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  • Using Pylint with Django

    - by rcreswick
    I would very much like to integrate pylint into the build process for my python projects, but I have run into one show-stopper: One of the error types that I find extremely useful--:E1101: *%s %r has no %r member*--constantly reports errors when using common django fields, for example: E1101:125:get_user_tags: Class 'Tag' has no 'objects' member which is caused by this code: def get_user_tags(username): """ Gets all the tags that username has used. Returns a query set. """ return Tag.objects.filter( ## This line triggers the error. tagownership__users__username__exact=username).distinct() # Here is the Tag class, models.Model is provided by Django: class Tag(models.Model): """ Model for user-defined strings that help categorize Events on on a per-user basis. """ name = models.CharField(max_length=500, null=False, unique=True) def __unicode__(self): return self.name How can I tune Pylint to properly take fields such as objects into account? (I've also looked into the Django source, and I have been unable to find the implementation of objects, so I suspect it is not "just" a class field. On the other hand, I'm fairly new to python, so I may very well have overlooked something.) Edit: The only way I've found to tell pylint to not warn about these warnings is by blocking all errors of the type (E1101) which is not an acceptable solution, since that is (in my opinion) an extremely useful error. If there is another way, without augmenting the pylint source, please point me to specifics :) See here for a summary of the problems I've had with pychecker and pyflakes -- they've proven to be far to unstable for general use. (In pychecker's case, the crashes originated in the pychecker code -- not source it was loading/invoking.)

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  • Django apache-wsgi configuration problem

    - by omat
    Hi, I am trying to get my Django project running on the production server. I setup the environment using pip, so it is identical to the development environment where everything is running fine. The only difference is that I don't use virtualenv on production, because this project is the only one that is going to run on production. Also on production, there is an Nginx reverse proxy to serve static content, and passes dynamic requests to Apache2. The Apache wsgi file is as follows: import sys, os sys.path.append('/home/project/src') os.environ['DJANGO_SETTINGS_MODULE'] = 'settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() When I access the server, I get an import error: ImproperlyConfigured: Error importing middleware middleware: "cannot import name UserProfile" Which refers to the middleware.py under src/ folder which is referred by the settings. But I can import both the middleware and the UserProfile class from within ./manage.py shell prompt. It seems like a problem with paths in wsgi file but I cannot see what. The directory structure is: /home/project /home/project/src (which contains the settings.py, middleware.py and app folders) /home/apache/apache.wsgi Any help is greatly appreciated. Thanks, oMat

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  • problems doing the Django tutorial on Dreamhost using Passengers

    - by Pietro Speroni
    I have looked and I could not find this question before, and it surprises me. I am reasonably proficient in Python, and I used Dreamhost for a number of years. Now I would like to learn Django. They are finally supporting it using Passenger. Which I do not know what is. Following the instructions on Dreamhost I installed Django. Then I started following the tutorial 01. This went well, except that I could not start the server (this in the tutorial) since the code was live on dreamhost. At the time this did not seem to make any difference. Then when I went on the second part of the tutorial I had to access the admin site. And it worked well going to myurl/admin/ , as it should. But here the problems started. According to the tutorial (here) I have to add a file in the poll application and then restart the server. But I never started the server in the first place, my code is running live on the web... but when I add a file the website the admin acts as if it does not see it. Probably dreamhost has started its own server, and I don't know how to restart it. But I assume this is going to be a common problem when you run django on dreamhost. Every time you add a file you will have to tell the server to consider it. So what should I do to let the server know about it? Thanks, Pietro

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  • On-Demand thumbnail creation with django and nginx

    - by sharjeel
    I want to generate thumbnails of images on the fly. My site is built with django and deployed using nginx which serves all the static content and communicates with django/apache using reverse proxy. Right now, for every image in my site, I generate all required sizes of thumbnails on-hand and deliver them when required. The problem is that whenever I change the size of a thumbnail, I have to regenerate all of them (and they are tons). However now I'd like to generate the thumbnail the first time it is accessed and later on nginx would deliver the same file over n over. If I delete that thumbnail file because of lesser accesses, it should get generated automatically the next time. Thumbnails in my case also have watermarks which require some computation logic of my application so a webserver thumbnail module might not work very well. The size of the thumbnail can be embedded in the URL. So http://www.example.com/thumbnail/abc_320x240.jpg gets the 320x240 size of the thumbnail. The approach I'm looking right now is to let nginx lookup the file and if it doesn't exist, forward the query to my django application which would create the thumbnail and send either the response or a redirect string. However I'm not sure about the concurrency issues and any other issues which might pop up later. What is the appropriate way to achieve this?

