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  • Django 1.4 dependency when packaging a Precise application

    - by Caustic
    I am trying to package a program I wrote that depends on Django 1.4.1 in Ubuntu 12.04. As Django 1.4.1 isn't available in Precise I am wondering if it is best to: Package up Django 1.4.1 and drop it in my ppa OR write a script that wgets Django at build time and installs. OR Something better that I haven't thought of. I am still inexperienced with packaging and would appreciate some advice Thanks

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  • Correct permissions for /var/www and wordpress

    - by dpbklyn
    Hello and thank you in advance! I am relatively new to ubuntu, so please excuse the newbie-ness of this question... I have set up a LAMP server (ubuntu server 11.10) and I have access via SSH and to the "it works" page from a web browser from inside my network (via ip address) and from outside using dyndns. I have a couple of projects in development with some outside developers and I want to use this server as a development server for testing and for client approvals. We have some Wordpress projects that sit in subdirectories in /var/www/wordpress1 /var/www/wordpress2, etc. I cannot access these sub directories from a browser in order to set up WP--or (I assume) to see the content on a browser. I get a 403 Forbidden error on my browser. I assume that this is a permissions problem. Can you please tell me the proper settings for the permissions to: 1) Allow the developers and me to read/write. 2) to allow WP set up and do its thing 3) Allow visitors to access the site(s) via the web. I should also mention that the subfolder are actually simlinks to folder on another internal hdd--I don't think this will make a difference, but I thought I should disclose. Since I am a newbie to ubuntu, step-by-step directions are greatly appreciated! Thank you for taking the time! dp total 12 drwxr-xr-x 2 root root 4096 2012-07-12 10:55 . drwxr-xr-x 13 root root 4096 2012-07-11 20:02 .. lrwxrwxrwx 1 root root 43 2012-07-11 20:45 admin_media -> /root/django_src/django/contrib/admin/media -rw-r--r-- 1 root root 177 2012-07-11 17:50 index.html lrwxrwxrwx 1 root root 14 2012-07-11 20:42 media -> /hdd/web/media lrwxrwxrwx 1 root root 18 2012-07-12 10:55 wordpress -> /hdd/web/wordpress Here is the result of using chown -R www-data:www-data /var/www total 12 drwxr-xr-x 2 www-data www-data 4096 2012-07-12 10:55 . drwxr-xr-x 13 root root 4096 2012-07-11 20:02 .. lrwxrwxrwx 1 www-data www-data 43 2012-07-11 20:45 admin_media -> /root/django_src/django/contrib/admin/media -rw-r--r-- 1 www-data www-data 177 2012-07-11 17:50 index.html lrwxrwxrwx 1 www-data www-data 14 2012-07-11 20:42 media -> /hdd/web/media lrwxrwxrwx 1 www-data www-data 18 2012-07-12 10:55 wordpress -> /hdd/web/wordpress I am still unable to access via browser...

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  • unittest import error with virtualenv + google-app-engine-django

