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  • Quering distinct values throught related model

    - by matheus.emm
    Hi! I have a simple one-to-many (models.ForeignKey) relationship between two of my model classes: class TeacherAssignment(models.Model): # ... some fields year = models.CharField(max_length=4) class LessonPlan(models.Model): teacher_assignment = models.ForeignKey(TeacherAssignment) # ... other fields I'd like to query my database to get the set of distinct years of TeacherAssignments related to at least one LessonPlan. I'm able to get this set using Django query API if I ignore the relation to LessonPlan: class TeacherAssignment(models.Model): # ... model's fields def get_years(self): year_values = self.objects.all().values_list('year').distinct().order_by('-year') return [yv[0] for yv in year_values if len(yv[0]) == 4] Unfortunately I don't know how to express the condition that the TeacherAssignment must be related to at least one LessonPlan. Any ideas how I'd be able to write the query? Thanks in advance.

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  • Passing session data to ModelForm inside of ModelAdmin

    - by theactiveactor
    I'm trying to initialize the form attribute for MyModelAdmin class inside an instance method, as follows: class MyModelAdmin(admin.ModelAdmin): def queryset(self, request): MyModelAdmin.form = MyModelForm(request.user) My goal is to customize the editing form of MyModelForm based on the current session. When I try this however, I keep getting an error (shown below). Is this the proper place to pass session data to ModelForm? If so, then what may be causing this error? TypeError at ... Exception Type: TypeError Exception Value: issubclass() arg 1 must be a class Exception Location: /usr/lib/pymodules/python2.6/django/forms/models.py in new, line 185

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  • Is it ok to hardcode dynamic links in a permanent view?

    - by meder
    Let's say I wanted to showcase 2-3 clickable buttons on my homepage which will be there permanently. These are links to the css, html, and javascript tag listing pages. Is it fine to just hardcode href=/tags/css and href=/tags/html right in my django templates/view? I won't change them for at least a year or so, meaning I don't think I need to add a column to the tags table to distinguish them - is this common or should I try to make it somewhat dynamic? These tags are in a table but so are 1000 other tags.

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  • Saving related model objects

    - by iHeartDucks
    I have two related models (one to many) in my django app and When I do something like this ObjBlog = Blog() objBlog.name = 'test blog' objEntry1 = Entry() objEntry1.title = 'Entry one' objEntry2 = Entry() objEntry2.title = 'Entry Two' objBlog.entry_set.add(objEntry1) objBlog.entry_set.add(objEntry2) I get an error which says "null value in column and it violates the foreign key not null constraint". None of my model objects have been saved. Do I have to save the "objBlog" before I could set the entries? I was hoping I could call the save method on objBlog to save it all. NOTE: I am not creating a blog engine and this is just an example.

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  • Overwrite queryset which builds filter sidebar

    - by cw
    Hi, I'm writing a hockey database/manager. So I have the following models: class Team(models.Model): name = models.CharField(max_length=60) class Game(models.Model): home_team = models.ForeignKey(Team,related_name='home_team') away_team = models.ForeignKey(Team,related_name='away_team') class SeasonStats(models.Model): team = models.ForeignKey(Team) Ok, so my problem is the following. There are a lot of teams, but Stats are just managed for my Club. So if I use "list_display" in the admin backend, I'd like to modify/overwrite the queryset which builds the sidebar for filtering, to just display our home teams as a filter option. Is this somehow possible in Django? I already made a custom form like this class SeasonPlayerStatsAdminForm(forms.ModelForm): team = forms.ModelChoiceField(Team.objects.filter(club__home=True)) So now just the filtering is missing. Any ideas?

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  • Select distinct users with referrals

    - by Mark
    I have a bunch of Users. Since Django doesn't really let me extend the default User model, they each have Profiles. The Profiles have a referred_by field (a FK to User). I'm trying to get a list of Users with = 1 referral. Here's what I've got so far Profile.objects.filter(referred_by__isnull=False).values_list('referred_by', flat=True) Which gives me a list of IDs of the users who have referrals... but I need it to be distinct, and I want the User object, not their ID. Or better yet, it would be nice if it could return the number of referrals a user has. Any ideas?

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  • Enable export to XML via HTTP on a large number of models with child relations

    - by Vasil
    I've a large number of models (120+) and I would like to let users of my application export all of the data from them in XML format. I looked at django-piston, but I would like to do this with minimum code. Basically I'd like to have something like this: GET /export/applabel/ModelName/ Would stream all instances of ModelName in applabel together with it's tree of related objects . I'd like to do this without writing code for each model. What would be the best way to do this?

