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  • Prepopulate drop-box according to another drop-box choice in Django Admin

    - by onorua
    I have models like this: class User(models.Model): Switch = models.ForeignKey(Switch, related_name='SwitchUsers') Port = models.ForeignKey(Port) class Switch(models.Model): Name = models.CharField(max_length=50) class Port(models.Model): PortNum = models.PositiveIntegerField() Switch = models.ForeignKey(Switch, related_name = "Ports") When I'm in Admin interface and choose Switch from Switches available, I would like to have Port prepopulated accordingly with Ports from the related Switch. As far as I understand I need to create some JS script to prepopulate it. Unfortunately I don't have this experience, and I would like to keep things simple as it possible and don't rewrite all Django admin interface. Just add this functionality for one Field. Could you please help me with my problem? Thank you.

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  • etree.findall: 'OR'-lookup?

    - by piquadrat
    I want to find all stylesheet definitions in a XHTML file with lxml.etree.findall. This could be as simple as elems = tree.findall('link[@rel="stylesheet"]') + tree.findall('style') But the problem with CSS style definitions is that the order matters, e.g. <link rel="stylesheet" type="text/css" href="/media/css/first.css" /> <style>body:{font-size: 10px;}</style> <link rel="stylesheet" type="text/css" href="/media/css/second.css" /> if the contents of the style tag is applied after the rules in the two link tags, the result may be completely different from the one where the rules are applied in order of definition. So, how would I do a lookup that inlcudes both link[@rel="stylesheet"] and style?

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  • What could cause Django to start failing its own tests after an OS and Django reinstall?

    - by Macha
    I had to reinstall my OS, and so, I reinstalled django 1.1. Since reinstalling, when I run tests in my app, I get several failures from django.contrib.auth. Logs: http://dpaste.com/178153/ I asked on #django, and no one is too sure what the cause of the errors are. Some of my own code fails its tests, because it's not fully written yet, but that shouldn't cause django to fail it's core tests... I have included django.contrib.admin, which was mentioned as a possible cause.

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  • How do I code this relationship in SQLAlchemy?

    - by Martin Del Vecchio
    I am new to SQLAlchemy (and SQL, for that matter). I can't figure out how to code the idea I have in my head. I am creating a database of performance-test results. A test run consists of a test type and a number (this is class TestRun below) A test suite consists the version string of the software being tested, and one or more TestRun objects (this is class TestSuite below). A test version consists of all test suites with the given version name. Here is my code, as simple as I can make it: from sqlalchemy import * from sqlalchemy.ext.declarative import declarative_base from sqlalchemy.orm import relationship, backref, sessionmaker Base = declarative_base() class TestVersion (Base): __tablename__ = 'versions' id = Column (Integer, primary_key=True) version_name = Column (String) def __init__ (self, version_name): self.version_name = version_name class TestRun (Base): __tablename__ = 'runs' id = Column (Integer, primary_key=True) suite_directory = Column (String, ForeignKey ('suites.directory')) suite = relationship ('TestSuite', backref=backref ('runs', order_by=id)) test_type = Column (String) rate = Column (Integer) def __init__ (self, test_type, rate): self.test_type = test_type self.rate = rate class TestSuite (Base): __tablename__ = 'suites' directory = Column (String, primary_key=True) version_id = Column (Integer, ForeignKey ('versions.id')) version_ref = relationship ('TestVersion', backref=backref ('suites', order_by=directory)) version_name = Column (String) def __init__ (self, directory, version_name): self.directory = directory self.version_name = version_name # Create a v1.0 suite suite1 = TestSuite ('dir1', 'v1.0') suite1.runs.append (TestRun ('test1', 100)) suite1.runs.append (TestRun ('test2', 200)) # Create a another v1.0 suite suite2 = TestSuite ('dir2', 'v1.0') suite2.runs.append (TestRun ('test1', 101)) suite2.runs.append (TestRun ('test2', 201)) # Create another suite suite3 = TestSuite ('dir3', 'v2.0') suite3.runs.append (TestRun ('test1', 102)) suite3.runs.append (TestRun ('test2', 202)) # Create the in-memory database engine = create_engine ('sqlite://') Session = sessionmaker (bind=engine) session = Session() Base.metadata.create_all (engine) # Add the suites in version1 = TestVersion (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = TestVersion (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = TestVersion (suite3.version_name) version3.suites.append (suite3) session.add (suite3) session.commit() # Query the suites for suite in session.query (TestSuite).order_by (TestSuite.directory): print "\nSuite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) # Query the versions for version in session.query (TestVersion).order_by (TestVersion.version_name): print "\nVersion %s has %d test suites:" % (version.version_name, len (version.suites)) for suite in version.suites: print " Suite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) The output of this program: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 1 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Version v1.0 has 1 test suites: Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This is not correct, since there are two TestVersion objects with the name 'v1.0'. I hacked my way around this by adding a private list of TestVersion objects, and a function to find a matching one: versions = [] def find_or_create_version (version_name): # Find existing for version in versions: if version.version_name == version_name: return (version) # Create new version = TestVersion (version_name) versions.append (version) return (version) Then I modified my code that adds the records to use it: # Add the suites in version1 = find_or_create_version (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = find_or_create_version (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = find_or_create_version (suite3.version_name) version3.suites.append (suite3) session.add (suite3) Now the output is what I want: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 2 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This feels wrong to me; it doesn't feel right that I am manually keeping track of the unique version names, and manually adding the suites to the appropriate TestVersion objects. Is this code even close to being correct? And what happens when I'm not building the entire database from scratch, as in this example. If the database already exists, do I have to query the database's TestVersion table to discover the unique version names? Thanks in advance. I know this is a lot of code to wade through, and I appreciate the help.