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  • Django sitemap intermittent www

    - by Jen Z
    The automatic sitemap for my Django site fluctuates between including the www on urls and leaving it out (I'm aiming to have it in all the time). This has ramifications in google not indexing my pages properly so I'm trying to narrow down what would be causing this issue. I have set PREPEND_WWW = True and my site record in the sites framework is set to include the www e.g. it's set to www.example.com as opposed to example.com. I'm using memcached but pages should expire from the cache after 48 hours so I wouldn't have thought that would be causing the issue? You can see the problem in effect at http://www.livingspaceltd.co.uk/sitemap.xml (refresh the page a few times). My sitemaps setup is fairly prosaic so I'm doubtful that that is the issue, but in case it's something obvious I'm missing here's the code: ***urls.py*** sitemaps = { 'subpages': Subpages_Sitemap, 'standalone_pages': Standalone_Sitemap, 'categories': Categories_Sitemap, } urlpatterns = patterns('', (r'^sitemap\.xml$', 'django.contrib.sitemaps.views.sitemap', {'sitemaps': sitemaps}), ... ***sitemaps.py*** # -*- coding: utf-8 -*- from django_ls.livingspace.models import Page, Category, Standalone_Page, Subpage from django.contrib.sitemaps import Sitemap class Subpages_Sitemap(Sitemap): changefreq = "monthly" priority = 0.4 def items(self): return Subpage.objects.filter(restricted_to__isnull=True) class Standalone_Sitemap(Sitemap): changefreq = "weekly" priority = 1 def items(self): return Standalone_Page.objects.all() class Categories_Sitemap(Sitemap): changefreq = "weekly" priority = 0.7 def items(self): return Category.objects.all()

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  • Django does not load internal .css files

    - by Rubén Jiménez
    I have created a Django project in local which runs without any kind of problem. But, after an annoying and difficult Cherokee + uWSGI installation on Amazon AWS, my project does not show Django .css internal files. http://f.cl.ly/items/2Q2W3I3R0X1n2X3v0q2P/django_error.jpg <-- /Admin/ looks like The image is a screen of my /admin/, which should have a different style, but .css files are not loaded. [pid: 23206|app: 0|req: 19/19] 83.49.10.217 () {56 vars in 1121 bytes} [Sun Apr 15 05:50:24 2012] GET /static/admin/css/base.css = generated 2896 bytes in 6 msecs (HTTP/1.1 404) 1 headers in 51 bytes (1 switches on core 0) [pid: 23206|app: 0|req: 20/20] 83.49.10.217 () {56 vars in 1125 bytes} [Sun Apr 15 05:50:24 2012] GET /static/admin/css/login.css = generated 2899 bytes in 5 msecs (HTTP/1.1 404) 1 headers in 51 bytes (1 switches on core 0) This is a log from Cherokee. I don't understand why it is looking for the .css files in that path. Cherokee should be searching the files in Django original directory so i didn't change .css files in my project. Any advice? Thanks a lot.

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  • inserting a form to session raises picklingerror - django

    - by shanyu
    I receive an exception when I add a form to the session: PicklingError: Can't pickle <class 'django.utils.functional.__proxy__'>: attribute lookup django.utils.functional.__proxy__ failed The form includes a few simple fields and has some javascript attached to a widget. It might be that Django forms cannot be pickled at all, but the exception seems to point to unicode lazy translation. To test further, I have also tried to insert only the form errors (an errordict) to the session and received the same error. I appreciate some help here, thanks in advance. EDIT: Here's why I insert a form into the session: I have an app that has a form. This form is rendered by a template tag in another app. When posted, if the form is valid, no problem, I do stuff and redirect to "next". However if it is not valid, I want to go back to the posting page to show errors. Recall that the comments app in this case redirects to an intermediate "hey, please fix the errors" page. I am trying to avoid this, and hence redirect back to the posting page with the form and its errors in the session that the template tag will render.

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  • Django website on Apache with wsgi failing

    - by notagain
    I have a website I've built in django that I'm trying to get working on our corporate Apache server (on debian) for our intranet at my workplace. Unfortunately, Apache keeps returning server errors whenever I try to navigate to my site. Although I can navigate to the statics folder. My Apache config and wsgi script look like the following... lbirdf.wsgi import os import sys sys.path.append('/home/lbi/rdfweb/web') sys.path.append('/home/lbi/rdfweb/lbirdf') os.environ['DJANGO_SETTINGS_MODULE'] = 'lbirdf.settings_production' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() Apache config Listen 8080 <VirtualHost *:8080> ServerName server1 WSGIScriptAlias /rdfweb /home/lbi/rdfweb/web/lbirdf/apache/lbirdf.wsgi Alias /statics /home/lbi/rdfweb/web/lbirdf/statics Alias /admin_media /home/lbi/rdfweb/web/lbirdf/admin_media <Directory /home/lbi/rdfweb/web/lbirdf/apache> Order allow,deny Allow from all </Directory> <Directory /home/lbi/rdfweb/web/lbirdf/admin_media> Order allow,deny Allow from all </Directory> </VirtualHost> Any ideas on where I might be going wrong?