    - by Ray Yun
    I'm working with google-app-engine-django + zipped django. Just running "python manage.py test" succeeded without error. But with virtualenv, test was failed with "import unittest error". same error with Django 1.1. - OSX 10.5.6 - google-app-engine-django (r101 via svn) : r100 was failed with launcher 1.3.0 - GoogleAppLauncher 1.3.0 - Django 1.1 & 1.1.1 (zipped) : both failed - virtualenv 1.4.5 - virtualenvwrapper 1.24 Error Message: (django_appengine)Reiot:warclouds Reiot$ python manage.py test WARNING:root:Could not read datastore data from /var/folders/UZ/UZ1vQeLFH2ShHk4kIiLcFk+++TI/-Tmp-/django_google-app-engine-django.datastore INFO:root:zipimporter('/Volumes/data/Documents/warclouds/django.zip', 'django/core/serializers/') .WARNING:root:Can't open zipfile /Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg: IOError: [Errno 13] file not accessible: '/Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg' WARNING:root:Can't open zipfile /Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg: IOError: [Errno 13] file not accessible: '/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg' ERROR:root:Exception encountered handling request Traceback (most recent call last): File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 3177, in _HandleRequest self._Dispatch(dispatcher, self.rfile, outfile, env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 3120, in _Dispatch base_env_dict=env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 515, in Dispatch base_env_dict=base_env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2379, in Dispatch self._module_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2289, in ExecuteCGI reset_modules = exec_script(handler_path, cgi_path, hook) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2185, in ExecuteOrImportScript exec module_code in script_module.__dict__ File "/Volumes/data/Documents/warclouds/main.py", line 28, in <module> from appengine_django import InstallAppengineHelperForDjango File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1914, in load_module return self.FindAndLoadModule(submodule, fullname, search_path) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1816, in FindAndLoadModule description) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1767, in LoadModuleRestricted description) File "/Volumes/data/Documents/warclouds/appengine_django/__init__.py", line 44, in <module> import unittest ImportError: No module named unittest INFO:root:"GET / HTTP/1.1" 500 - INFO:root:zipimporter('/Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg', '') INFO:root:zipimporter('/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg', '') F........................................................... ====================================================================== FAIL: a request to the default page works in the dev_appserver ---------------------------------------------------------------------- Traceback (most recent call last): File "/Volumes/data/Documents/warclouds/appengine_django/tests/integration_test.py", line 176, in testBasic self.assertEquals(rv.status_code, 200) AssertionError: 500 != 200 I also tried with console import but it was ok. > which python /Users/Reiot/.virtualenvs/django_appengine/bin/python > python >>> import unittest Here is my environments: $ mkvirtualenv --no-site-packages no-django $ mkvirtualenv --no-site-packages django-1.1 $ mkvirtualenv --no-site-packages django-1.1.1 (django-1.1)$ easy_install Django-1.1.tar (django-1.1.1)$ easy_install Django-1.1.1.tar $ mkdir google-app-engine-django-svn $ cp -r google-app-engine-django-svn google-app-engine-django-svn-django-1.1 // copy appropriate django.zip $ cp -r google-app-engine-django-svn google-app-engine-django-svn-django-1.1.1 // copy appropriate django.zip

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  • Insufficient permissions when calling flickr.auth.oauth.checkToken

    - by Designer 17
    This is a follow up on another question I had asked on stackoverflow a day or so ago. I'm working on trying to call flickr.people.getPhotos... but no matter what I do I keep getting this... jsonFlickrApi({"stat":"fail", "code":99, "message":"Insufficient permissions. Method requires read privileges; none granted."}); but if you were to look at my "Apps You're Using" page (on flickr) you'd see this. So, even though I've authorized the max permissions... flickr says I don't have any granted!? I even used flickr.auth.oauth.checkToken to double check that my access token was right, this was the value returned; jsonFlickrApi({"oauth":{"token":{"_content":"my-access-token"}, "perms":{"_content":"delete"}, "user":{"nsid":"my-user-nsid", "username":"designerseventeen", "fullname":"Designer Seventeen"}}, "stat":"ok"}) Here's how I'm attempting to call flickr.people.getPhotos... <?php // Attempt to call flickr.people.getPhotos $method = "flickr.people.getPhotos"; $format = 'json'; $nsid = 'my-user-nsid'; $sig_string = "{$api_secret}api_key{$api_key}format{$format}method{$method}user_id{$nsid}"; $api_sig = md5( $sig_string ); $flickr_call = "http://api.flickr.com/services/rest/?"; $url = "method=" . $method; $url .= "&api_key=" . $api_key; $url .= "&user_id=" . $nsid; $url .= "&format=" . $format; $url .= "&api_sig=" . $api_sig; $url = $flickr_call . $url; $results = file_get_contents( $url ); $rsp_arr = explode( '&',$results ); print "<pre>"; print_r($rsp_arr); print "</pre>"; I am officially stumped... and in need of help. Thanks!

