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  • Django Upload form to S3 img and form validation

    - by citadelgrad
    I'm fairly new to both Django and Python. This is my first time using forms and upload files with django. I can get the uploads and saves to the database to work fine but it fails to valid email or check if the users selected a file to upload. I've spent a lot of time reading documentation trying to figure this out. Thanks! views.py def submit_photo(request): if request.method == 'POST': def store_in_s3(filename, content): conn = S3Connection(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY) bucket = conn.create_bucket(AWS_STORAGE_BUCKET_NAME) mime = mimetypes.guess_type(filename)[0] k = Key(bucket) k.key = filename k.set_metadata("Content-Type", mime) k.set_contents_from_file(content) k.set_acl('public-read') if imghdr.what(request.FILES['image_url']): qw = request.FILES['image_url'] filename = qw.name image = filename content = qw.file url = "http://bpd-public.s3.amazonaws.com/" + image data = {image_url : url, user_email : request.POST['user_email'], user_twittername : request.POST['user_twittername'], user_website : request.POST['user_website'], user_desc : request.POST['user_desc']} s = BeerPhotos(data) if s.is_valid(): #import pdb; pdb.set_trace() s.save() store_in_s3(filename, content) return HttpResponseRedirect(reverse('photos.views.thanks')) return s.errors else: return errors else: form = BeerPhotoForm() return render_to_response('photos/submit_photos.html', locals(),context_instance=RequestContext(request) forms.py class BeerPhotoForm(forms.Form): image_url = forms.ImageField(widget=forms.FileInput, required=True,label='Beer',help_text='Select a image of no more than 2MB.') user_email = forms.EmailField(required=True,help_text='Please type a valid e-mail address.') user_twittername = forms.CharField() user_website = forms.URLField(max_length=128,) user_desc = forms.CharField(required=True,widget=forms.Textarea,label='Description',) template.html <div id="stylized" class="myform"> <form action="." method="post" enctype="multipart/form-data" width="450px"> <h1>Photo Submission</h1> {% for field in form %} {{ field.errors }} {{ field.label_tag }} {{ field }} {% endfor %} <label><span>Click here</span></label> <input type="submit" class="greenbutton" value="Submit your Photo" /> </form> </div>

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  • Invalidating Memcached Keys on save() in Django

    - by Zack
    I've got a view in Django that uses memcached to cache data for the more highly trafficked views that rely on a relatively static set of data. The key word is relatively: I need invalidate the memcached key for that particular URL's data when it's changed in the database. To be as clear as possible, here's the meat an' potatoes of the view (Person is a model, cache is django.core.cache.cache): def person_detail(request, slug): if request.is_ajax(): cache_key = "%s_ABOUT_%s" % settings.SITE_PREFIX, slug # Check the cache to see if we've already got this result made. json_dict = cache.get(cache_key) # Was it a cache hit? if json_dict is None: # That's a negative Ghost Rider person = get_object_or_404(Person, display = True, slug = slug) json_dict = { 'name' : person.name, 'bio' : person.bio_html, 'image' : person.image.extra_thumbnails['large'].absolute_url, } cache.set(cache_key) # json_dict will now exist, whether it's from the cache or not response = HttpResponse() response['Content-Type'] = 'text/javascript' response.write(simpljson.dumps(json_dict)) # Make sure it's all properly formatted for JS by using simplejson return response else: # This is where the fully templated response is generated What I want to do is get at that cache_key variable in it's "unformatted" form, but I'm not sure how to do this--if it can be done at all. Just in case there's already something to do this, here's what I want to do with it (this is from the Person model's hypothetical save method) def save(self): # If this is an update, the key will be cached, otherwise it won't, let's see if we can't find me try: old_self = Person.objects.get(pk=self.id) cache_key = # Voodoo magic to get that variable old_key = cache_key.format(settings.SITE_PREFIX, old_self.slug) # Generate the key currently cached cache.delete(old_key) # Hit it with both barrels of rock salt # Turns out this doesn't already exist, let's make that first request even faster by making this cache right now except DoesNotExist: # I haven't gotten to this yet. super(Person, self).save() I'm thinking about making a view class for this sorta stuff, and having functions in it like remove_cache or generate_cache since I do this sorta stuff a lot. Would that be a better idea? If so, how would I call the views in the URLconf if they're in a class?