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  • Django Development Environment Setup Questions

    - by Ross Peoples
    Hello, I'm trying to set up a good development environment for a Django project that I will be working on from two different physical locations. I have two Mac machines, one at home and one at work that I do most of my development on. I currently host a Ubuntu virtual machine on one of the machines to host the Django environemnt, install DropBox on it, and edit source code from my Mac. When I save the code file, the changes get synced over DropBox to the Ubuntu VM and the Django development server automatically restarts because of the change. This method has worked well in the past, but I am starting to use DropBox for a lot of other things now and don't want all of that to be downloaded on every virtual machine I use. Plus, I want to start using Eclipse + PyDev to be able to debug code and have code completion. Currently, I use TextEdit which is great, but doesn't support debugging or completion. So what are my options? I thought about setting up a Parallels VM on a thumb drive that has my entire environment on it (Eclipse included), but that has its own problems. Any other thoughts?

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  • JavaScript cookie value can't be retrieved in Django

    - by Boris Rusev
    I am trying to build a web site in both English and Bulgarian using the Django framework. My idea is the user should click on a button, the page will reload and the language will be changed. This is how I am trying to do it: In my html I hava a the button tag <button id='btn' onclick="changeLanguage();" type="button"> ... </button> An excerpt from cookies.js: function changeLanguage() { if (getCookie('language') == 'EN') { document.getElementById('btn').innerHTML = getCookie('language'); setCookie("language", 'BG'); } else { document.getElementById('btn').innerHTML = getCookie('language'); setCookie("language", 'EN'); } } function setCookie(sName, sValue, oExpires, sPath, sDomain, bSecure) { var sCookie = sName + "=" + encodeURIComponent(sValue); if (oExpires) { sCookie += "; expires=" + oExpires.toGMTString(); } if (sPath) { sCookie += "; path=" + sPath; } if (sDomain) { sCookie += "; domain=" + sDomain; } if (bSecure) { sCookie += "; secure"; } document.cookie = sCookie; } And in my views.py file this is the situation @base def index(request): if request.session['language'] == 'EN': return """<b>%s</b>""" % "Home" else request.session['language'] == 'BG': return """<b>%s</b>""" % "??????" So I know that my JS changes the value of the language cookie but I think Django doesn't get that. On the other hand when I set and get the cookie in my Python code again the cookie is set. My question is whether there is a way to make JS and Django work together - JavaScript sets the cookie value and Python only reads it when asked and takes adequate actions? Thank you.

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  • Django: Serving a Download in a Generic View

    - by TheLizardKing
    So I want to serve a couple of mp3s from a folder in /home/username/music. I didn't think this would be such a big deal but I am a bit confused on how to do it using generic views and my own url. urls.py url(r'^song/(?P<song_id>\d+)/download/$', song_download, name='song_download'), The example I am following is found in the generic view section of the Django documentations: http://docs.djangoproject.com/en/dev/topics/generic-views/ (It's all the way at the bottom) I am not 100% sure on how to tailor this to my needs. Here is my views.py def song_download(request, song_id): song = Song.objects.get(id=song_id) response = object_detail( request, object_id = song_id, mimetype = "audio/mpeg", ) response['Content-Disposition'= "attachment; filename=%s - %s.mp3" % (song.artist, song.title) return response I am actually at a loss of how to convey that I want it to spit out my mp3 instead of what it does now which is to output a .mp3 with all of the current pages html contained. Should my template be my mp3? Do I need to setup apache to serve the files or is Django able to retrieve the mp3 from the filesystem(proper permissions of course) and serve that? If it do need to configure Apache how do I tell Django that? Thanks in advanced. These files are all on the HD so I don't need to "generate" anything on the spot and I'd like to prevent revealing the location of these files if at all possible. A simple /song/1234/download would be fantastic.

