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  • django : Serving static files through nginx

    - by PlanetUnknown
    I'm using apache+mod_wsgi for django. And all css/js/images are served through nginx. For some odd, reason when others/friends/colleagues try accessing the site, jquery/css is not getting loaded for them, hence the page looks jumbled up. My html files use code like this - <link rel="stylesheet" type="text/css" href="http://x.x.x.x:8000/css/custom.css"/> <script type="text/javascript" src="http://1x.x.x.x:8000/js/custom.js"></script> My nginx configuration in sites-available is like this - server { listen 8000; server_name localhost; access_log /var/log/nginx/aa8000.access.log; error_log /var/log/nginx/aa8000.error.log; location / { index index.html index.htm; } location /static/ { autoindex on; root /opt/aa/webroot/; } } There is a directory /opt/aa/webroot/static/ which have corresponding css & js directories. The odd thing is that the pages show fine when I access them. I have cleared my cache/etc, but the page loads fine for me, from various browsers. Also, I don't see any 404 any error in the nginx log files. Actually the logs for nginx are not getting refreshed at all. I restarted the nginx server using root, is that incorrect ? There is a user www-data defined in the nginx configuration file. Any pointers would be great.

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  • Django query: Count and Group BY

    - by Tyler Lane
    I have a query that I'm trying to figure the "django" way of doing it: I want to take the last 100 calls from Call. Which is easy: calls = Call.objects.all().order_by('-call_time')[:100] However the next part I can't find the way to do it via django's ORM. I want to get a list of the call_types and the number of calls each one has WITHIN that previous queryset i just did. Normally i would do a query like this: "SELECT COUNT(id),calltype FROM call WHERE id IN ( SELECT id FROM call ORDER BY call_time DESC LIMIT 100 ) GROUP BY calltype;" I can't seem to find the django way of doing this particular query. Here are my 2 models: class Call( models.Model ): call_time = models.DateTimeField( "Call Time", auto_now = False, auto_now_add = False ) description = models.CharField( max_length = 150 ) response = models.CharField( max_length = 50 ) event_num = models.CharField( max_length = 20 ) report_num = models.CharField( max_length = 20 ) address = models.CharField( max_length = 150 ) zip_code = models.CharField( max_length = 10 ) geom = models.PointField(srid=4326) calltype = models.ForeignKey(CallType) objects = models.GeoManager() class CallType( models.Model ): name = models.CharField( max_length = 50 ) description = models.CharField( max_length = 150 ) active = models.BooleanField() time_init = models.DateTimeField( "Date Added", auto_now = False, auto_now_add = True ) objects = models.Manager()

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  • Django and mod_python config

    - by Peter
    My Django project is placed in /www/host1/htdocs/my/project, www and my are links to other actual folders. Apache has mod_python enabled. I have a .htaccess in project folder: SetHandler python-program PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE project.settings PythonDebug On PythonOption django.root /my/project PythonPath "['/www/host1/htdocs/my/project'] + sys.path" I suppose my site should be accessible from http://host1/my/project, but I see the following error: ImportError: Could not import settings 'project.settings' (Is it on sys.path? Does it have syntax errors?): No module named project.settings Can somebody give any suggestions?

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  • django app organization

    - by iHeartDucks
    I have been reading some django tutorial and it seems like all the view functions have to go in a file called "views.py" and all the models go in "models.py". I fear that I might end up with a lot of view functions in my view.py file and the same is the case with models.py. Is my understanding of django apps correct? Django apps lets us separate common functionality into different apps and keep the file size of views and models to a minimum? For example: My project can contain an app for recipes (create, update, view, and search) and a friend app, the comments app, and so on. Can I still move some of my view functions to a different file? So I only have the CRUD in one single file?

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  • Django doctests in views.py

    - by Brian M. Hunt
    The Django documentation on tests states: For a given Django application, the test runner looks for doctests in two places: The models.py file. You can define module-level doctests and/or a doctest for individual models. It's common practice to put application-level doctests in the module docstring and model-level doctests in the model docstrings. A file called tests.py in the application directory -- i.e., the directory that holds models.py. This file is a hook for any and all doctests you want to write that aren't necessarily related to models. Out of curiosity I'd like to know why Django's testrunner is limited to the doctests in models.py, but more practically I'd like to know how one could expand the testrunner's doctests to include (for example) views.py and other modules when running manage.py test. I'd be grateful for any input. Thank you. Brian

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  • integrating django-paypal

