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  • yahoo media player not working.

    - by luca590
    I have a yahoo media player embedded in my webpage. I am currently using Ruby on Rails to create/edit my web page. When i click the play button next to a track the YMP waits a while and then goes to the next track without playing the first one. I then get a warning on my second (last) track that its file could not be found. Does anyone has a better recommendation for an audio player or a way to fix this one?

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  • yahoo media player not working

    - by luca590
    I have a yahoo media player embedded in my webpage. I am currently using Ruby on Rails to create/edit my web page. When i click the play button next to a track the YMP waits a while and then goes to the next track without playing the first one. I then get a warning on my second (last) track that its file could not be found. Does anyone has a better recommendation for an audio player or a way to fix this one?

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  • Useful links

    - by Madhan ayyasamy
    Difference between STI vs Polymorphic associationsSTI vs PolymorphicDifference between habtm vs has_many :throughhabtm vs throughCapistrano Guide linkCapistrano guideRails application without database stuffclass Car < ActiveRecord::Baseself.abstract = trueendAnother link: rails without databaseNamed scope useful linkNamed scopeDifference between http and https verbhttp vs httpsRails 2.3 useful guide websiterails 2.3 guide

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  • Serialized values or separate table, which is more efficient?

    - by Aravind
    I have a Rails model email_condition_string with a word column in it. Now I have another model called request_creation_email_config with the following columns admin_filter_group:references vendor_service:references email_condition_string:references email_condition_string has many request_creation_email_config and request_creation_email_config belongs to email_condition_string. Instead of this model a colleague of mine is suggesting that strong the word inside the same model as comma separated values is efficient than storing as a separate model. Is that alright?

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  • nginx with ssl: I get a 403 and log "directory index of '...dir...' is forbidden" log message. works fine with unencrypted connection

    - by user72464
    As mentioned in the title, I had nginx working fine with my rails app, until I tried to add the ssl server. The unencrypted connection still works but the ssl always returns me a 403 page with the following line in the error log: directory index of "/home/user/rails/" is forbidden, client: [my ip], server: _, request: "GET / HTTP/1.1", host: "[server ip]" Below my nginx.conf server block: server { listen 80; listen 443 ssl; ssl_certificate /etc/ssl/server.crt; ssl_certificate_key /etc/ssl/server.key; client_max_body_size 4G; keepalive_timeout 5; root /home/user/rails; try_files $uri/index.html $uri.html $uri @app; location @app { proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_set_header X-Forwarded-Proto $scheme; proxy_set_header Host $http_host; proxy_redirect off; proxy_pass http://0.0.0.0:8080; } error_page 500 502 503 504 /500.html; location = /500.html { root /home/user/rails; } } the /home/user/rails directory and it's parent have all read to all rights. and they belong to the user nginx. the certificate and key file have the following rights: -rw-r--r-- 1 nginx root 830 Nov 8 09:09 server.crt -rw--w---- 1 nginx root 887 Nov 8 09:09 server.key any clue?

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  • connection to apache server switches sockets connection

    - by Newben
    I have just post a question but I post an other one because the problem is not the one I had in thought when asking the latter. So, I am running some rails app on osx, when I run rails s, everything works fine. If I shut down the apache server (mamp) and if I run rails s again, I have this message Can't connect to local MySQL server through socket '/Applications/MAMP/tmp/mysql/mysql.sock', which for sure is normal. For info, my mamp server is running, and the connection must pass through /Applications/MAMP/Library/bin/mysql, so I aliased it by setting in my bash profile : alias mysql="/Applications/MAMP/Library/bin/mysql" Now, when I launch a rails generate command type, I get this message : /$root/vendor/bundle/ruby/2.0.0/gems/mysql2-0.3.11/lib/mysql2/client.rb:44:in `connect': Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) (Mysql2::Error) So how it can be ?

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  • Is Apache ReverseProxy to Passenger Standalone an acceptable production deployment?