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  • Ajax request not receiving xml from Django

    - by amougeot
    I have a Django server which handles requests to a URL which will return some HTML for use in an image gallery. I can navigate to the URL and the browser will display the HTML that is returned, but I can't get that same HTML by doing an AJAX call (using jQuery) to the same URL. This is the view that generates the response: def gallery_images(request, gallery_name): return render_to_response('galleryimages.html', {'images': get_images_of_gallery(gallery_name)}, mimetype='text/xml') This is the 'galleryimages.html' template: {% for image in images %} <div id="{{image.name}}big"> <div class="actualImage" style="background-image:url({{image.image.name}});"> <h1>{{image.caption|safe}}</h1> </div> </div> {% endfor %} This is the jQuery call I am making: $("#allImages").load("http://localhost:8000/galleryimages/Web"); However, this loads nothing into my #allImages div. I've used firebug and ran jQuery's Ajax method .get("http://localhost:8000/galleryimages/Web") and firebug says that the response text is completely empty. When I check my Django server log, this is the entry I see for when I navigate to the URL manually, through my browser: [16/Jan/2010 17:34:10] "GET /galleryimages/Web HTTP/1.1" 200 215 This is the entry in the server log for when I make the AJAX call: [16/Jan/2010 17:36:19] "OPTIONS /galleryimages/Web HTTP/1.1" 200 215 Why does the AJAX request not get the xml that my Django page is serving?

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  • Django | passing form values

    - by MMRUser
    I want to create a user sign up process that requires two different forms with the same data one (1st form) is for filling out the data and other one (2nd form) is for displaying the filled data as a summery (before actually saving the data) so then user can view what he/she has filled up... my problem is that how do I pass 1st form's data in to the 2nd one .. I have used the basic Django form manipulation mechanism and passed the form field values to the next form using Django template tags.. if request.method == 'POST': form = Users(request.POST) if form.is_valid(): cd = form.cleaned_data try: name = cd['fullName'] email = cd['emailAdd'] password1 = cd['password'] password2 = cd['password2'] phoneNumber = cd['phoneNumber'] return render_to_response('signup2.html', {'name': name, 'email': email, 'password1': password1, 'password2': password2, 'phone': phone, 'pt': phoneType}) except Exception, ex: return HttpResponse("Error %s" % str(ex)) and from the second from I just displayed those field values using tags and also used hidden fields in order to submit the form with values, like this: <label for="">Email:</label> {{ email }} <input type="hidden" id="" name="email" class="width250" value="{{ email }}" readonly /> It works nicely from the out look, but the real problem is that if someone view the source of the html he can simply get the password even hackers can get through this easily. So how do I avoid this issue.. and I don't want to use Django session since this is just a simple sign up process and no other interactions involved. Thanks.

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  • Sort queryset by a generic foreign key (django)?

    - by thornomad
    I am using Django's comment framework which utilizes generic foreign keys. Question: How do I sort a given model's queryset by their comment count using the generic foreign key lookup? Reading the django docs on the subject it says one needs to calculate them not using the aggregation API: Django's database aggregation API doesn't work with a GenericRelation. [...] For now, if you need aggregates on generic relations, you'll need to calculate them without using the aggregation API. The only way I can think of, though, would be to iterate through my queryset, generate a list with content_type and object_id's for each item, then run a second queryset on the Comment model filtering by this list of content_type and object_id ... sort the objects by the count, then re-create a new queryset in this order by pulling the content_object for each comment ... This just seems wrong and I'm not even sure how to pull it off. Ideas? Someone must have done this before. I found this post online but it requires me handwriting SQL -- is that really necessary ?

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  • How to make my view better to save Django

    - by user558251
    Hy guys sorry for this post but i need help with my application, i need optimize my view. I have 5 models, how i can do this? def save(request): # get the request.POST in content if request.POST: content = request.POST dicionario = {} # create a dict to get the values in content for key,value in content.items(): # get my fk Course.objects if key == 'curso' : busca_curso = Curso.objects.get(id=value) dicionario.update({key:busca_curso}) else: dicionario.update({key:value}) #create the new teacher Professor.objects.create(**dicionario) my questions are? 1 - How i can do this function in a generic way? Can I pass a variable in a %s to create and get? like this way ? foo = "Teacher" , bar = "Course" def save(request, bar, foo): if request post: ... if key == 'course' : get_course = (%s.objects.get=id=value) %bar ... (%s.objects.create(**dict)) %foo ??? i tried do this in my view but don't work =/, can somebody help me to make this work ? Thanks

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  • Django rewrites URL as IP address in browser - why?

