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  • Iterating dictionary indexes in django templates

    - by unclaimedbaggage
    Hi folks...I have a dictionary with embedded objects, which looks something like this: notes = { 2009: [<Note: Test note>, <Note: Another test note>], 2010: [<Note: Third test note>, <Note: Fourth test note>], } I'm trying to access each of the note objects inside a django template, and having a helluva time navigating to them. In short, I'm not sure how to extract by index in django templating. Current template code is: <h3>Notes</h3> {% for year in notes %} {{ year }} # Works fine {% for note in notes.year %} {{ note }} # Returns blank {% endfor %} {% endfor %} If I replace {% for note in notes.year %} with {% for note in notes.2010 %} things work fine, but I need that '2010' to be dynamic. Any suggestions much appreciated.

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  • Pass errors in Django using HttpResponseRedirect

    - by JPC
    I know that HttpResponseRedirect only takes one parameter, a URL. But there are cases when I want to redirect with an error message to display. I was reading this post: How to pass information using an http redirect (in Django) and there were a lot of good suggestions. I don't really want to use a library that I don't know how works. I don't want to rely on messages which, according to the Django docs, is going to be removed. I thought about using sessions. I also like the idea of passing it in a URL, something like: return HttpResponseRedirect('/someurl/?error=1') and then having some map from error code to message. Is it good practice to have a global map-like structure which hard codes in these error messages or is there a better way? Or should I just use a session EDIT: I got it working using a session. Is that a good practice to put things like this in the session?

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  • Writing custom Django fields and widgets

    - by hekevintran
    Django has very good documentation that describes how to write custom database fields and custom template tags and filters. I cannot find the document that describes how to write custom form fields and widgets. Does this document exist? The way I've been able to write custom form fields and widgets is by reading the Django source code and imitating what I see there. I know that there are still things about implementing fields and widgets that I do not completely understand because I have not read any high level document that describes their interfaces.

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  • Django: Extending User Model - Inline User fields in UserProfile

    - by Jack Sparrow
    Is there a way to display User fields under a form that adds/edits a UserProfile model? I am extending default Django User model like this: class UserProfile(models.Model): user = models.OneToOneField(User, unique=True) about = models.TextField(blank=True) I know that it is possible to make a: class UserProfileInlineAdmin(admin.TabularInline): and then inline this in User ModelAdmin but I want to achieve the opposite effect, something like inverse inlining, displaying the fields of the model pointed by the OneToOne Relationship (User) in the page of the model defining the relationship (UserProfile). I don't care if it would be in the admin or in a custom view/template. I just need to know how to achieve this. I've been struggling with ModelForms and Formsets, I know the answer is somewhere there, but my little experience in Django doesn't allow me to come up with the solution yet. A little example would be really helpful!

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  • Writing custom Django form fields and widgets

    - by hekevintran
    Django has very good documentation that describes how to write custom database fields and custom template tags and filters. I cannot find the document that describes how to write custom form fields and widgets. Does this document exist? The way I've been able to write custom form fields and widgets is by reading the Django source code and imitating what I see there. I know that there are still things about implementing fields and widgets that I do not completely understand because I have not read any high level document that describes their interfaces.

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  • Django refresh page if change data by other user

    - by Fran Sobrino
    I have a test django app. In one page the test show the same question to all users. I'd like that when a user answers correctly, send a signal to other active user's browser to refresh to the next question. I have been learning about signals in django I learning work with them but I don't now how send the "refresh signal" to client browser. I think that it can do with a javascript code that check if a certain value (actual question) change and if change reload the page but I don't know this language and the information that I find was confused. Can anybody help me? Many Thanks.