    - by davetron5000
    I have the need to deploy Rails 3 apps, using RVM and gemsets, and am expecting “public” traffic (i.e. this is not an internal-only app). I also must use Apache as the public interface to my app. I understand that Passenger Standalone can help accomplish the rails/RVM end, and I have successfully set it up in my development environment. My question is how viable this setup is for a production deployment. Is deploying via Apache configured to ReverseProxy to my passenger-powered Rails app going to create problems? Since I'm designing the production deployment now, I want to understand if I should spend the additional time to set up Passenger connected to Apache and have that Passenger communicate with Passenger Standalone instance running my Rails app. So, I'm looking for one of I guess three answers: Apache Reverse Proxy to Passenger Standalone will be generally fine You should not use the Apache/Passenger Standalone configuration, but set up Passenger on the Apache side as well Your entire setup is just Wrong, please RTFM (and include link to "FM")

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  • Ubuntu in Virtualbox - web server very slow when using local IP address

    - by Lenny Marnham
    I'm using Ubuntu (Lucid Lynx) to learn Ruby On Rails. I'm running Ubuntu in VirtualBox (the host is Windows 7 Ultimate), using bridged networking. When I run my Rails app and point the browser at it using localhost:3000, the app responds immediately and my page is rendered in a second or two. However, if I use 10.0.0.5:3000 (where 10.0.0.5 is my IP address reported using ifconfig), the response from my rails app is incredibly slow - maybe 30 seconds or more for the server to respond and render the page. This happens in both Firefox and Chrome. Also, when I hit the Rails app from the host (to test it in IE), I get the same slooooooow response. Any ideas what might be going on? I've tried it with two different routers, and on two different networks (work and home) with the same result. Cheers all.

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  • django image upload forms

    - by gramware
    I am having problems with django forms and image uploads. I have googled, read the documentations and even questions ere, but cant figure out the issue. Here are my files my models class UserProfile(User): """user with app settings. """ DESIGNATION_CHOICES=( ('ADM', 'Administrator'), ('OFF', 'Club Official'), ('MEM', 'Ordinary Member'), ) onames = models.CharField(max_length=30, blank=True) phoneNumber = models.CharField(max_length=15) regNo = models.CharField(max_length=15) designation = models.CharField(max_length=3,choices=DESIGNATION_CHOICES) image = models.ImageField(max_length=100,upload_to='photos/%Y/%m/%d', blank=True, null=True) course = models.CharField(max_length=30, blank=True, null=True) timezone = models.CharField(max_length=50, default='Africa/Nairobi') smsCom = models.BooleanField() mailCom = models.BooleanField() fbCom = models.BooleanField() objects = UserManager() #def __unicode__(self): # return '%s %s ' % (User.Username, User.is_staff) def get_absolute_url(self): return u'%s%s/%s' % (settings.MEDIA_URL, settings.ATTACHMENT_FOLDER, self.id) def get_download_url(self): return u'%s%s/%s' % (settings.MEDIA_URL, settings.ATTACHMENT_FOLDER, self.name) ... class reports(models.Model): repID = models.AutoField(primary_key=True) repSubject = models.CharField(max_length=100) repRecepients = models.ManyToManyField(UserProfile) repPoster = models.ForeignKey(UserProfile,related_name='repposter') repDescription = models.TextField() repPubAccess = models.BooleanField() repDate = models.DateField() report = models.FileField(max_length=200,upload_to='files/%Y/%m/%d' ) deleted = models.BooleanField() def __unicode__(self): return u'%s ' % (self.repSubject) my forms from django import forms from django.http import HttpResponse from cms.models import * from django.contrib.sessions.models import Session from django.forms.extras.widgets import SelectDateWidget class UserProfileForm(forms.ModelForm): class Meta: model= UserProfile exclude = ('designation','password','is_staff', 'is_active','is_superuser','last_login','date_joined','user_permissions','groups') ... class reportsForm(forms.ModelForm): repPoster = forms.ModelChoiceField(queryset=UserProfile.objects.all(), widget=forms.HiddenInput()) repDescription = forms.CharField(widget=forms.Textarea(attrs={'cols':'50', 'rows':'5'}),label='Enter Report Description here') repDate = forms.DateField(widget=SelectDateWidget()) class Meta: model = reports exclude = ('deleted') my views @login_required def reports_media(request): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) if request.user.is_staff== True: repmedform = reportsForm(request.POST, request.FILES) if repmedform.is_valid(): repmedform.save() repmedform = reportsForm(initial = {'repPoster':user.id,}) else: repmedform = reportsForm(initial = {'repPoster':user.id,}) return render_to_response('staffrepmedia.html', {'repfrm':repmedform, 'rep_media': reports.objects.all()}) else: return render_to_response('reports_&_media.html', {'rep_media': reports.objects.all()}) ... @login_required def settingchng(request): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) form = UserProfileForm(instance = user) if request.method == 'POST': form = UserProfileForm(request.POST, request.FILES, instance = user) if form.is_valid(): form.save() return HttpResponseRedirect('/settings/') else: form = UserProfileForm(instance = user) if request.user.is_staff== True: return render_to_response('staffsettingschange.html', {'form': form}) else: return render_to_response('settingschange.html', {'form': form}) ... @login_required def useradd(request): if request.method == 'POST': form = UserAddForm(request.POST,request.FILES ) if form.is_valid(): password = request.POST['password'] request.POST['password'] = set_password(password) form.save() else: form = UserAddForm() return render_to_response('staffadduser.html', {'form':form}) Example of my templates {% if form.errors %} <ol> {% for field in form %} <H3 class="title"> <p class="error"> {% if field.errors %}<li>{{ field.errors|striptags }}</li>{% endif %}</p> </H3> {% endfor %} </ol> {% endif %} <form method="post" id="form" action="" enctype="multipart/form-data" class="infotabs accfrm"> {{ repfrm.as_p }} <input type="submit" value="Submit" /> </form>