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  • Handling permissions in a MVP application

    - by Chathuranga
    In a windows forms payroll application employing MVP pattern (for a small scale client) I'm planing user permission handling as follows (permission based) as basically its implementation should be less complicated and straight forward. NOTE : System could be simultaneously used by few users (maximum 3) and the database is at the server side. This is my UserModel. Each user has a list of permissions given for them. class User { string UserID { get; set; } string Name { get; set; } string NIC {get;set;} string Designation { get; set; } string PassWord { get; set; } List <string> PermissionList = new List<string>(); bool status { get; set; } DateTime EnteredDate { get; set; } } When user login to the system it will keep the current user in memory. For example in BankAccountDetailEntering view I control the controller permission as follows. public partial class BankAccountDetailEntering : Form { bool AccountEditable {get; set;} private void BankAccountDetailEntering_Load(object sender, EventArgs e) { cmdEditAccount.enabled = false; OnLoadForm (sender, e); // Event fires... If (AccountEditable ) { cmdEditAccount.enabled=true; } } } In this purpose my all relevant presenters (like BankAccountDetailPresenter) should aware of UserModel as well in addition to the corresponding business Model it is presenting to the View. class BankAccountDetailPresenter { BankAccountDetailEntering _View; BankAccount _Model; User _UserModel; DataService _DataService; BankAccountDetailPresenter( BankAccountDetailEntering view, BankAccount model, User userModel, DataService dataService ) { _View=view; _Model = model; _UserModel = userModel; _DataService = dataService; WireUpEvents(); } private void WireUpEvents() { _View.OnLoadForm += new EventHandler(_View_OnLoadForm); } private void _View_OnLoadForm(Object sender, EventArgs e) { foreach(string s in _UserModel.PermissionList) { If( s =="CanEditAccount") { _View.AccountEditable =true; return; } } } public Show() { _View.ShowDialog(); } } So I'm handling the user permissions in the presenter iterating through the list. Should this be performed in the Presenter or View? Any other more promising ways to do this? Thanks.

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  • What permissions are required for SET IDENTITY_INSERT ON?

    - by AaronBertrand
    SQL Server 2000's SET IDENTITY_INSERT ON topic says: Execute permissions default to the sysadmin fixed server role, and the db_owner and db_ddladmin fixed database roles, and the object owner. While the SET IDENTITY_INSERT topic for SQL Server 2005 (and up) says: User must own the object, or be a member of the sysadmin fixed server role, or the db_owner and db_ddladmin fixed database roles. This was clearly adapted from the 2000 books online and re-written by someone who misinterpreted "db_owner...(read more)

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  • Help recovering broken OS (permissions issue)

    - by Guandalino
    (At the bottom there is an important update.) I was doing experiments in order to backup a remote account to my local system, Ubuntu 12.04 LTS. I'm not confident with duplicity and probably, due to wrong syntax, some local files have been replaced with remote files. This is just a supposition, I'm not sure this is the real cause of OS corruption. The corruption happened after experimenting with backups, so I think I did something wrong at this regard. I was aware there was a problem when I tried to access a command using sudo: $ sudo ls sudo: unable to open /etc/sudoers: Permission denied sudo: no valid sudoers sources found, quitting sudo: unable to initialize policy plugin This is how /etc/sudoers looks like: $ ls -ald /etc/sudoers -r--r----- 1 root root 788 Oct 2 18:30 /etc/sudoers At this point I tried to reboot and now this is the message I get: The system is running in low graphics mode. Your screen, graphics card and input device settings could not be detected correctly. You will need to configure these yourself. I tried to follow the wizard to configure these settings, but without luck (the system prevents me going on when I press "Next"). The thing that makes me a bit less worried is that all the data on the disk seems readable and I'm able to access them using a live cd. I run memtest and RAM seems to be OK. Do you have any idea about how to recover my system? I'm very glad to provide further information, just let me know what info could be helpful. UPDATE. The issue is about wrong permissions and this is how I discovered: I mounted the root partition of the broken OS on /mnt/broken/ (live CD) and did ls /mnt/broken/. I got a permission denied error, while I expected to have the directory listing. I had to do sudo ls /mnt/broken/ and this worked. Thus without having root permission via sudo it's impossible to access the root of broken os. The current output of ls -ld /mnt/broken/ is: drwxr-x--- 29 1000 812 4096 2012-12-08 21:58 /mnt/broken Any thoughts on how to restore the old (working) set of permissions?