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  • Joining different models in Django

    - by Andrew Roberts
    Let's say I have this data model: class Workflow(models.Model): ... class Command(models.Model): workflow = models.ForeignKey(Workflow) ... class Job(models.Model): command = models.ForeignKey(Command) ... Suppose somewhere I want to loop through all the Workflow objects, and for each workflow I want to loop through its Commands, and for each Command I want to loop through each Job. Is there a way to structure this with a single query? That is, I'd like Workflow.objects.all() to join in its dependent models, so I get a collection that has dependent objects already cached, so workflows[0].command_set.get() doesn't produce an additional query. Is this possible?

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  • how to make my method running on the template of google-app-engine..

    - by zjm1126
    the model is : class someModel(db.Model): name = db.StringProperty() def name_is_sss(self): return self.name=='sss' the view is : a=someModel() a.name='sss' path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog/a.html')) self.response.out.write(template.render(path, {'a':a})) and the html is : {{ a.name_is_sss }} the page shows : True so i want to make it more useful, and like this: the model: class someModel(db.Model): name = db.StringProperty() def name_is_x(self,x): return self.name==x the html is : {% a.name_is_x 'www'%} or {{ a.name_is_x 'www'}} but the error is : TemplateSyntaxError: Invalid block tag: 'a.name_is_x' or TemplateSyntaxError: Could not parse the remainder: 'www' so how to make my method running thanks

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  • How to repeatedly show a Dialog with PyGTK / Gtkbuilder?

    - by Julian
    I have created a PyGTK application that shows a Dialog when the user presses a button. The dialog is loaded in my __init__ method with: builder = gtk.Builder() builder.add_from_file("filename") builder.connect_signals(self) self.myDialog = builder.get_object("dialog_name") In the event handler, the dialog is shown with the command self.myDialog.run(), but this only works once, because after run() the dialog is automatically destroyed. If I click the button a second time, the application crashes. I read that there is a way to use show() instead of run() where the dialog is not destroyed, but I feel like this is not the right way for me because I would like the dialog to behave modally and to return control to the code only after the user has closed it. Is there a simple way to repeatedly show a dialog using the run() method using gtkbuilder? I tried reloading the whole dialog using the gtkbuilder, but that did not really seem to work, the dialog was missing all child elements (and I would prefer to have to use the builder only once, at the beginning of the program). [SOLUTION] As pointed out by the answer below, using hide() does the trick. But one has to take care that the dialog is in fact destroyed if one does not catch its "delete-event". A simple example that works is: import pygtk import gtk class DialogTest: def rundialog(self, widget, data=None): self.dia.show_all() result = self.dia.run() def destroy(self, widget, data=None): gtk.main_quit() def closedialog(self, widget, data=None): self.dia.hide() return True def __init__(self): self.window = gtk.Window(gtk.WINDOW_TOPLEVEL) self.window.connect("destroy", self.destroy) self.dia = gtk.Dialog('TEST DIALOG', self.window, gtk.DIALOG_MODAL | gtk.DIALOG_DESTROY_WITH_PARENT) self.dia.vbox.pack_start(gtk.Label('This is just a Test')) self.dia.connect("delete-event", self.closedialog) self.button = gtk.Button("Run Dialog") self.button.connect("clicked", self.rundialog, None) self.window.add(self.button) self.button.show() self.window.show() if __name__ == "__main__": testApp = DialogTest() gtk.main()