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  • Django form linking 2 models by many to many field.

    - by Ed
    I have two models: class Actor(models.Model): name = models.CharField(max_length=30, unique = True) event = models.ManyToManyField(Event, blank=True, null=True) class Event(models.Model): name = models.CharField(max_length=30, unique = True) long_description = models.TextField(blank=True, null=True) I want to create a form that allows me to identify the link between the two models when I add a new entry. This works: class ActorForm(forms.ModelForm): class Meta: model = Actor The form includes both name and event, allowing me to create a new Actor and simultaneous link it to an existing Event. On the flipside, class EventForm(forms.ModelForm): class Meta: model = Event This form does not include an actor association. So I am only able to create a new Event. I can't simultaneously link it to an existing Actor. I tried to create an inline formset: EventFormSet = forms.models.inlineformset_factory(Event, Actor, can_delete = False, extra = 2, form = ActorForm) but I get an error <'class ctg.dtb.models.Actor'> has no ForeignKey to <'class ctg.dtb.models.Event'> This isn't too surprising. The inlineformset worked for another set of models I had, but this is a different example. I think I'm going about it entirely wrong. Overall question: How can I create a form that allows me to create a new Event and link it to an existing Actor?

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  • What is the most elegant way to access current_user from the models? or why is it a bad idea?

    - by TheLindyHop
    So, I've implemented some permissions between my users and the objects the users modify.. and I would like to lessen the coupling between the views/controllers with the models (calling said permissions). To do that, I had an idea: Implementing some of the permission functionality in the before_save / before_create / before_destroy callbacks. But since the permissions are tied to users (current_user.can_do_whatever?), I didn't know what to do. This idea may even increase coupling, as current_user is specifically controller-level. The reason why I initially wanted to do this is: All over my controllers, I'm having to check if a user has the ability to save / create / destroy. So, why not just return false upon save / create / destroy like rails' .save already does, and add an error to the model object and return false, just like rails' validations? Idk, is this good or bad? is there a better way to do this?

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  • Ruby on Rails - A Competent Web Development Application

    Written in Ruby programming language, Ruby on Rails is one of the most frequently used web application development framework. Often termed as RoR or Rails, it is an open source web development framework which is basically an object oriented programming language encouraging simple development, complete and potent web applications encompassing rich interactivity and functionality.