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  • Formatting inline many-to-many related models presented in django admin

    - by Jonathan
    I've got two django models (simplified): class Product(models.Model): name = models.TextField() price = models.IntegerField() class Invoice(models.Model): company = models.TextField() customer = models.TextField() products = models.ManyToManyField(Product) I would like to see the relevant products as a nice table (of product fields) in an Invoice page in admin and be able to link to the individual respective Product pages. My first thought was using the admin's inline - but django used a select box widget per related Product. This isn't linked to the Product pages, and also as I have thousands of products, and each select box independently downloads all the product names, it quickly becomes unreasonably slow. So I turned to using ModelAdmin.filter_horizontal as suggested here, which used a single instance of a different widget, where you have a list of all Products and another list of related Products and you can add\remove products in the later from the former. This solved the slowness, but it still doesn't show the relevant Product fields, and it ain't linkable. So, what should I do? tweak views? override ModelForms? I Googled around and couldn't find any example of such code...

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  • Math on Django Templates

    - by Leandro Abilio
    Here's another question about Django. I have this code: views.py cursor = connections['cdr'].cursor() calls = cursor.execute("SELECT * FROM cdr where calldate > '%s'" %(start_date)) result = [SQLRow(cursor, r) for r in cursor.fetchall()] return render_to_response("cdr_user.html", {'calls':result }, context_instance=RequestContext(request)) I use a MySQL query like that because the database is not part of a django project. My cdr table has a field called duration, I need to divide that by 60 and multiply the result by a float number like 0.16. Is there a way to multiply this values using the template tags? If not, is there a good way to do it in my views? My template is like this: {% for call in calls %} <tr class="{% cycle 'odd' 'even' %}"><h3> <td valign="middle" align="center"><h3>{{ call.calldate }}</h3></td> <td valign="middle" align="center"><h3>{{ call.disposition }}</h3></td> <td valign="middle" align="center"><h3>{{ call.dst }}</h3></td> <td valign="middle" align="center"><h3>{{ call.billsec }}</h3></td> <td valign="middle" align="center">{{ (call.billsec/60)*0.16 }}</td></h3> </tr> {% endfor %} The last is where I need to show the value, I know the "(call.billsec/60)*0.16" is impossible to be done there. I wrote it just to represent what I need to show.

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  • Django ModelForm is giving me a validation error that doesn't make sense

    - by River Tam
    I've got a ModelForm based on a Picture. class Picture(models.Model): name = models.CharField(max_length=100) pub_date = models.DateTimeField('date published') tags = models.ManyToManyField('Tag', blank=True) content = models.ImageField(upload_to='instaton') def __unicode__(self): return self.name class PictureForm(forms.ModelForm): class Meta: model = Picture exclude = ('pub_date','tags') That's the model and the ModelForm, of course. def submit(request): if request.method == 'POST': # if the form has been submitted form = PictureForm(request.POST) if form.is_valid(): return HttpResponseRedirect('/django/instaton') else: form = PictureForm() # blank form return render_to_response('instaton/submit.html', {'form': form}, context_instance=RequestContext(request)) That's the view (which is being correctly linked to by urls.py) Right now, I do nothing when the form submits. I just check to make sure it's valid. If it is, I forward to the main page of the app. <form action="/django/instaton/submit/" method="post"> {% csrf_token %} {{ form.as_p }} <input type="submit" value"Submit" /> </form> And there's my template (in the correct location). When I try to actually fill out the form and just validate it, even if I do so correctly, it sends me back to the form and says "This field is required" between Name and Content. I assume it's referring to Content, but I'm not sure. What's my problem? Is there a better way to do this?