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  • Problem bounding name to a class in Django

    - by martinthenext
    Hello! I've got a view function that has to decide which form to use depending on some conditions. The two forms look like that: class OpenExtraForm(forms.ModelForm): class Meta: model = Extra def __init__(self, *args, **kwargs): super(OpenExtraForm, self).__init__(*args, **kwargs) self.fields['opening_challenge'].label = "lame translation" def clean_opening_challenge(self): challenge = self.cleaned_data['opening_challenge'] if challenge is None: raise forms.ValidationError('??????? ???, ??????????? ?????? ???. ???????????') return challenge class HiddenExtraForm(forms.ModelForm): class Meta: model = Extra exclude = ('opening_challenge') def __init__(self, *args, **kwargs): super(HiddenExtraForm, self).__init__(*args, **kwargs) The view code goes like that: @login_required def manage_extra(request, extra_id=None, hidden=False): if not_admin(request.user): raise Http404 if extra_id is None: # Adding a new extra extra = Extra() if hidden: FormClass = HiddenExtraForm else: FormClass = OpenExtraForm else: # Editing an extra extra = get_object_or_404(Extra, pk=extra_id) if extra.is_hidden(): FromClass = HiddenExtraForm else: FormClass = OpenExtraForm if request.POST: form = FormClass(request.POST, instance=extra) if form.is_valid(): form.save() return HttpResponseRedirect(reverse(view_extra, args=[extra.id])) else: form = FormClass(instance=extra) return render_to_response('form.html', { 'form' : form, }, context_instance=RequestContext(request) ) The problem is somehow if extra.is_hidden() returns True, the statement FromClass = HiddenExtraForm doesn't work. I mean, in all other conditions that are used in the code it works fine: the correct Form classes are intantiated and it all works. But if extra.is_hidden(), the debugger shows that the condition is passed and it goes to the next line and does nothing! As a result I get a UnboundLocalVar error which says FormClass hasn't been asssigned at all. Any ideas on what's happening?

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  • How to create instances of a class from a static method?

    - by Pierre
    Hello. Here is my problem. I have created a pretty heavy readonly class making many database calls with a static "factory" method. The goal of this method is to avoid killing the database by looking in a pool of already-created objects if an identical instance of the same object (same type, same init parameters) already exists. If something was found, the method will just return it. No problem. But if not, how may I create an instance of the object, in a way that works with inheritance? >>> class A(Object): >>> @classmethod >>> def get_cached_obj(self, some_identifier): >>> # Should do something like `return A(idenfier)`, but in a way that works >>> class B(A): >>> pass >>> A.get_cached_obj('foo') # Should do the same as A('foo') >>> A().get_cached_obj('foo') # Should do the same as A('foo') >>> B.get_cached_obj('bar') # Should do the same as B('bar') >>> B().get_cached_obj('bar') # Should do the same as B('bar') Thanks.

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  • Process Name with inotify tool

    - by Priyanka
    Hi ALl Is there a way to find out which process is causing a particular event to occur? In the inotify-tool version 3.13, in struct_kernel , pid,uid and gid are used but it is not handled in the source code. Is there any latest patch available for the same. Please let me know.

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  • Using Range Function

    - by Michael Alexander Riechmann
    My goal is to make a program that takes an input (Battery_Capacity) and ultimately spits out a list of the (New_Battery_Capacity) and the Number of (Cycle) it takes for it ultimately to reach maximum capacity of 80. Cycle = range (160) Charger_Rate = 0.5 * Cycle Battery_Capacity = float(raw_input("Enter Current Capacity:")) New_Battery_Capacity = Battery_Capacity + Charger_Rate if Battery_Capacity < 0: print 'Battery Reading Malfunction (Negative Reading)' elif Battery_Capacity > 80: print 'Battery Reading Malfunction (Overcharged)' elif float(Battery_Capacity) % 0.5 !=0: print 'Battery Malfunction (Charges Only 0.5 Interval)' while Battery_Capacity >= 0 and Battery_Capacity < 80: print New_Battery_Capacity I was wondering why my Cycle = range(160) isn't working in my program?