    - by ramdaz
    Hi I am trying to integrate Django-Paypal onto a site, where once the customer makes a payment I need to send a dynamic URL from which the user can download some specific information. I am registering all users to a URL which allows them to buy the document. The URL which can only be accessed by a registered user with a verified email ids using django-registration, allows the user to connect to paypal and make payment. How do I capture the signal and verify which user has made the payment for which product he/she is purchased, based on this information 1) I need to know two information, which user made the payment, and for which product did he or she make the payment? 2) Only if I have these information can I send the right UR:L by email. Any help appreciated. I am not very sure how django-signals work. What all details does payment_was_successful signal return? I am using IPN

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  • Testing Django web app with Hudson and Selenium

    - by ycseattle
    This might be a newbie question for Hudson. I am trying to setup Selenium testing for my Django website in my Hudson CI server. The question is, the Hudson will use subversion to checkout my Django code into its own path, how do I "deploy" the code into the same server for testing? This is not a question about deploying django, but instead how to access the source file in hudson workspace. Most tutorials/blogs is about building and running tests, but I couldn't find useful information about how to setup the web application on the server to run the test against. 1) Should I write some shell script to copy the source files from the hudson workspace? Is there an environment variable to use to access the workspace? 2) Is there a tutorial on how to grab web app files in hudson workspace and deploy them? I am sure this apply for other technologies like PHP as well. Thanks!

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  • Dynamic model choice field in django formset using multiple select elements

    - by Aryeh Leib Taurog
    I posted this question on the django-users list, but haven't had a reply there yet. I have models that look something like this: class ProductGroup(models.Model): name = models.CharField(max_length=10, primary_key=True) def __unicode__(self): return self.name class ProductRun(models.Model): date = models.DateField(primary_key=True) def __unicode__(self): return self.date.isoformat() class CatalogItem(models.Model): cid = models.CharField(max_length=25, primary_key=True) group = models.ForeignKey(ProductGroup) run = models.ForeignKey(ProductRun) pnumber = models.IntegerField() def __unicode__(self): return self.cid class Meta: unique_together = ('group', 'run', 'pnumber') class Transaction(models.Model): timestamp = models.DateTimeField() user = models.ForeignKey(User) item = models.ForeignKey(CatalogItem) quantity = models.IntegerField() price = models.FloatField() Let's say there are about 10 ProductGroups and 10-20 relevant ProductRuns at any given time. Each group has 20-200 distinct product numbers (pnumber), so there are at least a few thousand CatalogItems. I am working on formsets for the Transaction model. Instead of a single select menu with the several thousand CatalogItems for the ForeignKey field, I want to substitute three drop-down menus, for group, run, and pnumber, which uniquely identify the CatalogItem. I'd also like to limit the choices in the second two drop-downs to those runs and pnumbers which are available for the currently selected product group (I can update them via AJAX if the user changes the product group, but it's important that the initial page load as described without relying on AJAX). What's the best way to do this? As a point of departure, here's what I've tried/considered so far: My first approach was to exclude the item foreign key field from the form, add the substitute dropdowns by overriding the add_fields method of the formset, and then extract the data and populate the fields manually on the model instances before saving them. It's straightforward and pretty simple, but it's not very reusable and I don't think it is the right way to do this. My second approach was to create a new field which inherits both MultiValueField and ModelChoiceField, and a corresponding MultiWidget subclass. This seems like the right approach. As Malcolm Tredinnick put it in a django-users discussion, "the 'smarts' of a field lie in the Field class." The problem I'm having is when/where to fetch the lists of choices from the db. The code I have now does it in the Field's __init__, but that means I have to know which ProductGroup I'm dealing with before I can even define the Form class, since I have to instantiate the Field when I define the form. So I have a factory function which I call at the last minute from my view--after I know what CatalogItems I have and which product group they're in--to create form/formset classes and instantiate them. It works, but I wonder if there's a better way. After all, the field should be able to determine the correct choices much later on, once it knows its current value. Another problem is that my implementation limits the entire formset to transactions relating to (CatalogItems from) a single ProductGroup. A third possibility I'm entertaining is to put it all in the Widget class. Once I have the related model instance, or the cid, or whatever the widget is given, I can get the ProductGroup and construct the drop-downs. This would solve the issues with my second approach, but doesn't seem like the right approach.

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  • Django on Dreamhost - testing/sand box environment

    - by elequ
    I've been using webfaction for all my django needs for the last couple of years but have had a high traffic site fall in my lap that dreamhost are probably better suited to handling. To set up and experiment with a site with webfaction there are your [user].webfactional.com accounts. Which is sweet. Equivalently Dreamhost also offers [name].dreamhosters.com Yet this doesn't seem to work with my database and this doesn't seem to be playing nice with setting up django or passenger_wsgi. So I'm wondering if I'm interpreting the documentation correctly to suggest that in order to make a site that depends on passenger it needs to be running from an active domain. The documentation is really implicit, I'm baffled. Has anyone set up a testing/sand box environment for django using dreamhost, or know how to?