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  • Plesk file permissions - Apache/PHP conflicting with user accounts.

    - by hfidgen
    Hiya, I'm building a Drupal site which performs various automatic disk operations using the apache user (id=40). The problem is that the site was set up on a subdomain belonging to user ID 10001 (ie my main FTP account) so the filesystem belongs to that user ID. So I keep getting errors like this: warning: move_uploaded_file() [function.move-uploaded-file]: SAFE MODE Restriction in effect. The script whose uid is 10001 is not allowed to access /var/www/vhosts/domain.com/httpdocs/sites/default/files/images/user owned by uid 48 in /var/www/vhosts/domain.com/httpdocs/includes/file.inc on line 579. I've tried changing the apache group in httpd.conf to apache:psacln, psacln being the default group for all web users but that's not helped. The situation now is: ..../files/images/ = 777 and chown = ftplogin:psacln ..../files/images/user = 775 and chown = apache:psacln ..../files/tmp = 777 and chown = ftplogin:psacln So apparently uid 40 and 10001 both have permissions to write to any of the 3 directories involved, but still can't. Am i missing something here? Can anyone help? Thanks!

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  • Overwrite clean method in Django Custom Forms

    - by John
    Hi I have wrote a custom widget class AutoCompleteWidget(widgets.TextInput): """ widget to show an autocomplete box which returns a list on nodes available to be tagged """ def render(self, name, value, attrs=None): final_attrs = self.build_attrs(attrs, name=name) if not self.attrs.has_key('id'): final_attrs['id'] = 'id_%s' % name if not value: value = '[]' jquery = u""" <script type="text/javascript"> $("#%s").tokenInput('%s', { hintText: "Enter the word", noResultsText: "No results", prePopulate: %s, searchingText: "Searching..." }); $("body").focus(); </script> """ % (final_attrs['id'], reverse('ajax_autocomplete'), value) output = super(AutoTagWidget, self).render(name, "", attrs) return output + mark_safe(jquery) class MyForm(forms.Form): AutoComplete = forms.CharField(widget=AutoCompleteWidget) this widget uses a jquery function which autocompletes a word based on entries from the database. You can preset its initial values by setting prePopulate to a json string in the form ['name': 'some name', 'id': 'some id'] I do this by setting the inital value of the form field to this json string jquery_string = ['name': 'some name', 'id': 'some id'] form = MyForm(initial={'AutoComplete':jquery_string}) When submitting the form the the value of AutoComplete is returned as a comma seperated list of the selected ids e.g. 12,45,43,66 which if what I want. However if there is an error in the form, for example a required field has not been entered the value of the AutoComplete field is now 12,45,43,66 and not the json string which it requires. What is the best way to solve this. I was thinking about overwriting the clean method in the form class but I'm not sure how to find out if any other element has returned an error. e.g. if forms.errors form.cleaned_date['autocomplete'] = json string return form.cleaned_data Thanks

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  • django admin: Add a "remove file" field for Image- or FileFields

    - by w-
    I was hunting around the net for a way to easily allow users to blank out imagefield/filefields they have set in the admin. I found this http://www.djangosnippets.org/snippets/894/ What was really interesting to me here was the code posted in the comment by rfugger remove_the_file = forms.BooleanField(required=False) def save(self, *args, **kwargs): object = super(self.__class__, self).save(*args, **kwargs) if self.cleaned_data.get('remove_the_file'): object.the_file = '' return object When i try to use this in my own form I basically added this to my admin.py which already had a BlahAdmin class BlahModelForm(forms.ModelForm): class Meta: model = Blah remove_img01 = forms.BooleanField(required=False) def save(self, *args, **kwargs): object = super(self.__class__, self).save(*args, **kwargs) if self.cleaned_data.get('remove_img01'): object.img01 = '' return object when i run it I get this error maximum recursion depth exceeded while calling a Python object at this line object = super(self.__class__, self).save(*args, **kwargs) When i think about it for a bit, it seems obvious that it is just infinitely calling itself causing the error. My problem is i can't figure out what is the correct way i should be doing this. Any suggestions? thanks

