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  • Efficient way to combine results of two database queries.

    - by ensnare
    I have two tables on different servers, and I'd like some help finding an efficient way to combine and match the datasets. Here's an example: From server 1, which holds our stories, I perform a query like: query = """SELECT author_id, title, text FROM stories ORDER BY timestamp_created DESC LIMIT 10 """ results = DB.getAll(query) for i in range(len(results)): #Build a string of author_ids, e.g. '1314,4134,2624,2342' But, I'd like to fetch some info about each author_id from server 2: query = """SELECT id, avatar_url FROM members WHERE id IN (%s) """ values = (uid_list) results = DB.getAll(query, values) Now I need some way to combine these two queries so I have a dict that has the story as well as avatar_url and member_id. If this data were on one server, it would be a simple join that would look like: SELECT * FROM members, stories WHERE members.id = stories.author_id But since we store the data on multiple servers, this is not possible. What is the most efficient way to do this? Thanks.

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  • Can't iterate over nestled dict in django

    - by fredrik
    Hi, Im trying to iterate over a nestled dict list. The first level works fine. But the second level is treated like a string not dict. In my template I have this: {% for product in Products %} <li> <p>{{ product }}</p> {% for partType in product.parts %} <p>{{ partType }}</p> {% for part in partType %} <p>{{ part }}</p> {% endfor %} {% endfor %} </li> {% endfor %} It's the {{ part }} that just list 1 char at the time based on partType. And it seams that it's treated like a string. I can however via dot notation reach all dict but not with a for loop. The current output looks like this: Color C o l o r Style S ..... The Products object looks like this in the log: [{'product': <models.Products.Product object at 0x1076ac9d0>, 'parts': {u'Color': {'default': u'Red', 'optional': [u'Red', u'Blue']}, u'Style': {'default': u'Nice', 'optional': [u'Nice']}, u'Size': {'default': u'8', 'optional': [u'8', u'8.5']}}}] What I trying to do is to pair together a dict/list for a product from a number of different SQL queries. The web handler looks like this: typeData = Products.ProductPartTypes.all() productData = Products.Product.all() langCode = 'en' productList = [] for product in productData: typeDict = {} productDict = {} for type in typeData: typeDict[type.typeId] = { 'default' : '', 'optional' : [] } productDict['product'] = product productDict['parts'] = typeDict defaultPartsData = Products.ProductParts.gql('WHERE __key__ IN :key', key = product.defaultParts) optionalPartsData = Products.ProductParts.gql('WHERE __key__ IN :key', key = product.optionalParts) for defaultPart in defaultPartsData: label = Products.ProductPartLabels.gql('WHERE __key__ IN :key AND partLangCode = :langCode', key = defaultPart.partLabelList, langCode = langCode).get() productDict['parts'][defaultPart.type.typeId]['default'] = label.partLangLabel for optionalPart in optionalPartsData: label = Products.ProductPartLabels.gql('WHERE __key__ IN :key AND partLangCode = :langCode', key = optionalPart.partLabelList, langCode = langCode).get() productDict['parts'][optionalPart.type.typeId]['optional'].append(label.partLangLabel) productList.append(productDict) logging.info(productList) templateData = { 'Languages' : Settings.Languges.all().order('langCode'), 'ProductPartTypes' : typeData, 'Products' : productList } I've tried making the dict in a number of different ways. Like first making a list, then a dict, used tulpes anything I could think of. Any help is welcome! Bouns: If someone have an other approach to the SQL quires, that is more then welcome. I feel that it kinda stupid to run that amount of quires. What is happening that each product part has a different label base on langCode. ..fredrik

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  • How to manage feeds with subclassed object in Django 1.2?