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  • foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = 15, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT c.*, ci.* FROM category c JOIN category_info ci ON (c.node_id = ci.node_id) WHERE c.type_id = 15 AND c.parent_id = 50 AND ci.type_id = 1 Any help is GREATLY appreciated. Regards

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  • Django - foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = type_g.type_id, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT n.*, l.* FROM objtree_nodes n JOIN objtree_labels l ON (n.node_id = l.node_id) WHERE n.type_id = 15 AND n.parent_id = 50 AND l.type_id = 1 Any help is GREATLY appreciated. Regards

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  • Django JSON serializable error

    - by Hulk
    With the following code below, There is an error saying File "/home/user/web_pro/info/views.py", line 184, in headerview, raise TypeError("%r is not JSON serializable" % (o,)) TypeError: <lastname: jerry> is not JSON serializable In the models code header(models.Model): firstname = models.ForeignKey(Firstname) lastname = models.ForeignKey(Lastname) In the views code headerview(request): header = header.objects.filter(created_by=my_id).order_by(order_by)[offset:limit] l_array = [] l_array_obj = [] for obj in header: l_array_obj = [obj.title, obj.lastname ,obj.firstname ] l_array.append(l_array_obj) dictionary_l.update({'Data': l_array}) ; return HttpResponse(simplejson.dumps(dictionary_l), mimetype='application/javascript') what is this error and how to resolve this? thanks..

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  • Use the Django ORM in a standalone script (again)

    - by Rishabh Manocha
    I'm trying to use the Django ORM in some standalone screen scraping scripts. I know this question has been asked before, but I'm unable to figure out a good solution for my particular problem. I have a Django project with defined models. What I would like to do is use these models and the ORM in my scraping script. My directory structure is something like this: project scrape #scraping scripts ... test.py web django_project settings.py ... #Django files I tried doing the following in project/scrape/test.py: print os.path.join(os.path.abspath('..'), 'web', 'django_project') sys.path.append(os.path.join(os.path.abspath('..'), 'web', 'django_project')) print sys.path print "-------" os.environ['DJANGO_SETTINGS_MODULE'] = 'django_project.settings' #print os.environ from django_project.myapp.models import MyModel print MyModel.objects.count() However, I get an ImportError when I try to run test.py: Traceback (most recent call last): File "test.py", line 12, in <module> from django_project.myapp.models import MyModel ImportError: No module named django_project.myapp.models One solution I found around this problem is to create a symbolic link to ../web/govcheck in the scrape folder: :scrape rmanocha$ ln -s ../web/govcheck ./govcheck With this, I can then run test.py just fine. However, this seems like a hack, and more importantly, is not very portable (I will have to create this symbolic link everywhere I run this code). So, I was wondering if anyone has any better solutions for my problem?

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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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  • Select a subset of foreign key elements in inlineformset_factory in Django

    - by Enis Afgan
    Hello, I have a model with two foreign keys: class Model1(models.Model): model_a = models.ForeignKey(ModelA) model_b = models.ForeignKey(ModelB) value = models.IntegerField() Then, I create an inline formset class, like so: an_inline_formset = inlineformset_factory(ModelA, Model1, fk_name="model_a") and then instantiate it, like so: a_formset = an_inline_formset(request.POST, instance=model_A_object) Once this formset gets rendered in a template/page, there is ChoiceField associated with the model_b field. The problem I'm having is that the elements in the resulting drop down menu include all of the elements found in ModelB table. I need to select a subset of those based on some criteria from ModelB. At the same time, I need to keep the reference to the instance of model_A_object when instantiating inlineformset_factory and, therefore, I can't just use this example. Any suggestions?

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  • Rails: Multiple "has_many through" for the two same models?

    - by neezer
    Can't wrap my head around this... class User < ActiveRecord::Base has_many :fantasies, :through => :fantasizings has_many :fantasizings, :dependent => :destroy end class Fantasy < ActiveRecord::Base has_many :users, :through => :fantasizings has_many :fantasizings, :dependent => :destroy end class Fantasizing < ActiveRecord::Base belongs_to :user belongs_to :fantasy end ... which works fine for my primary relationship, in that a User can have many Fantasies, and that a Fantasy can belong to many Users. However, I need to add another relationship for liking (as in, a User "likes" a Fantasy rather than "has" it... think of Facebook and how you can "like" a wall-post, even though it doesn't "belong" to you... in fact, the Facebook example is almost exactly what I'm aiming for). I gathered that I should make another association, but I'm kinda confused as to how I might use it, or if this is even the right approach. I started by adding the following: class Fantasy < ActiveRecord::Base ... has_many :users, :through => :approvals has_many :approvals, :dependent => :destroy end class User < ActiveRecord::Base ... has_many :fantasies, :through => :approvals has_many :approvals, :dependent => :destroy end class Approval < ActiveRecord::Base belongs_to :user belongs_to :fantasy end ... but how do I create the association through Approval rather than through Fantasizing? If someone could set me straight on this, I'd be much obliged!