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  • int() error in django views

    - by Hulk
    def displaydata(request): response_dict = {} offset = int(request.GET.get('iDisplayStart')) There is an error as, int() argument must be a string or a number at the above said line (i.e,`request.GET.get('iDisplayStart')) And in the template code, $(document).ready(function() { $.ajaxSetup({ cache: false }); oTable = $('#qp_table').dataTable( { "aoColumns": [ {"sWidth": "5%" }, {"sWidth": "35%" }, {"sWidth": "27%" }, {"sWidth": "15%"}, { "bSortable": false, "sWidth": "0%"}, {"bSortable": false, "sWidth": "0%"} ], "aaSorting": [[0, 'asc']], "bProcessing": true, "bServerSide": true, "sAjaxSource": "/diaplaydata/", "bJQueryUI": true, "sPaginationType": "full_numbers", "bFilter": false, "oLanguage" : { "sZeroRecords": "No data found", "sProcessing" : "Fetching Data" } });

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  • django templates array assignment

    - by Hulk
    The following is in views: rows=query.evaluation_set.all() row_arr = [] for row in rows: row_arr.append(row.row_details) dict.update({'row_arr' : row_arr ,'col_arr' : col_arr}) return render_to_response('valuemart/show.html',context_instance=RequestContext(request,{'dict': dict})) How to extract the row_Arr array in the templates in javascript and list out all its values.row_Arr contains data of a column <script> var row_arr = '{{dict.row_arr}}'; //extract values here </script> Thanks..

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  • django views getid

    - by Hulk
    class host(models.Model): emp = models.ForeignKey(getname) def __unicode__(self): return self.topic In views there is the code as, real =[] for emp in my_emp: real.append(host.objects.filter(emp=emp.id)) This above results only the values of emp,My question is that how to get the ids along with emp values. Thanks..

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  • django-admin: creating,saving and relating a m2m model

    - by pastylegs
    I have two models: class Production(models.Model): gallery = models.ManyToManyField(Gallery) class Gallery(models.Model): name = models.CharField() I have the m2m relationship in my productions admin, but I want that functionality that when I create a new Production, a default gallery is created and the relationship is registered between the two. So far I can create the default gallery by overwriting the productions save: def save(self, force_insert=False, force_update=False): if not ( Gallery.objects.filter(name__exact="foo").exists() ): g = Gallery(name="foo") g.save() self.gallery.add(g) This creates and saves the model instance (if it doesn't already exist), but I don't know how to register the relationship between the two?

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  • Django Formset management-form validation error