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  • Best Approach to process images in Django

    - by primalpop
    I've have an application with Android front end and Django as the back end. As part of the answers here, I'm confused over the approach which I should take to send images to Django Server. I've 2 options at my disposal as Piro pointed out there. 1) Sending images as Multi Part entity 2) Send image as a String after encoding it using Base 64. So I am considering the approach that would make it easy to be processed by Django. The images are small in size (<200kb) and number (<10). Any suggestions or pointers are most welcome.

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  • django : nginx : jquery css not being served

    - by PlanetUnknown
    I'm using apache+mod_wsgi for django. And all css/js/images are served through nginx. For some odd, reason when others/friends/colleagues try accessing the site, jquery/css is not getting loaded for them, hence the page looks jumbled up. My html files use code like this - <link rel="stylesheet" type="text/css" href="http://x.x.x.x:8000/css/custom.css"/> <script type="text/javascript" src="http://1x.x.x.x:8000/js/custom.js"></script> My nginx configuration in sites-available is like this - server { listen 8000; server_name localhost; access_log /var/log/nginx/aa8000.access.log; error_log /var/log/nginx/aa8000.error.log; location / { index index.html index.htm; } location /static/ { autoindex on; root /opt/aa/webroot/; } } There is a directory /opt/aa/webroot/static/ which have corresponding css & js directories. The odd thing is that the pages show fine when I access them. I have cleared my cache/etc, but the page loads fine for me, from various browsers. Also, I don't see any 404 any error in the nginx log files. Actually the logs for nginx are not getting refreshed at all. I restarted the nginx server using root, is that incorrect ? There is a user www-data defined in the nginx configuration file. Any pointers would be great.

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  • django admin how to limit selectbox values

    - by SledgehammerPL
    model: class Store(models.Model): name = models.CharField(max_length = 20) class Admin: pass def __unicode__(self): return self.name class Stock(Store): products = models.ManyToManyField(Product) class Admin: pass def __unicode__(self): return self.name class Product(models.Model): name = models.CharField(max_length = 128, unique = True) parent = models.ForeignKey('self', null = True, blank = True, related_name='children') (...) def __unicode__(self): return self.name mptt.register(Product, order_insertion_by = ['name']) admin.py: from bar.drinkstore.models import Store, Stock from django.contrib import admin admin.site.register(Store) admin.site.register(Stock) Now when I look at admin site I can select any product from the list. But I'd like to have a limited choice - only leaves. In mptt class there's function: is_leaf_node() -- returns True if the model instance is a leaf node (it has no children), False otherwise. But I have no idea how to connect it I'm trying to make a subclass: in admin.py: from bar.drinkstore.models import Store, Stock from django.contrib import admin admin.site.register(Store) class StockAdmin(admin.ModelAdmin): def queryset(self, request): qs = super(StockAdmin, self).queryset(request).filter(ihavenoideawhatfilter) admin.site.register(Stock, StockAdmin) but I'm not sure if it's right way, and what filter set.

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  • QAbstractTableModel as a model for one QTableView and few QListViews

    - by ??????
    community. Briefly. I wrote usual model over QAbstractTableModel and using it in usual way for QTableView. But I think I need to use some columns of this model for the few QListViews in QWizard to fill main table in the right way (for user). For example: use the column2 as the QListView's model on the page1 of the wizard; column3 for page2 for its QListView etc. Please, help me to understand just two things: Am I on the right way? If yes then how can I make it simply and explicitly?

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  • Specialization hierarchy in a domain-model

    - by devoured elysium
    I'm trying to make the domain model of a management system. I have the following kinds of persons in this system: employee manager top mananger I decided to define a User, from where employee, manager and top manager will specialize from. What I don't know is what kind of specialization hierarchy I should choose from. I thought of two ways: or Which might be preferable and why? As a long time coder, every time I try to do a domain-model, I have to fight against the idea of trying to think in how I'm going to code this. From what I've understood, I should not think about those matters in the domain-model, only in object relationships. I don't have to think of code duplication or any of these kind of details here, so I can't really pick any of the options over the other. Thanks

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  • Hosting django backend for iPhone / Android app

    - by Ashok Fernandez
    I am looking to make an iPhone / Android app for my university using the Appcelerator Titanium framework. The app will rely heavily on a server backend which will pull information from other sites, figuring out what is relevant to the user then deliver the content. Some of the information is individual to the user (calendar data), other bits are updates frequently but are shared (bus timetables) and others are static and the same for everyone (magazine articles). I was going to use django as I am fairly proficent in python so I thought it would save time. My question is, which hosting services do you recommend to host the server backend? I am expecting about 9000 people to use the app with very random spikes in traffic, but unfortunately I have very little to go on at this stage. I have heard a lot about Webfaction, is it suitable for something like this or am I likely to need something bigger? I don't really want to fork out for a VPS at this stage. What about Amazons EC2? Would that be more suitable than Webfaction? Sorry for the fairly open ended question, Im sort of new to this so I open to all suggestions.