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  • Errors in Decimal Calcs within def clean method?

    - by allanhenderson
    I'm attempting a few simple calculations in a def clean method following validation (basically spitting out a euro conversion of retrieved uk product price on the fly). I keep getting a TypeError. Full error reads: Cannot convert {'product': , 'invoice': , 'order_discount': Decimal("0.00"), 'order_price': {...}, 'order_adjust': None, 'order_value': None, 'DELETE': False, 'id': 92, 'quantity': 8} to Decimal so I guess django is passing through the entire cleaned_data form to Decimal method. Not sure where I'm going wrong - the code I'm working with is: def clean_order_price(self): cleaned_data = self.cleaned_data data = self.data order_price = cleaned_data.get("order_price") if not order_price: try: existing_price = ProductCostPrice.objects.get(supplier=data['supplier'], product_id=cleaned_data['product'], is_latest=True) except ProductCostPrice.DoesNotExist: existing_price = None if not existing_price: raise forms.ValidationError('No match found, please enter new price') else: if data['invoice_type'] == 1: return existing_price.cost_price_gross elif data['invoice_type'] == 2: exchange = EuroExchangeRate.objects.latest('exchange_date') calc = exchange.exchange_rate * float(existing_price.cost_price_gross) calc = Decimal(str(calc)) return calc return cleaned_data If the invoice is of type 2 (a euro invoice) then the system should grab the latest exchange rate and apply that to the matching UK pound price pulled through to get euro result. Should performing a decimal conversion be a problem within def clean method? Thanks

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  • Elegant solution for multiple forms on single page

    - by NFicano
    I'm building a web application (in Django) that will accept a search criteria and display a report - once the user is satisfied with the results, save both the criteria and a reference to these objects back to the database. The problem I'm having is finding an elegant solution for having 2 forms: Display (GET) the results of their criteria. Enter in some descriptions, and save (POST) everything back to the database. I'm leaning towards AJAX for the GET stuff and a POST for the save, but I wanted to make sure there wasn't a more elegant solution first.

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  • Indexing a method return (depending on Internationalization)

    - by Hedde
    Consider a django model with an IntegerField with some choices, e.g. COLORS = ( (0, _(u"Blue"), (1, _(u"Red"), (2, _(u"Yellow"), ) class Foo(models.Model): # ...other fields... color = models.PositiveIntegerField(choices=COLOR, verbose_name=_(u"color")) My current (haystack) index: class FooIndex(SearchIndex): text = CharField(document=True, use_template=True) color = CharField(model_attr='color') def prepare_color(self, obj): return obj.get_color_display() site.register(Product, ProductIndex) This obviously only works for keyword "yellow", but not for any (available) translations. Question: What's would be a good way to solve this problem? (indexing method returns based on the active language) What I have tried: I created a function that runs a loop over every available language (from settings) appending any translation to a list, evaluating this against the query, pre search. If any colors are matched it converts them backwards into their numeric representation to evaluate against obj.color, but this feels wrong.

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  • Not quite nested inlines?

    - by Lynden Shields
    Not quite sure what to call this, it's not quite nested inlines, but is probably related. I have a 3 level hierarchy of objects, A one-to-many B one-to-many C. Therefore, every C implicitly also belongs to an A. class A(models.Model): stuff = models.CharField("Stuff", max_length=50) class B(models.Model): a = models.ForeignKey(A) class C(models.Model): b = models.ForeignKey(B) I would like all C's that belong to an A to be listed on the admin page for A in an in-line. They do not have to show which B they belong to on the same page. Is this possible or is it the same problem as nested inlines anyway? If it's possible, how do I do it? I'm using django 1.3

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  • Celery tasks not works with gevent