    - by Mitch
    I am using django, nginx and apache. When I access my site with a URL (e.g., http://www.foo.com/) what appears in my browser address is the IP address with admin appended (e.g., http://123.45.67.890/admin/). When I access the site by IP, it is redirected as expected by django's urls.py (e.g., http://123.45.67.890/ - http://123.45.67.890/accounts/login/?next=/) I would like to have the name URL act the same way as the IP. That is, if the URL goes to a new view, the host in the browser address should remain the same and not change to the IP address. Where should I be looking to fix this? My files: ; cpa.com (apache) NameVirtualHost *:8080 <VirtualHost *:8080> AddOutputFilterByType DEFLATE text/html text/plain text/xml text/css text/javascript application/javascript application/x-javascript BrowserMatch ^Mozilla/4 gzip-only-text/html BrowserMatch ^Mozilla/4\.0[678] no-gzip BrowserMatch \bMSIE !no-gzip !gzip-only-text/htm DocumentRoot /path/to/root ServerName www.foo.com <IfModule mod_rpaf.c> RPAFenable On RPAFsethostname On RPAFproxy_ips 127.0.0.1 </IfModule> <Directory /public/static> AllowOverride None AddHandler mod_python .py PythonHandler mod_python.publisher </Directory> Alias / /dj <Location /> SetHandler python-program PythonPath "['/usr/lib/python2.5/site-packages/django', '/usr/lib/python2.5/site-packages/django/forms'] + sys.path" PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE dj.settings PythonDebug On </Location> </VirtualHost> ; ; ports.conf (apache) Listen 127.0.0.1:8080 ; ; cpa.conf (nginx) server { listen 80; server_name www.foo.com; location /static { root /var/public; index index.html; } location /cpa/js { root /var/public/js; } location /cpa/css { root /var/public/css; } location /djmedia { alias "/usr/lib/python2.5/site-packages/django/contrib/admin/media/"; } location / { include /etc/nginx/proxy.conf; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_pass http://127.0.0.1:8080; } } ; ; proxy.conf (nginx) proxy_redirect off; proxy_set_header Host $host; proxy_set_header X-Real-IP $remote_addr; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; client_max_body_size 10m; client_body_buffer_size 128k; proxy_connect_timeout 90; proxy_send_timeout 90; proxy_read_timeout 500; proxy_buffers 32 4k;

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  • File uploads and client_max_body_size in nginx + gunicorn + django

    - by carlosescri
    I need to configure nginx + gunicorn to be able to upload files greater than the default max size in both servers. My nginx .conf file looks like this: server { # ... location / { proxy_pass_header Server; proxy_set_header Host $http_host; proxy_redirect off; proxy_set_header X-Real-IP $remote_addr; proxy_set_header X-Scheme $scheme; proxy_connect_timeout 60; proxy_pass http://localhost:8000/; } } The idea is to allow requests of 20M for two locations: /admin/path/to/upload?param=value /installer/other/path/to/upload?param=value I've tried to add location directives at the same level than the one I've pasted here (getting 404 errors) and also tried to add them inside the location / directive (getting 413 Entity Too Large errors). My location directives look like these in their simplest form: location /admin/path/to/upload/ { client_max_body_size 20M; } location /installer/other/path/to/upload/ { client_max_body_size 20M; } But they don't work (actually I tested lots of combinations and I'm desperate thinking about this. Please, help If you can: What settings do I need to set to make this work? Thank you so much!

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  • permission errors with python/django

    - by tipu
    Error can be seen here: http://djaffry.selfip.com:8080/ If i go to the folder /srv/twingle/search and do ls -l I get -rwxrwxrwx 1 root root 65142784 May 26 20:28 words.db I gave it 777 access (absolutely unsafe, I know, but I thought it would atleast work) any idea what can be the permissions problem? Edit: A very strange problem is that the code doesn't crash once every few refreshes.. then goes back to crashing

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  • Django Aggregation Across Reverse Relationship

    - by Tom
    Given these two models: class Profile(models.Model): user = models.ForeignKey(User, unique=True, verbose_name=_('user')) about = models.TextField(_('about'), blank=True) zip = models.CharField(max_length=10, verbose_name='zip code', blank=True) website = models.URLField(_('website'), blank=True, verify_exists=False) class ProfileView(models.Model): profile = models.ForeignKey(Profile) viewer = models.ForeignKey(User, blank=True, null=True) created = models.DateTimeField(auto_now_add=True) I want to get all profiles sorted by total views. I can get a list of profile ids sorted by total views with: ProfileView.objects.values('profile').annotate(Count('profile')).order_by('-profile__count') But that's just a dictionary of profile ids, which means I then have to loop over it and put together a list of profile objects. Which is a number of additional queries and still doesn't result in a QuerySet. At that point, I might as well drop to raw SQL. Before I do, is there a way to do this from the Profile model? ProfileViews are related via a ForeignKey field, but it's not as though the Profile model knows that, so I'm not sure how to tie the two together. As an aside, I realize I could just store views as a property on the Profile model and that may turn out to be what I do here, but I'm still interested in learning how to better use the Aggregation functions.