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  • Django crash without message

    - by schneck
    Hi there, I have a Django-project which was running fine, until I made "some changes" I don't remember exactly. Since that point, on every request, my project crashes silently: $ ./manage.py runserver -v2 Validating models... 0 errors found Django version 1.1.1, using settings 'src.settings' Development server is running at http://127.0.0.1:8000/ Quit the server with CONTROL-C. After I requested a page (admin oder frontend), it returns to the prompt. I did not find any other option to get verbose output than -v2 - is there any logfile I can use? I'm using Django 1.1.1 on a Mac OS X 10.6 with virtualenv, Python 2.6 Thanks a lot.

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  • Django: Admin with multiple sites & languages

    - by lazerscience
    Hi everybody! I'm supposed to build some Django apps, that allow you to administer multiple sites through one backend. The contrib.sites framework is quite perfect for my purposes. I can run multiple instances of manage.py with different settings for each site; but how should django's admin deal with different settings for different sites, eg. if they have different sets of languages, a different (default) language? So there are some problem s to face if you have to work on objects coming from different sites in one admin... I think settings.ADMIN_FOR is supposed to be quite helpful for cases like this, but theres hardly any documentation about it and I think it's not really used in the actual Django version (?). So any ideas/solutions are welcome and much appreciated! Thanks a lot...

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  • django left join with null

    - by SledgehammerPL
    The model: class Product(models.Model): name = models.CharField(max_length = 128) def __unicode__(self): return self.name class Receipt(models.Model): name = models.CharField(max_length=128) components = models.ManyToManyField(Product, through='ReceiptComponent') class Admin: pass def __unicode__(self): return self.name class ReceiptComponent(models.Model): product = models.ForeignKey(Product) receipt = models.ForeignKey(Receipt) quantity = models.FloatField(max_length=9) unit = models.ForeignKey(Unit) def __unicode__(self): return unicode(self.quantity!=0 and self.quantity or '') + ' ' + unicode(self.unit) + ' ' + self.product.genitive The idea: there are a components on stock. I'd like to find out which recipes I can made with components which I have. It's not easy - but possible - I made a SQL view, which gets the solution. But I'm learning python and Django so I'd like to make it Django-style ;D The concept of solution: get the set of recipes which has at last one component: list_of_available_components = ReceiptComponent.objects.filter(product__in=list_of_available_products).distinct() list_of_related_receipts = Receipt.objects.filter(receiptcomponent__in = list_of_available_components).distinct() get recipes (from list_of_related_receipts) which has not at last one component list_of_incomplete_recipes = (SELECT * FROM drinkbook_receiptcomponent LEFT JOIN drinkstore_stock_products USING(product_id) WHERE drinkstore_stock_products.stock_id IS NULL AND receipt_id IN (SELECT receipt_id FROM drinkbook_receiptcomponent JOIN drinkstore_stock_products USING(product_id))) get recipes (from list_of_related_receipts) which are not in "list_of_incomplete_recipes"

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  • Rails - building an absolute url in a model's virtual attribute without url helper

    - by Nick
    I have a model that has paperclip attachments. The model might be used in multiple rails apps I need to return a full (non-relative) url to the attachment as part of a JSON API being consumed elsewhere. I'd like to abstract the paperclip aspect and have a simple virtual attribute like this: def thumbnail_url self.photo.url(:thumb) end This however only gives me the relative path. Since it's in the model I can't use the URL helper methods, right? What would be a good approach to prepending the application root url since I don't have helper support? I would like to avoid hardcoding something or adding code to the controller method that assembles my JSON. Thank you

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  • ASP.Net Layered app - Share Entity Data Model amongst layers

    - by Chris Klepeis
    How can I share the auto-generated entity data model (generated object classes) amongst all layers of my C# web app whilst only granting query access in the data layer? This uses the typical 3 layer approach: data, business, presentation. My data layer returns an IEnumerable<T> to my business layer, but I cannot return type T to the presentation layer because I do not want the presentation layer to know of the existence of the data layer - which is where the entity framework auto-generated my classes. It was recommended to have a seperate layer with just the data model, but I'm unsure how to seperate the data model from the query functionality the entity framework provides.