    - by gramware
    I have a form and a formset on my template. The problem is that the formset is throwing validation error claiming that the management form is "missing or has been tampered with". Here is my view @login_required def home(request): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) blogz = list(blog.objects.filter(deleted='0')) delblog = modelformset_factory(blog, exclude=('poster','date' ,'title','content')) if request.user.is_staff== True: staff = 1 else: staff = 0 staffis = 1 if request.method == 'POST': delblogformset = delblog(request.POST) if delblogformset.is_valid(): delblogformset.save() return HttpResponseRedirect('/home') else: delblogformset = delblog(queryset=blog.objects.filter( deleted='0')) blogform = BlogForm(request.POST) if blogform.is_valid(): blogform.save() return HttpResponseRedirect('/home') else: blogform = BlogForm(initial = {'poster':user.id}) blogs= zip(blogz,delblogformset.forms) paginator = Paginator(blogs, 10) # Show 25 contacts per page # Make sure page request is an int. If not, deliver first page. try: page = int(request.GET.get('page', '1')) except ValueError: page = 1 # If page request (9999) is out of range, deliver last page of results. try: blogs = paginator.page(page) except (EmptyPage, InvalidPage): blogs = paginator.page(paginator.num_pages) return render_to_response('home.html', {'user':user, 'blogform':blogform, 'staff': staff, 'staffis': staffis, 'blog':blogs, 'delblog':delblogformset}, context_instance = RequestContext( request )) my template {%block content%} <h2>Home</h2> {% ifequal staff staffis %} {% if form.errors %} <ul> {% for field in form %} <H3 class="title"> <p class="error"> {% if field.errors %}<li>{{ field.errors|striptags }}</li>{% endif %}</p> </H3> {% endfor %} </ul> {% endif %} <h3>Post a Blog to the Front Page</h3> <form method="post" id="form2" action="" class="infotabs accfrm"> {{ blogform.as_p }} <input type="submit" value="Submit" /> </form> <br> <br> {% endifequal %} <div class="pagination"> <span class="step-links"> {% if blog.has_previous %} <a href="?page={{ blog.previous_page_number }}">previous</a> {% endif %} <span class="current"> Page {{ blog.number }} of {{ blog.paginator.num_pages }}. </span> {% if blog.has_next %} <a href="?page={{ blog.next_page_number }}">next</a> {% endif %} </span> <form method="post" action="" class="usertabs accfrm"> {{delblog.management_form}} {% for b, form in blog.object_list %} <div class="blog"> <h3>{{b.title}}</h3> <p>{{b.content}}</p> <p>posted by <strong>{{b.poster}}</strong> on {{b.date}}</p> {% ifequal staff staffis %}<p>{{form.as_p}}<input type="submit" value="Delete" /></p>{% endifequal %} </div> {% endfor %} </form> {%endblock%}

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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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  • Django syncdb error

    - by Hulk
    /mysite/project4 class notes(models.Model): created_by = models.ForeignKey(User) detail = models.ForeignKey(Details) Details and User are in the same module i.e,/mysite/project1 In project1 models i have defined class User(): ...... class Details(): ...... When DB i synced there is an error saying Error: One or more models did not validate: project4: Accessor for field 'detail' clashes with related field . Add a related_name argument to the definition for 'detail'. How can this be solved.. thanks..

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  • Filter Queryset in Django inlineformset_factory

    - by Dave
    I am trying to use inlineformset_factory to generate a formset. My models are defined as: class Measurement(models.Model): subject = models.ForeignKey(Animal) experiment = models.ForeignKey(Experiment) assay = models.ForeignKey(Assay) values = models.CommaSeparatedIntegerField(blank=True, null=True) class Experiment(models.Model): date = models.DateField() notes = models.TextField(max_length = 500, blank=True) subjects= models.ManyToManyField(Subject) in my view i have: def add_measurement(request, experiment_id): experiment = get_object_or_404(Experiment, pk=experiment_id) MeasurementFormSet = inlineformset_factory(Experiment, Measurement, extra=10, exclude=('experiment')) if request.method == 'POST': formset = MeasurementFormSet(request.POST,instance=experiment) if formset.is_valid(): formset.save() return HttpResponseRedirect( experiment.get_absolute_url() ) else: formset = MeasurementFormSet(instance=experiment) return render_to_response("data_entry_form.html", {"formset": formset, "experiment": experiment }, context_instance=RequestContext(request)) but i want to restrict the Measurement.subject field to only subjects defined in the Experiment.subjects queryset. I have tried a couple of different ways of doing this but I am a little unsure what the best way to accomplish this is. I tried to over-ride the BaseInlineFormset class with a new queryset, but couldnt figure out how to correctly pass the experiment parameter.