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  • Default value for hidden field in Django model

    - by Daniel Garcia
    I have this Model: class Occurrence(models.Model): id = models.AutoField(primary_key=True, null=True) reference = models.IntegerField(null=True, editable=False) def save(self): self.collection = self.id super(Occurrence, self).save() I want for the reference field to be hidden and at the same time have the same value as id. This code works if the editable=True but if i want to hide it it doesnt change the value of reference. how can i fix that?

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  • Calling a method from within a django model save() override

    - by Jonathan
    I'm overriding a django model save() method. Within the override I'm calling another method of the same class and instance which calculates one of the instance's fields based on other fields of the same instance. class MyClass(models.Model): field1 = models.FloatField() field2 = models.FloatField() field3 = models.FloatField() def calculateField1(self) self.field1 = self.field2 + self.field3 def save(self, *args, **kwargs): self.calculateField1() super(MyClass, self).save(*args, **kwargs) The override method is called when I change the model in admin. Alas I've discovered that within calculateField1() field2 and field3 have the values of the instance from before I edited them in admin. If I enter the instance again in admin and save again, only then field1 receives the correct value as field2 and field3 are already updated. Is this the correct behavior on django's side? If yes, then how can I use the new values within calculateField1? I cannot implement the calculation within the save() as calculateField1() actually quite long and I need it to be called from elsewhere.

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  • CharField values disappearing after save (readonly field)

    - by jamida
    I'm implementing simple "grade book" application where the teacher would be able to update the grades w/o being allowed to change the students' names (at least not on the update grade page). To do this I'm using one of the read-only tricks, the simplest one. The problem is that after the SUBMIT the view is re-displayed with 'blank' values for the students. I'd like the students' names to re-appear. Below is the simplest example that exhibits this problem. (This is poor DB design, I know, I've extracted just the relevant parts of the code to showcase the problem. In the real example, student is in its own table but the problem still exists there.) models.py class Grade1(models.Model): student = models.CharField(max_length=50, unique=True) finalGrade = models.CharField(max_length=3) class Grade1OForm(ModelForm): student = forms.CharField(max_length=50, required=False) def __init__(self, *args, **kwargs): super(Grade1OForm,self).__init__(*args, **kwargs) instance = getattr(self, 'instance', None) if instance and instance.id: self.fields['student'].widget.attrs['readonly'] = True self.fields['student'].widget.attrs['disabled'] = 'disabled' def clean_student(self): instance = getattr(self,'instance',None) if instance: return instance.student else: return self.cleaned_data.get('student',None) class Meta: model=Grade1 views.py from django.forms.models import modelformset_factory def modifyAllGrades1(request): gradeFormSetFactory = modelformset_factory(Grade1, form=Grade1OForm, extra=0) studentQueryset = Grade1.objects.all() if request.method=='POST': myGradeFormSet = gradeFormSetFactory(request.POST, queryset=studentQueryset) if myGradeFormSet.is_valid(): myGradeFormSet.save() info = "successfully modified" else: myGradeFormSet = gradeFormSetFactory(queryset=studentQueryset) return render_to_response('grades/modifyAllGrades.html',locals()) template <p>{{ info }}</p> <form method="POST" action=""> <table> {{ myGradeFormSet.management_form }} {% for myform in myGradeFormSet.forms %} {# myform.as_table #} <tr> {% for field in myform %} <td> {{ field }} {{ field.errors }} </td> {% endfor %} </tr> {% endfor %} </table> <input type="submit" value="Submit"> </form>

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