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  • app_label in an abstract Django model

    - by rayan
    Hi all, I'm trying to get an abstract model working in Django and I hit a brick wall trying to set the related_name per the recommendation here: http://docs.djangoproject.com/en/dev/topics/db/models/#be-careful-with-related-name This is what my abstract model looks like: class CommonModel(models.Model): created_on = models.DateTimeField(editable=False) creared_by = models.ForeignKey(User, related_name="%(app_label)s_%(class)s_created", editable=False) updated_on = models.DateTimeField(editable=False) updated_by = models.ForeignKey(User, related_name="%(app_label)s_%(class)s_updated", editable=False) def save(self): if not self.id: self.created_on = datetime.now() self.created_by = user.id self.updated_on = datetime.now() self.updated_by = user.id super(CommonModel, self).save() class Meta: abstract = True My common model is in [project_root]/models.py. It is the parent object of this model, which is located in an app called Feedback [project_root]/feedback/models.py: from django.db import models from mediasharks.models import CommonModel class Feedback(CommonModel): message = models.CharField(max_length=255) request_uri = models.CharField(max_length=255) domain = models.CharField(max_length=255) feedback_type = models.IntegerField() Basically I'm trying to set up a common model so that I'll always be able to tell when and by whom database entries were created. When I run "python manage.py validate" I get this error message: KeyError: 'app_label' Am I missing something here?

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  • Django startup importing causes reverse to happen

    - by nicknack
    This might be an isolated problem, but figured I'd ask in case someone has thoughts on a graceful approach to address it. Here's the setup: -------- views.py -------- from django.http import HttpResponse import shortcuts def mood_dispatcher(request): mood = magic_function_to_guess_my_mood(request) return HttpResponse('Please go to %s' % shortcuts.MOODS.get(mood, somedefault)) ------------ shortcuts.py ------------ MOODS = # expensive load that causes a reverse to happen The issue is that shortcuts.py causes an exception to be thrown when a reverse is attempted before django is done building the urls. However, views.py doesn't yet need to import shortcuts.py (used only when mood_dispatcher is actually called). Obvious initial solutions are: 1) Import shortcuts inline (just not very nice stylistically) 2) Make shortcuts.py build MOODS lazily (just more work) What I ideally would like is to be able to say, at the top of views.py, "import shortcuts except when loading urls"

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  • how to do server side form validation for dynamic inputs with Django

    - by Satoru.Logic
    Hi, all. I am using django.forms.Form to validate form data in a survey applications. In a survey-creating form, a user can submit multiple questions that belong to the survey being created. Names for the question inputs are in the form of 'question_seq' , where seq is maintained using Javascript. Back in the server side, my code doesn't know before hand how many such questions will be submitted. Is there any way to do this with Django form so that the form can automatically recognizes the questions and validate them?

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  • Django - how to make ImageField/FileField optional?

    - by ilya
    class Product(models.Model): ... image = models.ImageField(upload_to = generate_filename, blank = True) When I use ImageField (blank=True) and do not select image into admin form, exception occures. In django code you can see this: class FieldFile(File): .... def _require_file(self): if not self: raise ValueError("The '%s' attribute has no file associated with it." % self.field.name) def _get_file(self): self._require_file() ... Django trac has ticket #13327 about this problem, but seems it can't be fixed soon. How to make these field optional?

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  • Django: Data corrupted after loading? (possible programmer error)

    - by Rosarch
    I may be loading data the wrong way. excerpt of data.json: { "pk": "1", "model": "myapp.Course", "fields": { "name": "Introduction to Web Design", "requiredFor": [9], "offeringSchool": 1, "pre_reqs": [], "offeredIn": [1, 5, 9] } }, I run python manage.py loaddata -v2 data: Installed 36 object(s) from 1 fixture(s) Then, I go to check the above object using the Django shell: >>> info = Course.objects.filter(id=1) >>> info.get().pre_reqs.all() [<Course: Intermediate Web Programming>] # WRONG! There should be no pre-reqs >>> from django.core import serializers >>> serializers.serialize("json", info) '[{"pk": 1, "model": "Apollo.course", "fields": {"pre_reqs": [11], "offeredIn": [1, 5, 9], "offeringSchool": 1, "name": "Introduction to Web Design", "requiredFor": [9]}}]' The serialized output of the model is not the same as the input that was given to loaddata. The output has a non-empty pre_req list, whereas the input's pre_reqs field is empty. What am I doing wrong?