    - by Novarg
    When i use celery + gevent for tasks that uses subprocess module i'm getting following stacktrace: Traceback (most recent call last): File "/home/venv/admin/lib/python2.7/site-packages/celery/task/trace.py", line 228, in trace_task R = retval = fun(*args, **kwargs) File "/home/venv/admin/lib/python2.7/site-packages/celery/task/trace.py", line 415, in __protected_call__ return self.run(*args, **kwargs) File "/home/webapp/admin/webadmin/apps/loggingquarantine/tasks.py", line 107, in release_mail_task res = call_external_script(popen_obj.communicate) File "/home/webapp/admin/webadmin/apps/core/helpers.py", line 42, in call_external_script return func_to_call(*args, **kwargs) File "/usr/lib64/python2.7/subprocess.py", line 740, in communicate return self._communicate(input) File "/usr/lib64/python2.7/subprocess.py", line 1257, in _communicate stdout, stderr = self._communicate_with_poll(input) File "/usr/lib64/python2.7/subprocess.py", line 1287, in _communicate_with_poll poller = select.poll() AttributeError: 'module' object has no attribute 'poll' My manage.py looks following (doing monkeypatch there): #!/usr/bin/env python from gevent import monkey import sys import os if __name__ == "__main__": if not 'celery' in sys.argv: monkey.patch_all() os.environ.setdefault("DJANGO_SETTINGS_MODULE", "webadmin.settings") from django.core.management import execute_from_command_line sys.path.append(".") execute_from_command_line(sys.argv) Is there a reason why celery tasks act like it wasn't patched properly? p.s. strange thing that my local setup on Macos works fine while i getting such exceptions under Centos (all package versions are the same, init and config scripts too)

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  • Using map() on a _set in a template?

    - by Stuart Grimshaw
    I have two models like this: class KPI(models.Model): """KPI model to hold the basic info on a Key Performance Indicator""" title = models.CharField(blank=False, max_length=100) description = models.TextField(blank=True) target = models.FloatField(blank=False, null=False) group = models.ForeignKey(KpiGroup) subGroup = models.ForeignKey(KpiSubGroup, null=True) unit = models.TextField(blank=True) owner = models.ForeignKey(User) bt_measure = models.BooleanField(default=False) class KpiHistory(models.Model): """A historical log of previous KPI values.""" kpi = models.ForeignKey(KPI) measure = models.FloatField(blank=False, null=False) kpi_date = models.DateField() and I'm using RGraph to display the stats on internal wallboards, the handy thing is Python lists get output in a format that Javascript sees as an array, so by mapping all the values into a list like this: def f(x): return float(x.measure) stats = map(f, KpiHistory.objects.filter(kpi=1) then in the template I can simply use {{ stats }} and the RGraph code sees it as an array which is exactly what I want. [87.0, 87.5, 88.5, 90] So my question is this, is there any way I can achieve the same effect using Django's _set functionality to keep the amount of data I'm passing into the template, up until now I've been passing in a single KPI object to be graphed but now I want to pass in a whole bunch so is there anything I can do with _set {{ kpi.kpihistory_set }} dumps the whole model out, but I just want the measure field. I can't see any of the built in template methods that will let me pull out just the single field I want. How have other people handled this situation?

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  • Using ftplib for multithread uploads

    - by Arty
    I'm trying to do multithread uploads, but get errors. I guessed that maybe it's impossible to use multithreads with ftplib? Here comes my code: class myThread (threading.Thread): def __init__(self, threadID, src, counter, image_name): self.threadID = threadID self.src = src self.counter = counter self.image_name = image_name threading.Thread.__init__(self) def run(self): uploadFile(self.src, self.image_name) def uploadFile(src, image_name): f = open(src, "rb") ftp.storbinary('STOR ' + image_name, f) f.close() ftp = FTP('host') # connect to host, default port ftp.login() # user anonymous, passwd anonymous@ dirname = "/home/folder/" i = 1 threads = [] for image in os.listdir(dirname): if os.path.isfile(dirname + image): thread = myThread(i , dirname + image, i, image ) thread.start() threads.append( thread ) i += 1 for t in threads: t.join() Get bunch of ftplib errors like raise error_reply, resp error_reply: 200 Type set to I If I try to upload one by one, everything works fine

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  • File Uploads with Turbogears 2

    - by William Chambers
    I've been trying to work out the 'best practices' way to manage file uploads with Turbogears 2 and have thus far not really found any examples. I've figured out a way to actually upload the file, but I'm not sure how reliable it us. Also, what would be a good way to get the uploaded files name? file = request.POST['file'] permanent_file = open(os.path.join(asset_dirname, file.filename.lstrip(os.sep)), 'w') shutil.copyfileobj(file.file, permanent_file) file.file.close() this_file = self.request.params["file"].filename permanent_file.close() So assuming I'm understanding correctly, would something like this avoid the core 'naming' problem? id = UUID. file = request.POST['file'] permanent_file = open(os.path.join(asset_dirname, id.lstrip(os.sep)), 'w') shutil.copyfileobj(file.file, permanent_file) file.file.close() this_file = file.filename permanent_file.close()