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  • Postfix SMTP auth not working with virtual mailboxes + SASL + Courier userdb

    - by Greg K
    So I've read a variety of tutorials and how-to's and I'm struggling to make sense of how to get SMTP auth working with virtual mailboxes in Postfix. I used this Ubuntu tutorial to get set up. I'm using Courier-IMAP and POP3 for reading mail which seems to be working without issue. However, the credentials used to read a mailbox are not working for SMTP. I can see from /var/log/auth.log that PAM is being used, does this require a UNIX user account to work? As I'm using virtual mailboxes to avoid creating user accounts. li305-246 saslauthd[22856]: DEBUG: auth_pam: pam_authenticate failed: Authentication failure li305-246 saslauthd[22856]: do_auth : auth failure: [user=fred] [service=smtp] [realm=] [mech=pam] [reason=PAM auth error] /var/log/mail.log li305-246 postfix/smtpd[27091]: setting up TLS connection from mail-pb0-f43.google.com[209.85.160.43] li305-246 postfix/smtpd[27091]: Anonymous TLS connection established from mail-pb0-f43.google.com[209.85.160.43]: TLSv1 with cipher ECDHE-RSA-RC4-SHA (128/128 bits) li305-246 postfix/smtpd[27091]: warning: SASL authentication failure: Password verification failed li305-246 postfix/smtpd[27091]: warning: mail-pb0-f43.google.com[209.85.160.43]: SASL PLAIN authentication failed: authentication failure I've created accounts in userdb as per this tutorial. Does Postfix also use authuserdb? What debug information is needed to help diagnose my issue? main.cf: # TLS parameters smtpd_tls_cert_file = /etc/ssl/certs/smtpd.crt smtpd_tls_key_file = /etc/ssl/private/smtpd.key smtpd_use_tls=yes smtpd_tls_session_cache_database = btree:${data_directory}/smtpd_scache smtp_tls_session_cache_database = btree:${data_directory}/smtp_scache # SMTP parameters smtpd_sasl_local_domain = smtpd_sasl_auth_enable = yes smtpd_sasl_security_options = noanonymous broken_sasl_auth_clients = yes smtpd_recipient_restrictions = permit_sasl_authenticated,permit_mynetworks,reject_unauth_destination smtp_tls_security_level = may smtpd_tls_security_level = may smtpd_tls_auth_only = no smtp_tls_note_starttls_offer = yes smtpd_tls_CAfile = /etc/ssl/certs/cacert.pem smtpd_tls_loglevel = 1 smtpd_tls_received_header = yes smtpd_tls_session_cache_timeout = 3600s tls_random_source = dev:/dev/urandom /etc/postfix/sasl/smtpd.conf pwcheck_method: saslauthd mech_list: plain login /etc/default/saslauthd START=yes PWDIR="/var/spool/postfix/var/run/saslauthd" PARAMS="-m ${PWDIR}" PIDFILE="${PWDIR}/saslauthd.pid" DESC="SASL Authentication Daemon" NAME="saslauthd" MECHANISMS="pam" MECH_OPTIONS="" THREADS=5 OPTIONS="-c -m /var/spool/postfix/var/run/saslauthd" /etc/courier/authdaemonrc authmodulelist="authuserdb" I've only modified one line in authdaemonrc and restarted the service as per this tutorial. I've added accounts to /etc/courier/userdb via userdb and userdbpw and run makeuserdb as per the tutorial. SOLVED Thanks to Jenny D for suggesting use of rimap to auth against localhost IMAP server (which reads userdb credentials). I updated /etc/default/saslauthd to start saslauthd correctly (this page was useful) MECHANISMS="rimap" MECH_OPTIONS="localhost" THREADS=0 OPTIONS="-c -m /var/spool/postfix/var/run/saslauthd -r" After doing this I got the following error in /var/log/auth.log: li305-246 saslauthd[28093]: auth_rimap: unexpected response to auth request: * BYE [ALERT] Fatal error: Account's mailbox directory is not owned by the correct uid or gid: li305-246 saslauthd[28093]: do_auth : auth failure: [user=fred] [service=smtp] [realm=] [mech=rimap] [reason=[ALERT] Unexpected response from remote authentication server] This blog post detailed a solution by setting IMAP_MAILBOX_SANITY_CHECK=0 in /etc/courier/imapd. Then restart your courier and saslauthd daemons for config changes to take effect. sudo /etc/init.d/courier-imap restart sudo /etc/init.d/courier-authdaemon restart sudo /etc/init.d/saslauthd restart Watch /var/log/auth.log while trying to send email. Hopefully you're good!