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  • Sorting related objects in the Django Admin form interface

    - by Carver
    I am looking to sort the related objects that show up when editing an object using the admin form. So for example, I would like to take the following object: class Person(models.Model): first_name = models.CharField( ... ) last_name = models.CharField( ... ) hero = models.ForeignKey( 'self', null=True, blank=True ) and edit the first name, last name and hero using the admin interface. I want to sort the objects as they show up in the drop down by last name, first name (ascending). How do I do that? Context I'm using Django v1.1. I started by looking for help in the django admin docs, but didn't find the solution As you can see in the example, the foreign key is pointing to itself, but I expect it would be the same as pointing to a different model object. Bonus points for being able to filter the related objects, too (eg~ only allow selecting a hero with the same first name)

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  • Django: Why Doesn't the Current URL Match any Patterns in urls.py

    - by austin_sherron
    I've found a few questions here related to my issue, but I haven't found anything that has helped me resolve my issue. I'm using Python 2.7.5 and Django 1.8.dev20140627143448. I have a view that's interacting with my database to delete objects, and it takes two arguments in addition to a request: def delete_data_item(request, dataclass_id, dataitem_id): form = AddDataItemForm(request.POST) data_set = get_object_or_404(DataClass, pk=dataclass_id) context = {'data_set': data_set, 'form': form} data_item = get_object_or_404(DataItem, pk=dataitem_id) data_item.delete() data_set.save() return HttpResponseRedirect(reverse('detail', args=(dataclass_id,))) The URL in myapp.urls.py looks something like this: url(r'^(?P<dataclass_id>[0-9]+)/(?P<dataitem_id>[0-9]+)/delete_data_item/$', views.delete_data_item, name='delete_data_item') and the portion of my template relevant to the view is: <a href="{% url 'delete_data_item' data_set.id data_item.id %}">DELETE</a> Whenever I click on the DELETE link, django tells me that the request URL: http://127.0.0.1:8000/myapp/5/%7B%%20url%20'delete_data_item'%20data_set.id%20data_item.id%20%%7D doesn't match any of my URL patterns. What am I missing? The URL on which the DELETE links exist is myapp/(<dataclass_id>[0-9]+)/

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  • Getting a KeyError in DB backend of django-digest

    - by rtmie
    I have just started to integrate django_digest into my app. As a start I have added the @httpdigest decorator to one of my views. If I try to connect to it I get a KeyError exception thrown in django_digest/backend/db.py . Depending on which db I configure I get a different KeyError in a different location. I am using Django 1.2.1, with MySql (also tested with sqlite). I am using the default values for all the settings options. As far as I can see I have followed all instructions but am struggling all day with this. I am using the repository versions of django-digest and python-digest. Any steer would be greatly appreciated. Tracebacks for sqlite and mysql below: with sqlite: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 248, in __call__ signals.request_finished.send(sender=self.__class__) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/dispatch/dispatcher.py", line 162, in send response = receiver(signal=self, sender=sender, **named) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 16, in close_connection _connection.close() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/sqlite3/base.py", line 186, in close if self.settings_dict['NAME'] != ":memory:": KeyError: 'NAME' with mysql: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 241, in __call__ response = self.get_response(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 142, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 166, in handle_uncaught_exception return debug.technical_500_response(request, *exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 80, in get_response response = middleware_method(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/middleware.py", line 13, in process_request if (not self._authenticator.authenticate(request) and File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/__init__.py", line 86, in authenticate partial_digest = self._account_storage.get_partial_digest(digest_response.username) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 97, in get_partial_digest cursor = get_connection().cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/__init__.py", line 75, in cursor cursor = self._cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/mysql/base.py", line 281, in _cursor if settings_dict['USER']: KeyError: 'USER'

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  • Google App Engine with local Django 1.1 gets Intermittent Failures