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  • Rendering ASP.NET MVC Razor Views outside of MVC revisited

    - by Rick Strahl
    Last year I posted a detailed article on how to render Razor Views to string both inside of ASP.NET MVC and outside of it. In that article I showed several different approaches to capture the rendering output. The first and easiest is to use an existing MVC Controller Context to render a view by simply passing the controller context which is fairly trivial and I demonstrated a simple ViewRenderer class that simplified the process down to a couple lines of code. However, if no Controller Context is available the process is not quite as straight forward and I referenced an old, much more complex example that uses my RazorHosting library, which is a custom self-contained implementation of the Razor templating engine that can be hosted completely outside of ASP.NET. While it works inside of ASP.NET, it’s an awkward solution when running inside of ASP.NET, because it requires a bit of setup to run efficiently.Well, it turns out that I missed something in the original article, namely that it is possible to create a ControllerContext, if you have a controller instance, even if MVC didn’t create that instance. Creating a Controller Instance outside of MVCThe trick to make this work is to create an MVC Controller instance – any Controller instance – and then configure a ControllerContext through that instance. As long as an HttpContext.Current is available it’s possible to create a fully functional controller context as Razor can get all the necessary context information from the HttpContextWrapper().The key to make this work is the following method:/// <summary> /// Creates an instance of an MVC controller from scratch /// when no existing ControllerContext is present /// </summary> /// <typeparam name="T">Type of the controller to create</typeparam> /// <returns>Controller Context for T</returns> /// <exception cref="InvalidOperationException">thrown if HttpContext not available</exception> public static T CreateController<T>(RouteData routeData = null) where T : Controller, new() { // create a disconnected controller instance T controller = new T(); // get context wrapper from HttpContext if available HttpContextBase wrapper = null; if (HttpContext.Current != null) wrapper = new HttpContextWrapper(System.Web.HttpContext.Current); else throw new InvalidOperationException( "Can't create Controller Context if no active HttpContext instance is available."); if (routeData == null) routeData = new RouteData(); // add the controller routing if not existing if (!routeData.Values.ContainsKey("controller") && !routeData.Values.ContainsKey("Controller")) routeData.Values.Add("controller", controller.GetType().Name .ToLower() .Replace("controller", "")); controller.ControllerContext = new ControllerContext(wrapper, routeData, controller); return controller; }This method creates an instance of a Controller class from an existing HttpContext which means this code should work from anywhere within ASP.NET to create a controller instance that’s ready to be rendered. This means you can use this from within an Application_Error handler as I needed to or even from within a WebAPI controller as long as it’s running inside of ASP.NET (ie. not self-hosted). Nice.So using the ViewRenderer class from the previous article I can now very easily render an MVC view outside of the context of MVC. Here’s what I ended up in my Application’s custom error HttpModule: protected override void OnDisplayError(WebErrorHandler errorHandler, ErrorViewModel model) { var Response = HttpContext.Current.Response; Response.ContentType = "text/html"; Response.StatusCode = errorHandler.OriginalHttpStatusCode; var context = ViewRenderer.CreateController<ErrorController>().ControllerContext; var renderer = new ViewRenderer(context); string html = renderer.RenderView("~/Views/Shared/GenericError.cshtml", model); Response.Write(html); }That’s pretty sweet, because it’s now possible to use ViewRenderer just about anywhere in any ASP.NET application, not only inside of controller code. This also allows the constructor for the ViewRenderer from the last article to work without a controller context parameter, using a generic view as a base for the controller context when not passed:public ViewRenderer(ControllerContext controllerContext = null) { // Create a known controller from HttpContext if no context is passed if (controllerContext == null) { if (HttpContext.Current != null) controllerContext = CreateController<ErrorController>().ControllerContext; else throw new InvalidOperationException( "ViewRenderer must run in the context of an ASP.NET " + "Application and requires HttpContext.Current to be present."); } Context = controllerContext; }In this case I use the ErrorController class which is a generic controller instance that exists in the same assembly as my ViewRenderer class and that works just fine since ‘generically’ rendered views tend to not rely on anything from the controller other than the model which is explicitly passed.While these days most of my apps use MVC I do still have a number of generic pieces in most of these applications where Razor comes in handy. This includes modules like the above, which when they error often need to display error output. In other cases I need to generate string template output for emailing or logging data to disk. Being able to render simply render an arbitrary View to and pass in a model makes this super nice and easy at least within the context of an ASP.NET application!You can check out the updated ViewRenderer class below to render your ‘generic views’ from anywhere within your ASP.NET applications. Hope some of you find this useful.ResourcesViewRenderer Class in Westwind.Web.Mvc Library (Github)Original ViewRenderer ArticleRazor Hosting Library (GitHub)Original Razor Hosting Article© Rick Strahl, West Wind Technologies, 2005-2013Posted in ASP.NET  MVC   Tweet !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="//platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() { var po = document.createElement('script'); po.type = 'text/javascript'; po.async = true; po.src = 'https://apis.google.com/js/plusone.js'; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(po, s); })();