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  • Two Google accounts in firefox for email/reader/openid

    - by deddebme
    I usually checking my personal gmail account account at work, and I have another gmail account for work/professional purpose. Now I am starting to see more sites using OpenID. The problem I am facing is that I want to check my gmail from firefox, but I want to use my work google account to login with OpenID website. Is there an easy to do so? Of course one way is to logout my personal account, login my work account, OpenID login to those sites. Second way is to use another browser for my personal gmail and firefox for work, but are there a better way because I hate using two browsers at the same time?

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  • Is there an OpenID demo server out there?

    - by billpg
    Hi everyone. I'm doing some experiements with adding OpenID to something I'm working on, and I'd like to test out a few providers. Is there a server out there that will go through the OpenID login process (same way that the StackOverflow group does) and tell me all the information the provider shows. I imagine it would work like... I go to example.com and type in https://www.google.com/accounts/o8/id example.com bounces me to google. I log in. Google asks me to confirm if I allow example.com access to everything. Google bounces me back to example.com example.com tells me my OpenID, email address, anything else it's got. Does such a thing please already exist?

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  • django views question

    - by Hulk
    In my django views i have the following def create(request): query=header.objects.filter(id=a)[0] a=query.criteria_set.all() logging.debug(a.details) I get an error saying 'QuerySet' object has no attribute 'details' in the debug statement .What is this error and what should be the correct statemnt to query this.And the model corresponding to this is as follows where as the models has the following: class header(models.Model): title = models.CharField(max_length = 255) created_by = models.CharField(max_length = 255) def __unicode__(self): return self.id() class criteria(models.Model): details = models.CharField(max_length = 255) headerid = models.ForeignKey(header) def __unicode__(self): return self.id() Thanks..

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  • Django url parameters

    - by Hulk
    How to pass two paramters in urls in django <script> url=/toolbox/display/" + id + "2"; window.location=url; </script> Also how is this handeled in urls.py (r'^display/(?P<rid>\d+)/(?P<param>\d+)/$', 'table_display'), In views, def table_display(request,rid,param): print param //This should print 2

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  • Django forms "not" using forms from models

    - by zubinmehta
    I have a form generated from various models and the various values filled go and sit in some other table. Hence, in this case I haven't used the inbuilt Django forms(i.e. I am not creating forms from models ). Now the data which is posted from the self made form is handled by view1 which should clean the data accordingly. How do I go about it and use the various functions clean and define validation errors (and preferably not do validation logic in the view itself!)

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  • Extended Django base-class with multiple instances

    - by Gijs
    I'm modeling a simple movie database using Django. models.py defines a base model Person. I extend Person into Actor and Director, which works as I imagined. Persons must be unique. When (in the Admin) I create an instance of Actor, and this person is also a Director, it won't save because of the unique = True. Any ideas how to solve this problem? (generic foreign keys?) Thx

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  • Django: How to get current user in admin forms

    - by lazerscience
    In Django's ModelAdmin I need to display forms customized according to the permissions an user has. Is there a way of getting the current user object into the form class, so that i can customize the form in its __init__ method? I think saving the current request in a thread local would be a possibility but this would be my last resort think I'm thinking it is a bad design approach....