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  • Writing good tests for Django applications

    - by Ludwik Trammer
    I've never written any tests in my life, but I'd like to start writing tests for my Django projects. I've read some articles about tests and decided to try to write some tests for an extremely simple Django app or a start. The app has two views (a list view, and a detail view) and a model with four fields: class News(models.Model): title = models.CharField(max_length=250) content = models.TextField() pub_date = models.DateTimeField(default=datetime.datetime.now) slug = models.SlugField(unique=True) I would like to show you my tests.py file and ask: Does it make sense? Am I even testing for the right things? Are there best practices I'm not following, and you could point me to? my tests.py (it contains 11 tests): # -*- coding: utf-8 -*- from django.test import TestCase from django.test.client import Client from django.core.urlresolvers import reverse import datetime from someproject.myapp.models import News class viewTest(TestCase): def setUp(self): self.test_title = u'Test title: bareksc' self.test_content = u'This is a content 156' self.test_slug = u'test-title-bareksc' self.test_pub_date = datetime.datetime.today() self.test_item = News.objects.create( title=self.test_title, content=self.test_content, slug=self.test_slug, pub_date=self.test_pub_date, ) client = Client() self.response_detail = client.get(self.test_item.get_absolute_url()) self.response_index = client.get(reverse('the-list-view')) def test_detail_status_code(self): """ HTTP status code for the detail view """ self.failUnlessEqual(self.response_detail.status_code, 200) def test_list_status_code(self): """ HTTP status code for the list view """ self.failUnlessEqual(self.response_index.status_code, 200) def test_list_numer_of_items(self): self.failUnlessEqual(len(self.response_index.context['object_list']), 1) def test_detail_title(self): self.failUnlessEqual(self.response_detail.context['object'].title, self.test_title) def test_list_title(self): self.failUnlessEqual(self.response_index.context['object_list'][0].title, self.test_title) def test_detail_content(self): self.failUnlessEqual(self.response_detail.context['object'].content, self.test_content) def test_list_content(self): self.failUnlessEqual(self.response_index.context['object_list'][0].content, self.test_content) def test_detail_slug(self): self.failUnlessEqual(self.response_detail.context['object'].slug, self.test_slug) def test_list_slug(self): self.failUnlessEqual(self.response_index.context['object_list'][0].slug, self.test_slug) def test_detail_template(self): self.assertContains(self.response_detail, self.test_title) self.assertContains(self.response_detail, self.test_content) def test_list_template(self): self.assertContains(self.response_index, self.test_title)

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  • Editting ForeignKey from "child" table

    - by profuel
    I'm programming on py with django. I have models: class Product(mymodels.Base): title = models.CharField() price = models.ForeignKey(Price) promoPrice = models.ForeignKey(Price, related_name="promo_price") class Price(mymodels.Base): value = models.DecimalField(max_digits=10, decimal_places=3) taxValue = models.DecimalField("Tax Value", max_digits=10, decimal_places=3) valueWithTax = models.DecimalField("Value with Tax", max_digits=10, decimal_places=3) I want to see INPUTs for both prices when editing product, but cannot find any possibility to do that. inlines = [...] works only from Price to Product, which is stupid in this case. Thanx for adnvance.

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  • GUI IDE with PyDev Eclipse

    - by gizgok
    I have 2 weeks to finish my final year project.I need a GUI IDE or a GUI framework compatible with PyDev and Eclipse. I cannot spend time learning something cause the functionality is yet to be completed.I'm looking for very simple GUI for a simulation game.

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  • How can I login in a website with Pyhon?

    - by Shady
    How can I do it? I was trying to enter some specified link (with urllib), but to do it, I need to log. I have this source from the site <form id="login-form" action="auth/login" method="post"> <div> <!--label for="rememberme">Remember me</label><input type="checkbox" class="remember" checked="checked" name="remember me" /--> <label for="email" id="email-label" class="no-js">Email</label> <input id="email-email" type="text" name="handle" value="" autocomplete="off" /> <label for="combination" id="combo-label" class="no-js">Combination</label> <input id="password-clear" type="text" value="Combination" autocomplete="off" /> <input id="password-password" type="password" name="password" value="" autocomplete="off" /> <input id="sumbitLogin" class="signin" type="submit" value="Sign In" /> It's possible?

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  • List Directories and get the name of the Directory

    - by chrissygormley
    Hello, I am trying to get the code to list all the directories in a folder, change directory into that folder and get the name of the current folder. The code I have so far is below and isn't working at the minute. I seem to be getting the parent folder name. import os for directories in os.listdir(os.getcwd()): dir = os.path.join('/home/user/workspace', directories) os.chdir(dir) current = os.path.dirname(dir) new = str(current).split("-")[0] print new I also have other files in the folder but I do not want to list them. I have tried the below code but I haven't got it working yet either. for directories in os.path.isdir(os.listdir(os.getcwd())): Can anyone see where I am going wrong? Thanks

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