain(self.choices, choices): TypeError: 'CmdString' object is not iterable* What is happening here? When I remove the "choices" kwarg in the "app_cmds" form field, I can print out the HTML, and the "select" field is there as expected, but I have no choices/options in the HTML. This is why I was trying to use the "choices" attribute. Now the documentation for the "choices" kwarg is pasted below, which says that this should be an iterable of 2-tuples, but seems to indicate it could be a list as well. The documentation here seems unclear, i.e an iterable of tuple of 2-tuples, I don't think it means this. How can I achieve what I am looking for here? > choices > An iterable _/(e.g., a list or tuple) /_of 2-tuples to use as choices > for this field. This argument accepts the > same formats as the choices argument to a model field. See the model > field reference documenta
Problem re-creating SQLite database with manage.py...
I'm trying to re-create my SQLite database from scratch. I am now
receiving the following error message from:
*(foobarr)jmarcedwards@jmelinux:~/git/django-nimbis/projects/dev$ python
./manage.py syncdb
Syncing...
Traceback (most recent call last):
File "./manage.py", line 11, in
execute_manager(settings)
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/core/management/__init__.py",
line 438, in execute_manager
utility.execute()
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/core/management/__init__.py",
line 379, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/core/management/base.py",
line 191, in run_from_argv
self.execute(*args, **options.__dict__)
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/core/management/base.py",
line 220, in execute
output = self.handle(*args, **options)
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/core/management/base.py",
line 351, in handle
return self.handle_noargs(**options)
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/South-0.7.3-py2.7.egg/south/management/commands/syncdb.py",
line 87, in handle_noargs
db.connection_init()
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/South-0.7.3-py2.7.egg/south/db/mysql.py",
line 38, in connection_init
cursor = self._get_connection().cursor()
File
"/home/jmarcedwards/foobarr/local/lib/python2.7/site-packages/django/db/backends/dummy/base.py",
line 15, in complain
raise ImproperlyConfigured("You haven't set the database ENGINE
setting yet.")
django.core.exceptions.ImproperlyConfigured: You haven't set the
database ENGINE setting yet.
(foobarr)jmarcedwards@jmelinux:~/git/django-nimbis/projects/dev$ ls
celerybeat-schedule fixtures __init__.pycmanage.py
requirements.txt settings.py static urls.pyc
env.sh __init__.py local_tests.py manage.pyc
run_server.sh settings.pyc urls.py
(foobarr)jmarcedwards@jmelinux:~/git/django-nimbis/projects/dev$
*
sys.path.insert(0, PROJDIR)
sys.path.insert(0, APPSDIR)
# sqlite database for testing
DATABASES = {
'default': {
'ENGINE': 'django.db.backends.sqlite3',
'NAME': os.path.join(PROJDIR, 'db.sqlite'),
}
}
SOUTH_DATABASE_ADAPTERS = {
'default': "south.db.mysql"
}
I am working from within the directory where my manage.py and
settings.py are located. It seems that perhaps my settings.py file is
not being picked up from the same directory. I'm getting the same error
inside of my Eclipse Django PyDev environment.
Are there any more direct, manual process for invoking the syncdb with
the settings.py and generating my database?
Regards, Marc
--
J. Marc Edwards
Lead Architect - Semiconductor Design Portals
Nimbis Services, Inc.
Skype: (919) 747-3775
Cell: (919) 345-1021
Fax: (919) 882-8602
marc.edwa...@nimbisservices.com
www.nimbisservices.com
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<>
Can I escape or delimit the "{{ }}" template braces?
I'd like to write a paragraph on my page that describes the {{ }}
syntax, but the template interpreter is interpreting the braces.
--
J. Marc Edwards
Lead Architect - Semiconductor Design Portals
Nimbis Services, Inc.
Skype: (919) 747-3775
Cell: (919) 345-1021
Fax: (919) 882-8602
marc.edwa...@nimbisservices.com
www.nimbisservices.com
--
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