    - by Jon Watte
    I'm using the Windows Launcher development environment for Google App Engine. I have downloaded Django 1.1.2 source, and un-tarrred the "django" subdirectory to live within my application directory (a peer of app.yaml) At the top of each .py source file, I do this: import settings import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' In my file settings.py (which lives at the root of the app directory, as well), I do this: DEBUG = True TEMPLATE_DIRS = ('html') INSTALLED_APPS = ('filters') import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' from google.appengine.dist import use_library use_library('django', '1.1') from django.template import loader Yes, this looks a bit like overkill, doesn't it? I only use django.template. I don't explicitly use any other part of django. However, intermittently I get one of two errors: 1) Django complains that DJANGO_SETTINGS_MODULE is not defined. 2) Django complains that common.html (a template I'm extending in other templates) doesn't exist. 95% of the time, these errors are not encountered, and they randomly just start happening. Once in that state, the local server seems "wedged" and re-booting it generally fixes it. What's causing this to happen, and what can I do about it? How can I even debug it? Here is the traceback from the error: Traceback (most recent call last): File "C:\code\kwbudget\edit_budget.py", line 34, in get self.response.out.write(t.render(template.Context(values))) File "C:\code\kwbudget\django\template\__init__.py", line 165, in render return self.nodelist.render(context) File "C:\code\kwbudget\django\template\__init__.py", line 784, in render bits.append(self.render_node(node, context)) File "C:\code\kwbudget\django\template\__init__.py", line 797, in render_node return node.render(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 71, in render compiled_parent = self.get_parent(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 66, in get_parent raise TemplateSyntaxError, "Template %r cannot be extended, because it doesn't exist" % parent TemplateSyntaxError: Template u'common.html' cannot be extended, because it doesn't exist And edit_budget.py starts with exactly the lines that I included up top. All templates live in a directory named "html" in my root directory, and "html/common.html" exists. I know the template engine finds them, because I start out with "html/edit_budget.html" which extends common.html. It looks as if the settings module somehow isn't applied (because that's what adds html to the search path for templates).

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  • gridview commandargument on buttonfield pagination used

    - by ClareBear
    Hello all, I am using c#.net I have a gridview which needs to contain a 'Use' button (appointmentId set as the commandargument). Source Code <asp:GridView ID="resultsReturned" runat="server" AllowPaging="True" AutoGenerateColumns="False" EnableSortingAndPagingCallbacks="True" OnPageIndexChanged="resultsReturned_PageIndexChanged" onrowcommand="resultsReturned_RowCommand"> <Columns> <asp:BoundField DataField="UserAppointmentId" HeaderText="App ID" /> <asp:BoundField DataField="UserBookingName" HeaderText="Booking Name" /> <asp:TemplateField> <ItemTemplate> <asp:Button runat="server" ID="UseButton" Text="Use" CommandName="Use" CommandArgument="UserAppointmentId" /> </ItemTemplate> </asp:TemplateField> </Columns> </asp:GridView> Code-Behind protected void resultsReturned_RowCommand(object sender, GridViewCommandEventArgs e) { if (e.CommandName == "Use") { correctAppointmentID.Value = (e.CommandArgument.ToString()); } } This is used for the pagination: private void BindAppointments() { var results = appointmentRepos.GetBookingIdBySearchCriteria(catgoryid, resultsReturned.PageIndex * resultsReturned.PageSize, -1); resultsReturned.DataSource = results; resultsReturned.DataBind(); } I am binding the appointments to the gridview within the PageLoad/search_Click This is the error I am receiving: Callbacks are not supported on TemplateField because some controls cannot update properly in a callback. Turn callbacks off on 'resultsReturned'. Thanks in advance for any help Clare

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  • Codeigniter Pagination: Run the Query Twice?

    - by Frank
    I'm using codeigniter and the pagination class. This is such a basic question, but I need to make sure I'm not missing something. In order to get the config items necessary to paginate results getting them from a MySQL database it's basically necessary to run the query twice is that right? In other words, you have to run the query to determine the total number of records before you can paginate. So I'm doing it like: Do this query to get number of results $this->db->where('something', $something); $query = $this->db->get('the_table_name'); $num_rows = $query->num_rows(); Then I'll have to do it again to get the results with the limit and offset. Something like: $this->db->where('something', $something); $this->db->limit($limit, $offset); $query = $this->db->get('the_table_name'); if($query->num_rows()){ foreach($query->result_array() as $row){ ## get the results here } } I just wonder if I'm actually doing this right in that the query always needs to be run twice? The queries I'm using are much more complex than what is shown above.