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  • queries in django

    - by Hulk
    How to query Employee to get all the address related to the employee, Employee.Add.all() doe not work.. class Employee(): Add = models.ManyToManyField(Address) parent = models.ManyToManyField(Parent, blank=True, null=True) class Address(models.Model): address_emp = models.CharField(max_length=512) description = models.TextField() def __unicode__(self): return self.name()

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  • Django and ImageField Question

    - by Hellnar
    Hello I have a such model: Foo (models.Model): slug = models.SlugField(unique=True) image = models.ImageField(upload_to='uploads/') I want to do two things with this: First of all, I want my image to be forced to resize to a specific width and height after the upload. I have tried this reading the documentation but seems to getting error: image = models.ImageField(upload_to='uploads/', height_field=258, width_field=425) Secondly, when adding an item via admin panel, I want my image's file name to be renamed as same as slug, if any issue arises (like if such named image already exists, add "_" to the end as it used to do. IE: My slug is i-love-you-guys , uploaded image such have i-love-you-guys.png at the end.

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  • Django QuerySet filter method returns multiple entries for one record

    - by Yaroslav
    Trying to retrieve blogs (see model description below) that contain entries satisfying some criteria: Blog.objects.filter(entries__title__contains='entry') The results is: [<Blog: blog1>, <Blog: blog1>] The same blog object is retrieved twice because of JOIN performed to filter objects on related model. What is the right syntax for filtering only unique objects? Data model: class Blog(models.Model): name = models.CharField(max_length=100) def __unicode__(self): return self.name class Entry(models.Model): title = models.CharField(max_length=100) blog = models.ForeignKey(Blog, related_name='entries') def __unicode__(self): return self.title Sample data: b1 = Blog.objects.create(name='blog1') e1 = Entry.objects.create(title='entry 1', blog=b1) e1 = Entry.objects.create(title='entry 2', blog=b1)

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  • Django and floatformat tag

    - by Hellnar
    Hello, I want to modify / change the way the floatformat works. By default it changes the input decimal as such: {{ 1.00|floatformat }} -> 1 {{ 1.50|floatformat }} -> 1.5 {{ 1.53|floatformat }} -> 1.53 I want to change this abit as such: If there is a floating part, it should keep the first 2 floating digits. If no floating (which means .00) it should simply cut out the floating part. IE: {{ 1.00|floatformat }} -> 1 {{ 1.50|floatformat }} -> 1.50 {{ 1.53|floatformat }} -> 1.53

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  • django: CheckboxMultiSelect problem with db queries

    - by xiackok
    firstly sorry for my bad english there is a simple model Person. That contains just languages: LANGUAGE_LIS = ( (1, 'English'), (2, 'Turkish'), (3, 'Spanish') ) class Person(models.Model): languages = models.CharField(max_length=100, choices=LANGUAGE_LIST) #languages is multi value (CheckBoxSelectMultiple) and here person_save_form: class person_save_form(forms.ModelForm): languages = forms.CharField(widget=forms.CheckBoxSelectMultiple(choices=LANGUAGE_LIST)) class Meta: model = Person it is ok. but how can i search persons for languages like "get persons who knows turkish and english" in the database (MySQL) record "languages" column seen like "[u'1', u'2']". but i want search persons like this: persons = Person.objects.filter(languages__in=request.POST.getlist('languages'))

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  • Django many to many queries

    - by Hulk
    In the following, How to get designation when querying Emp sc=Emp.objects.filter(pk=profile.emp.id)[0] sc.desg //this gives an error class Emp(models.Model): name = models.CharField(max_length=255, unique=True) address1 = models.CharField(max_length=255) city = models.CharField(max_length=48) state = models.CharField(max_length=48) country = models.CharField(max_length=48) desg = models.ManyToManyField(Designation) class Designation(models.Model): description = models.TextField() title = models.TextField() def __unicode__(self): return self.board

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  • Tricky model inheritance - Django

    - by RadiantHex
    Hi folks, I think this is a bit tricky, at least for me. :) So I have 4 models Person, Singer, Bassist and Ninja. Singer, Bassist and Ninja inherit from Person. The problem is that each Person can be any of its subclasses. e.g. A person can be a Singer and a Ninja. Another Person can be a Bassist and a Ninja. Another one can be all three. How should I organise my models? Help would be much appreciated!

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