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  • (2006, 'MySQL server has gone away') in WSGI django

    - by Stefano Borini
    I have a MySQL gone away with Django under WSGI. I found entries for this problem on stackoverflow, but nothing with Django specifically. Google does not help, except for workarounds (like polling the website every once in a while, or increasing the database timeout). Nothing definitive. Technically, Django and/or MySQLdb (I'm using the latest 1.2.3c1) should attempt a reconnect if the server hanged the connection, but this does not happen. How can I solve this issue without workarounds ?

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  • Django: Summing values

    - by Anry
    I have a two Model - Project and Cost. class Project(models.Model): title = models.CharField(max_length=150) url = models.URLField() manager = models.ForeignKey(User) class Cost(models.Model): project = models.ForeignKey(Project) cost = models.FloatField() date = models.DateField() I must return the sum of costs for each project. view.py: from mypm.costs.models import Project, Cost from django.shortcuts import render_to_response from django.db.models import Avg, Sum def index(request): #... return render_to_response('index.html',... How?

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  • Catching Oracle Errors in Django

    - by Dashdrum
    My Django app runs on an Oracle database. A few times a year, the database is unavailable because of a scheduled process or unplanned downtime. However, I can't see how to catch the error a give a useful message back to the requester. Instead, a 500 error is triggered, and I get an email (or hundreds) showing the exception. One example is: File "/opt/UDO/env/events/lib/python2.6/site-packages/django/db/backends/oracle/base.py", line 447, in _cursor self.connection = Database.connect(conn_string, **conn_params) DatabaseError: ORA-01035: ORACLE only available to users with RESTRICTED SESSION privilege I see a similar error with a different ORA number when the DB is down. Because the exception is thrown deep within the Django libraries, and can be triggered by any of my views or the built in admin views, I don't know where any exception trapping code would go. Any suggestions?

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  • django: How to make one form from multiple models containing foreignkeys

    - by Tim
    I am trying to make a form on one page that uses multiple models. The models reference each other. I am having trouble getting the form to validate because I cant figure out how to get the id of two of the models used in the form into the form to validate it. I used a hidden key in the template but I cant figure out how to make it work in the views My code is below: views: def the_view(request, a_id,): if request.method == 'POST': b_form= BForm(request.POST) c_form =CForm(request.POST) print "post" if b_form.is_valid() and c_form.is_valid(): print "valid" b_form.save() c_form.save() return HttpResponseRedirect(reverse('myproj.pro.views.this_page')) else: b_form= BForm() c_form = CForm() b_ide = B.objects.get(pk=request.b_id) id_of_a = A.objects.get(pk=a_id) return render_to_response('myproj/a/c.html', {'b_form':b_form, 'c_form':c_form, 'id_of_a':id_of_a, 'b_id':b_ide }) models class A(models.Model): name = models.CharField(max_length=256, null=True, blank=True) classe = models.CharField(max_length=256, null=True, blank=True) def __str__(self): return self.name class B(models.Model): aid = models.ForeignKey(A, null=True, blank=True) number = models.IntegerField(max_length=1000) other_number = models.IntegerField(max_length=1000) class C(models.Model): bid = models.ForeignKey(B, null=False, blank=False) field_name = models.CharField(max_length=15) field_value = models.CharField(max_length=256, null=True, blank=True) forms from mappamundi.mappa.models import A, B, C class BForm(forms.ModelForm): class Meta: model = B exclude = ('aid',) class CForm(forms.ModelForm): class Meta: model = C exclude = ('bid',) B has a foreign key reference to A, C has a foreign key reference to B. Since the models are related, I want to have the forms for them on one page, 1 submit button. Since I need to fill out fields for the forms for B and C & I dont want to select the id of B from a drop down list, I need to somehow get the id of the B form into the form so it will validate. I have a hidden field in the template, I just need to figure how to do it in the views

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  • How to pass parameters to report model in Reporting Services

    - by savras
    I'm developing report in RS that show top N customers based on some criteria. It also allows to select number of customers and period of time. Is it possible to do it by using report model? Thing that it seems to be difficult is how to pass parameters determined by user. Another thing that in my oppinion is very disappointing is that i cannot use SQL query as dataset query, because it uses odd and elaborate XML. Although report model items seem to map its fields to query or table fields. I m concerning using report models because i need to provide uniform data model (the same tables and fields) for more or less different database schemas. It would be very nice if somebody would explain what can be done with report models and what can not.

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