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  • Strategy for locale sensitive sort with pagination

    - by Thom Birkeland
    Hi, I work on an application that is deployed on the web. Part of the app is search functions where the result is presented in a sorted list. The application targets users in several countries using different locales (= sorting rules). I need to find a solution for sorting correctly for all users. I currently sort with ORDER BY in my SQL query, so the sorting is done according to the locale (or LC_LOCATE) set for the database. These rules are incorrect for those users with a locale different than the one set for the database. Also, to further complicate the issue, I use pagination in the application, so when I query the database I ask for rows 1 - 15, 16 - 30, etc. depending on the page I need. However, since the sorting is wrong, each page contains entries that are incorrectly sorted. In a worst case scenario, the entire result set for a given page could be out of order, depending on the locale/sorting rules of the current user. If I were to sort in (server side) code, I need to retrieve all rows from the database and then sort. This results in a tremendous performance hit given the amount of data. Thus I would like to avoid this. Does anyone have a strategy (or even technical solution) for attacking this problem that will result in correctly sorted lists without having to take the performance hit of loading all data? Tech details: The database is PostgreSQL 8.3, the application an EJB3 app using EJB QL for data query, running on JBoss 4.5.

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  • Total row count for pagination using JPA Criteria API

    - by ThinkFloyd
    I am implementing "Advanced Search" kind of functionality for an Entity in my system such that user can search that entity using multiple conditions(eq,ne,gt,lt,like etc) on attributes of this entity. I am using JPA's Criteria API to dynamically generate the Criteria query and then using setFirstResult() & setMaxResults() to support pagination. All was fine till this point but now I want to show total number of results on results grid but I did not see a straight forward way to get total count of Criteria query. This is how my code looks like: CriteriaBuilder builder = em.getCriteriaBuilder(); CriteriaQuery<Brand> cQuery = builder.createQuery(Brand.class); Root<Brand> from = cQuery.from(Brand.class); CriteriaQuery<Brand> select = cQuery.select(from); . . //Created many predicates and added to **Predicate[] pArray** . . select.where(pArray); // Added orderBy clause TypedQuery typedQuery = em.createQuery(select); typedQuery.setFirstResult(startIndex); typedQuery.setMaxResults(pageSize); List resultList = typedQuery.getResultList(); My result set could be big so I don't want to load my entities for count query, so tell me efficient way to get total count like rowCount() method on Criteria (I think its there in Hibernate's Criteria).

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  • wordpress -> showing custom data from child pages + pagination

    - by GaVrA
    Hello! You can see here what i am doing: http://www.arvag.net/otkrijte-svet/leto/ So its pulling 2 custom fields from child pages, one of them is just url for the image on the left, and the other field is that text showing on the right. Now, what i want to do is to add pagination there. The code i have now will just simply show all child pages, but i want to show only 5 child pages, so if user want to see 6th child page he would have to click on link for "Page 2" and so on. The code im using to display these child pages is this: <?php $pages = get_pages('child_of='.$post->ID.'&sort_column=post_title&sort_order=desc'); $count = 0; foreach($pages as $page) { $short_info = get_post_meta($page->ID,'info',true); $image = get_post_meta($page->ID,'slika',true); $count++; ?> <div class='preview_slika left'><img src="<?php echo $image ?>" alt="<?php echo $page->post_title ?>" /></div> <div class='preview_info right'> <h2><?php echo $page->post_title ?></h2> <p><?php echo $short_info ?></p> <a href="<?php echo get_page_link($page->ID) ?>">Više o >></a> </div> <div class='clear'></div> <?php } ?> So any idea what to do to get what i need?