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  • Django template-printing variables

    - by Hulk
    In django views def add(request): dict{} co_data = optarr dict.update({'co_data' : co_data}) logging.debug(co_data) return render_to_response('scheme/create.html',context_instance=RequestContext(request,{'dict': dict})) And data has the following string 1##2##3##4## And in the template when i say {{co_data}} it doesnt display the values.Please point out whats wrong in the code. Thanks..

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  • Django filter vs exclude

    - by Enrico
    Is there a difference between filter and exclude in django? If I have self.get_query_set().filter(modelField=x) and I want to add another criteria, is there a meaningful difference between to following two lines of code? self.get_query_set().filter(user__isnull=False, modelField=x) self.get_query_set().filter(modelField=x).exclude(user__isnull=True) is one considered better practice or are they the same in both function and performance?

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  • Django User model, adding function

    - by Hellnar
    Hello, I want to add a new function to the default User model of Django for retrieveing a related list of Model type. Such Foo model: class Foo(models.Model): owner = models.ForeignKey(User, related_name="owner") likes = models.ForeignKey(User, related_name="likes") ........ #at some view user = request.user foos= user.get_related_foo_models() Hwo can this be achieved ?

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  • How to store an integer leaded by zeros in django

    - by Oscar Carballal
    Hello, I'm trying to store a number in django that looks like this: 000001 My problem is that if I type this inside an IntegerField it gets converted to "1" without the leading zeros. I've tried also with a DecimalField with the same result. How can I store the leading zeros whithout using a CharField? (I need to manipulate that number in it's integer form)

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  • How to make custom join query with Django ?

    - by xRobot
    I have these 2 models: genre = ( ('D', 'Dramatic'), ('T', 'Thriller'), ('L', 'Love'), ) class Book(models.Model): title = models.CharField(max_length=100) genre = models.CharField(max_length=1, choices=genre) class Author(models.Model): user = models.ForeignKey(User, unique=True) born = models.DateTimeField('born') book = models.ForeignKey(Book) I need to retrieve first_name and last_name of all authors of dramatic's books. How can I do this in django ?

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  • Django: name of many to many items in the admin interface

    - by Adam
    I have a many to many field, which I'm displaying in the django admin panel. When I add multiple items, they all come up as "ASGGroup object" in the display selector. Instead, I want them to come up as whatever the ASGGroup.name field is set to. How do I do this? My models looks like: class Thing(Model): read_groups = ManyToManyField('ASGGroup', related_name="thing_read", blank=True) class ASGGroup(Model): name = CharField(max_length=63, null=True) But what I'm seeing the m2m widget display is:

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  • Accessing updated M2M fields in overriden save() in django's admin

    - by Jonathan
    I'd like to use the user updated values of a ManyToManyField in a model's overriden save() method when I save an instance in admin. It turns out that by design, django does not update the M2M field before calling save(), but only after the save() is complete as part of the form save... How can I access the new values of this field in the override save() ?

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  • Django: Inherit Permssions from abstract models?

    - by lazerscience
    Is it possible to inherit permissions from an abstract model in Django? I can not really find anything about that. For me this doesn't work! class PublishBase(models.Model): class Meta: abstract = True get_latest_by = 'created' permissions = (('change_foreign_items', "Can change other user's items"),) EDIT: Not working means it fails silently. Permission is not created, as it wouldn't exist on the models inheriting from this class.

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  • Django - problem with {% url facebook_xd_receiver %}

    - by Gaurav
    I'm using {% url facebook_xd_receiver %} in one of my HTML files. This works just fine when I run my project using the command python manage.py runserver But the same project stops running and gives me a "TemplateSyntaxError" at the line {% url facebook_xd_receiver %} Can anyone please tell me what could be the difference between the dev server run through the command line and the apache server. Is there anything I'm missing out on while configuring the Apache server? Or is it a Django problem?

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