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  • Pure Front end JavaScript with Web API versus MVC views with ajax

    - by eyeballpaul
    This was more a discussion for what peoples thoughts are these days on how to split a web application. I am used to creating an MVC application with all its views and controllers. I would normally create a full view and pass this back to the browser on a full page request, unless there were specific areas that I did not want to populate straight away and would then use DOM page load events to call the server to load other areas using AJAX. Also, when it came to partial page refreshing, I would call an MVC action method which would return the HTML fragment which I could then use to populate parts of the page. This would be for areas that I did not want to slow down initial page load, or areas that fitted better with AJAX calls. One example would be for table paging. If you want to move on to the next page, I would prefer it if an AJAX call got that info rather than using a full page refresh. But the AJAX call would still return an HTML fragment. My question is. Are my thoughts on this archaic because I come from a .net background rather than a pure front end background? An intelligent front end developer that I work with, prefers to do more or less nothing in the MVC views, and would rather do everything on the front end. Right down to web API calls populating the page. So that rather than calling an MVC action method, which returns HTML, he would prefer to return a standard object and use javascript to create all the elements of the page. The front end developer way means that any benefits that I normally get with MVC model validation, including client side validation, would be gone. It also means that any benefits that I get with creating the views, with strongly typed html templates etc would be gone. I believe this would mean I would need to write the same validation for front end and back end validation. The javascript would also need to have lots of methods for creating all the different parts of the DOM. For example, when adding a new row to a table, I would normally use the MVC partial view for creating the row, and then return this as part of the AJAX call, which then gets injected into the table. By using a pure front end way, the javascript would would take in an object (for, say, a product) for the row from the api call, and then create a row from that object. Creating each individual part of the table row. The website in question will have lots of different areas, from administration, forms, product searching etc. A website that I don't think requires to be architected in a single page application way. What are everyone's thoughts on this? I am interested to hear from front end devs and back end devs.

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  • Kendo UI Mobile with Knockout for Master-Detail Views