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  • Plupload - how to stop uploads individualy

    - by Manny Calavera
    Hello. I'm using plupload script with PHP and I am uploading using HTML5. I would like to stop uploads separately, from the list, not the whole process. Is this possible with plupload ? I think it would be useful to stop individual files. Thanks. EDIT: Let's just say that a user tries to upload 2 files and they are added to queue. After that, he clicks the upload button and the upload starts. At some point during the upload he realize that he doesn't really want to upload the second file, he needs to stop the 2nd upload and the 1st upload should continue. I can send an uploader.start(); on the plupload but it will stop all upload processes...and I need to stop just one. Thanks.

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  • hosting acounts for large uploads

    - by Phil Jackson
    Hi, im wondering if anyone knows of any host providers ( uk preferably ) that deals mostly with accepting large file uploads. Most hosts only let you push something like 1.5mb ( thats taking into account the connection and the max execution time ). What i am looking for is a host specificaly for storing files on. I was going to create an upload script onto my application which posted the file to the external host and then return back ( using headers so the user doen't even know they have left ). Does anyone know of a host for this?

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  • [Wordpress MU] Changing the uploads directory

    - by Pedro Reis
    Hi, I've looked everywhere and while there is solutions to change the uploads directory for all the blogs by changing this line in the wp-settings.php: define( "BLOGUPLOADDIR", WP_CONTENT_DIR . "/blogs.dir/{$wpdb->blogid}/files/" ); I can't find a way of changing the directory for each blog individually, something like: define( "BLOGUPLOADDIR", WP_CONTENT_DIR . "/blogs.dir/{$blog_name}/files/" ); But I have no idea how could I get the name of the blog from within the wp-settings.php as you can't use get_bloginfo('name'); outside of the template. Anybody with a solution for this?

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  • Tomcat application: Frequent OutOfMemory PermGen exception while image uploads

    - by rabbit
    Hi, I have a tomcat 6 application which I have set parameters of -Xms512m -Xmx1024m. I thought 1 GB of memory in a 4 GB RAM would be enough, but that is not the case. On application stop/start multiple times (from tomcat manager) and also on image uploads (sometimes) I run into the OutOfMemory PermGen space error and the site stops responding. Should I increase the memory still some more? Is there anything else that I can do to from the tomcat side so that it does not run into the PermGen space issue? Thanks in advance for pointers/tips etc.

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  • One table for uploads for multiple resources in CakePHP

    - by Mef
    Hi, I have CakePHP app in which I'd like to attach gallery to multiple resources. Let's say I've got artists, each one has own gallery. I've got articles, every article has some images attached to it and so on. Now I set up tables like this: Artists hasMany Artistimages, fields in artistimages table are: id, artist_id, filename, filetype, filesize etc. Articles hasMany Articleimages, fields in articleimages table are: id, article_id, filename, filetype, filesize etc. ...but this is not how it should be, I think. Is there possibility to have one table called for example uploads which will contain all images with foreign key pointing to resource its reffering to? How to tell CakePHP which image is coming from which resource?

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  • Is keeping uploads folder outside of 'public_html' enough to keep my application secure from malicio

    - by ecu
    Although I realise there are different approaches to securing upload process, I'm still confused when it comes to basic principles. I want to allow users to upload any kind of file they want, but keep my app secure. So my question is: Is it sufficient to store the files with their original names in 'uploads' folder outside 'webroot' and fetching them via some download.php script? If it't not secure enough, please point me in the right direction, or suggest what additional steps I should take to make it safe. Thank you.

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  • What's the best way to run Drupal and Django sites behind the same Varnish server?

    - by Alexis Bellido
    I have a high traffic website running with Drupal and Apache, five web servers behind a Varnish server load balancing. Let's say this site is example.com. I'm using five backends and a director like this in my default.vcl: director balancer round-robin { { .backend = web1; } { .backend = web2; } { .backend = web3; } { .backend = web4; } { .backend = web5; } } Now I'm working on a new Django project that will be a new section of this site running on example.com/new-section. After checking the documentation I found I can do something like this: sub vcl_recv { if (req.url ~ "^/new-section/") { set req.backend = newbackend; } else { set req.backend = default; } } That is, using a different backend for a subdirectory /new-section under the same domain. My question is, how do I make something like this work with my director and load balancing setup? I'm probably going to run two or more web servers (backends) with my new Django project, each one with a mix of Gunicorn, Nginx, and a few Python packages, and would like to put all of those in their own Varnish director to load balance. Is it possible to do use the above approach to decide which director to use?, like this: sub vcl_recv { if (req.url ~ "^/new-section/") { set req.director = newdirector; } else { set req.director = balancer; } } All suggestions welcome. Thanks!