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  • How do I implement Hibernate Pagination using a cursor (so the results stay consistent, despite new

    - by hunterae
    Hey all, Is there any way to maintain a database cursor using Hibernate between web requests? Basically, I'm trying to implement pagination, but the data that is being paged is consistently changing (i.e. new records are added into the database). We are trying to set it up such that when you do your initial search (returning a maximum of 5000 results), and you page through the results, those same records always appear on the same page (i.e. we're not continuously running the query each time next and previous page buttons are clicked). The way we're currently implementing this is by merely selecting 5000 (at most) primary keys from the table we're paging, storing those keys in memory, and then just using 20 primary keys at a time to fetch their details from the database. However, we want to get away from having to store these keys in memory and would much prefer a database cursor that we just keep going back to and moving backwards and forwards over the cursor to generate pages. I tried doing this with Hibernate's ScrollableResults but found that I could not call methods like next() and previous() would cause an exception if you within a different web request / Hibernate session (no surprise there). Is there any way to reattach a ScrollableResults object to a Session, much the same way you would reattach a detached database object to make it persistent? Are there any other approaches to implement this data paging with consistent paging results without caching the primary keys?

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  • why my pagination link doesnt appear ?

    - by udaya
    This is my script which i have used to paginate ,,The datas are restricted to 4 but the pagination link doesn't appear <? require_once ('Pager/Pager.php'); $connection = mysql_connect( "localhost" , "root" , "" ); mysql_select_db( "ssit",$connection); $result=mysql_query("SELECT dFrindName FROM tbl_friendslist", $connection); $row = mysql_fetch_array($result); $totalItems = $row['total']; $pager_options = array( 'mode' => 'Sliding', // Sliding or Jumping mode. See below. 'perPage' => 4, // Total rows to show per page 'delta' => 4, // See below 'totalItems' => $totalItems, ); $pager = Pager::factory($pager_options); echo $pager->links; list($from, $to) = $pager->getOffsetByPageId(); $from = $from - 1; $perPage = $pager_options['perPage']; $result = mysql_query("SELECT * FROM tbl_friendslist LIMIT 5 , $perPage",$connection); while($row = mysql_fetch_array($result)) { echo $row['dFrindName'].'</br>'; } ?>

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  • CakePHP pagination with HABTM models

    - by nickf
    I'm having some problems with creating pagination with a HABTM relationship. First, the tables and relationships: requests (id, to_location_id, from_location_id) locations (id, name) items_locations (id, item_id, location_id) items (id, name) So, a Request has a Location the request is coming from and a Location the Request is going to. For this question, I'm only concerned about the "to" location. Request --belongsTo--> Location* --hasAndBelongsToMany--> Item (* as "ToLocation") In my RequestController, I want to paginate all the Items in a Request's ToLocation. // RequestsController var $paginate = array( 'Item' => array( 'limit' => 5, 'contain' => array( "Location" ) ) ); // RequestController::add() $locationId = 21; $items = $this->paginate('Item', array( "Location.id" => $locationId )); And this is failing, because it is generating this SQL: SELECT COUNT(*) AS count FROM items Item WHERE Location.id = 21 I can't figure out how to make it actually use the "contain" argument of $paginate... Any ideas?

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  • django & postgres linux hosting (with SSH access) recommendations

    - by Justin Grant
    We're looking for a good place to host our custom Django app (a fork of OSQA) and its postgresql backend. Requirements include: Linux Python 2.6 or (ideally) Python 2.7 Django 1.2 Postgres 8.4 or later DB backup/restore handled by the hoster, not us OS & dev-platform-stack patching/maintenance handled by the hoster, not us SSH access (so we can pull source code from GitHub, so we can install python eggs, etc.) ability to set up cron jobs (e.g. to send out dail email updates) ability to send up to 10K emails/day good performance (not ganged up with a zillion other sites on one CPU, not starved for RAM) FTP or SCP access to web logs dedicated public IP SSL support Costs under $1000/month for a relatively small site (<5M pageviews/month) Good customer service We already have a prototype site running on EC2 on top of a Bitnami DjangoStack. The problem is that we have to patch the OS, patch postgres, etc. We'd really prefer a platform-as-a-service (PaaS) offering, like Heroku offers for Rails apps, where all we need to worry about is deploying our code instead of worrying about system software patching and maintenance. Google App Engine is closest to what we're looking for, but they don't offer relational DB access (not yet at least). Anyone have a recommendation?

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