    - by Steve Michelotti
    Lately I’ve been playing with Kendo UI Mobile to build iPhone apps. It’s similar to jQuery Mobile in that they are both HTML5/JavaScript based frameworks for buildings mobile apps. The primary thing that drew me to investigate Kendo UI was its innate ability to adaptively render a native looking app based on detecting the device it’s currently running on. In other words, it will render to look like a native iPhone app if it’s running on an iPhone and it will render to look like a native Droid app if it’s running on a Droid. This is in contrast to jQuery Mobile which looks the same on all devices and, therefore, it can never quite look native for whatever device it’s running on. My first impressions of Kendo UI were great. Using HTML5 data-* attributes to define “roles” for UI elements is easy, the rendering looked great, and the basic navigation was simple and intuitive. However, I ran into major confusion when trying to figure out how to “correctly” build master-detail views. Since I was already very family with KnockoutJS, I set out to use that framework in conjunction with Kendo UI Mobile to build the following simple scenario: I wanted to have a simple “Task Manager” application where my first screen just showed a list of tasks like this:   Then clicking on a specific task would navigate to a detail screen that would show all details of the specific task that was selected:   Basic navigation between views in Kendo UI is simple. The href of an <a> tag just needs to specify a hash tag followed by the ID of the view to navigate to as shown in this jsFiddle (notice the href of the <a> tag matches the id of the second view):   Direct link to jsFiddle: here. That is all well and good but the problem I encountered was: how to pass data between the views? Specifically, I need the detail view to display all the details of whichever task was selected. If I was doing this with my typical technique with KnockoutJS, I know exactly what I would do. First I would create a view model that had my collection of tasks and a property for the currently selected task like this: 1: function ViewModel() { 2: var self = this; 3: self.tasks = ko.observableArray(data); 4: self.selectedTask = ko.observable(null); 5: } Then I would bind my list of tasks to the unordered list - I would attach a “click” handler to each item (each <li> in the unordered list) so that it would select the “selectedTask” for the view model. The problem I found is this approach simply wouldn’t work for Kendo UI Mobile. It completely ignored the click handlers that I was trying to attach to the <a> tags – it just wanted to look at the href (at least that’s what I observed). But if I can’t intercept this, then *how* can I pass data or any context to the next view? The only thing I was able to find in the Kendo documentation is that you can pass query string arguments on the view name you’re specifying in the href. This enabled me to do the following: Specify the task ID in each href – something like this: <a href=”#taskDetail?id=3></a> Attach an “init method” (via the “data-show” attribute on the details view) that runs whenever the view is activated Inside this “init method”, grab the task ID passed from the query string to look up the item from my view model’s list of tasks in order to set the selected task I was able to get all that working with about 20 lines of JavaScript as shown in this jsFiddle. If you click on the Results tab, you can navigate between views and see the the detail screen is correctly binding to the selected item:   Direct link to jsFiddle: here.   With all that being done, I was very happy to get it working with the behavior I wanted. However, I have no idea if that is the “correct” way to do it or if there is a “better” way to do it. I know that Kendo UI comes with its own data binding framework but my preference is to be able to use (the well-documented) KnockoutJS since I’m already familiar with that framework rather than having to learn yet another new framework. While I think my solution above is probably “acceptable”, there are still a couple of things that bug me about it. First, it seems odd that I have to loop through my items to *find* my selected item based on the ID that was passed on the query string - normally, with Knockout I can just refer directly to my selected item from where it was used. Second, it didn’t feel exactly right that I had to rely on the “data-show” method of the details view to set my context – normally with Knockout, I could just attach a click handler to the <a> tag that was actually clicked by the user in order to set the “selected item.” I’m not sure if I’m being too picky. I know there are many people that have *way* more expertise in Kendo UI compared to me – I’d be curious to know if there are better ways to achieve the same results.

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  • Is windows a "second class citizen" in the django community?

    - by Daniel Upton
    I'm currently doing R&D for a web application which we plan to host ourselves initially and then allow customers to self host. My task has been evaluating web frameworks to see which would give us the biggest productivity initially and ease of maintence while also allowing us to easily support deployment to customer controlled environments. Our team has experience with ASP.NET (MVC and Webforms) and Ruby on Rails. Our experience with rails is that windows deployment is a very taboo subject and any questions on IRC or SO are met with knee jerk "why not linux" responses.. However in this case our target market may be running windows or linux servers. Is this also the case in django land? Is it possible with rubbish performance? Is it possible with lost of pain? Is it seen as reasonable and not treated as a completely stupid idea for not wanting to run linux?

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  • The layout page "~/Views/Shared/_Layout.cshtml" could not be found

    - by Rei Brazilva
    I got this error and I can't figure out what is going on. I am positive the _layout.cshtml resides in the shared folder and for the sake of trying things out, I moved to the Home folder and it then told that the Views/Home/_Layout.cshtml couldn't be found there either. So now I'm thinking the problem is in the call for this file for some reason. I'm not going to pretend I know ASP.NET MVC4, so please when you answer, explain it as you would to someone who is not familiar with the system at all. Believe it or not, this error came from tutorial #1 ha ha Here's the code to show that I did code it right: @{ ViewBag.Title = "Home Page"; Layout = "~/Views/Shared/_Layout.cshtml"; } And here is a picture of the location p.s. I did my research, Google has nothing and there is another question here but it was asked on 2008 with MVC3 which is completely different I am running ASP.NET MVC4 on Azure Thanks

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  • BizTalk 2010 - BAM Portal - No Views to Display

    - by Stuart Brierley
    Our latest BizTalk Server 2010 development project is utilising BizTalk as the integration ring around a new and sizable implementaion of Dynamics AX 2012. With this project we have decided to use BAM to monitor the processes within our various new applications.Although I have been specialising in BizTalk for around 9 years, this is my first time using BAM so it is an interesting process to be going through.Recently when deploying a solution I was attempting to check the BAM Portal to see that the View that I had created was properly deployed and that the Activity I was populating was being surfaced in the Portal as expected. Initially I was presented with the message "No view to display" in the "My Views" area of the BAM Portal landing page.This was because you need to set the permissions on the views that you want to see from the command line using the bm.exe tool:bm.exe add-account -AccountName:YourServerOrDomain\YourUsername -View:YourViewThis tool can be found in the BAM folder at the BizTalk installation location:C:\Program Files (x86)\Microsoft BizTalk Server 2010\Tracking

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