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  • Go for Zend framework or Django for a modular web application?

    - by dr. squid
    I am using both Zend framework and Django, and they both have they strengths and weakness, but they are both good framworks in their own way. I do want to create a highly modular web application, like this example: modules: Admin cms articles sections ... ... ... I also want all modules to be self contained with all confid and template files. I have been looking into a way to solve this is zend the last days, but adding one omer level to the module setup doesn't feel right. I am sure this could be done, but should I? I have also included Doctrine to my zend application that could give me even more problems in my module setup! When we are talking about Django this is easy to implement (Easy as in concept, not in implementation time or whatever) and a great way to create web apps. But one of the downsides of Django is the web hosing part. There are some web hosts offering Django support, but not that many.. So then I guess the question is what have the most value; rapid modular development versus hosting options! Well, comments are welcome! Thanks

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  • Django: How can I identify the calling view from a template?

    - by bryan
    Short version: Is there a simple, built-in way to identify the calling view in a Django template, without passing extra context variables? Long (original) version: One of my Django apps has several different views, each with its own named URL pattern, that all render the same template. There's a very small amount of template code that needs to change depending on the called view, too small to be worth the overhead of setting up separate templates for each view, so ideally I need to find a way to identify the calling view in the template. I've tried setting up the views to pass in extra context variables (e.g. "view_name") to identify the calling view, and I've also tried using {% ifequal request.path "/some/path/" %} comparisons, but neither of these solutions seems particularly elegant. Is there a better way to identify the calling view from the template? Is there a way to access to the view's name, or the name of the URL pattern? Update 1: Regarding the comment that this is simply a case of me misunderstanding MVC, I understand MVC, but Django's not really an MVC framework. I believe the way my app is set up is consistent with Django's take on MVC: the views describe which data is presented, and the templates describe how the data is presented. It just happens that I have a number of views that prepare different data, but that all use the same template because the data is presented the same way for all the views. I'm just looking for a simple way to identify the calling view from the template, if this exists. Update 2: Thanks for all the answers. I think the question is being overthought -- as mentioned in my original question, I've already considered and tried all of the suggested solutions -- so I've distilled it down to a "short version" now at the top of the question. And right now it seems that if someone were to simply post "No", it'd be the most correct answer :) Update 3: Carl Meyer posted "No" :) Thanks again, everyone.

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  • What's the next steps for moving from appengine to full django?

    - by tomcritchlow
    Hey guys, I'm super new to programming and I've been using appengine to help me learn python and general coding. I'm getting better quickly and I'm loving it all the way :) Appengine was awesome for allowing me to just dive into writing my app and getting something live that works (see http://www.7bks.com/). But I'm realising that the longer I continue to learn on appengine the more I'm constraining myself and locking myself into a single system. I'd like to move to developing on full django (since django looks super cool!). What are my next steps? To give you a feel for my level of knowledge: I'm not a unix user I'm not familiar with command line controls (I still use appengine/python completely via the appengine SDK) I've never programmed in anything other than python, anywhere other than appengine I know the word SQL, but don't know what MySQL is really or how to use it. So, specifically: What are the skills I need to learn to get up and running with full django/python? If I'm going to host somewhere else I suppose I'll need to learn some sysadmin type skills (maybe even unix?). Is there anywhere that offers easy hosting (like appengine) but that supports django? I hear such great things about heroku I'm considering switching to RoR and going there I appreciate that I'm likely not quite ready to move away from appengine just yet but I'm a fiercely passionate learner (http://www.7bks.com/blog/179001) and would love it if I knew all the steps I needed to learn so I could set about learning them. At the moment, I don't even know what the steps are I need to learn! Thank you very much. Sorry this isn't a specific programming question but I've looked around and haven't found a good how-to for someone of my level of experience and I think others would appreciate a good roadmap for the things we need to learn to get up and running. Thanks, Tom PS - if anyone is in London and fancies showing me the ropes in person that would be super awesome :)

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