    - by Matteo
    Hi, I'm trying to generate a feed rss from a model like this one, selecting all the Entry objects: from django.db import models from django.contrib.sites.models import Site from django.contrib.auth.models import User from imagekit.models import ImageModel import datetime class Entry(ImageModel): date_pub = models.DateTimeField(default=datetime.datetime.now) author = models.ForeignKey(User) via = models.URLField(blank=True) comments_allowed = models.BooleanField(default=True) icon = models.ImageField(upload_to='icon/',blank=True) class IKOptions: spec_module = 'journal.icon_specs' cache_dir = 'icon/resized' image_field = 'icon' class Post(Entry): title = models.CharField(max_length=200) description = models.TextField() slug = models.SlugField(unique=True) def __unicode__(self): return self.title class Photo(Entry): alt = models.CharField(max_length=200) description = models.TextField(blank=True) original = models.ImageField(upload_to='photo/') class IKOptions: spec_module = 'journal.photo_specs' cache_dir = 'photo/resized' image_field = 'original' def __unicode__(self): return self.alt class Quote(Entry): blockquote = models.TextField() cite = models.TextField(blank=True) def __unicode__(self): return self.blockquote When I use the render_to_response in my views I simply call: def get_journal_entries(request): entries = Entry.objects.all().order_by('-date_pub') return render_to_response('journal/entries.html', {'entries':entries}) And then I use a conditional template to render the right snippets of html: {% extends "base.html" %} {% block main %} <hr> {% for entry in entries %} {% if entry.post %}[...]{% endif %}[...] But I cannot do the same with the Feed Framework in django 1.2... Any suggestion, please?

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  • CherryPy always returning HTTP 200 [closed]

    - by DarkArctic
    I'm having a bit of a problem when browsing to a non-existent resource. I get a response code of 200 instead of 404. I'm using the MethodDispatcher and I have a class that overloads the __getattr__ method to instantiate a resource if a child exists or to return AttributeError if one doesn't. My class is always returning the AttributeError correctly, but the data I actually get is always from the last good resource. Here's a simplified (except for __getattr__) version of my class: class BaseResource(object): exposed = True def __init__(self, name): self.children = [] # Pretend this has child resources def __getattr__(self, name): if name in self._children: uuid, application, obj_type, server = self._children[name] try: resource = getattr(app[application], obj_type) except AttributeError as e: raise cherrypy.HTTPError(500, e) return resource(uuid) else: raise AttributeError('Child with name \'{}\' could not be found.'.format(name)) def GET(self): cherrypy.log.error('*** {} not found, raising AttributeError'.format(name)) return 'GET request for {}'.format(self._name) So fetching I get the following when I browse to the following resources: http://localhost:8000/users - This resource exists, so it returns it correctly. http://localhost:8000/users/fake - This returns the "users" resource giving an HTTP 200. http://localhost:8000/users/fake/reallyfake - This returns the "users" resource again. So my question is, where can I start looking to find out why my code isn't returning a 404 for a non-existent resource. I'm sure I've done something wrong, but I'm not sure what. Whatever I did wrong I've undone and I'm now getting a 404 returned correctly. I'm sorry I can't give any detail on what the issue was, but I'm honestly not sure what I did.

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  • get the list and input from one function and run them in different function

    - by rookie
    i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code def menu(x,l): print (x) if x == 1: return make_table(l) if x == 2: y= input("enter a row (as a number) or a column (as an uppercase letter") if y in [ "1",'2','3']: print("Minmum is:",minimum(y,l)) if x== 3: print ('bye') def main(): bad_filename = True l =[] while bad_filename == True: try: filename = input("Enter the filename: ") fp = open(filename, "r") for f_line in fp: f_str=f_line.strip() f_str=f_str.split(',') for unit_str in f_str: unit=float(unit_str) l.append(unit) bad_filename = False except IOError: print("Error: The file was not found: ", filename) #print(l) condition=True while condition==True: print('1- open\n','2- maximum') x=input("Enter the choice") menu(x,l) main() from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you

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  • Online job-searching is tedious. Help me automate it.

    - by ehsanul
    Many job sites have broken searches that don't let you narrow down jobs by experience level. Even when they do, it's usually wrong. This requires you to wade through hundreds of postings that you can't apply for before finding a relevant one, quite tedious. Since I'd rather focus on writing cover letters etc., I want to write a program to look through a large number of postings, and save the URLs of just those jobs that don't require years of experience. I don't require help writing the scraper to get the html bodies of possibly relevant job posts. The issue is accurately detecting the level of experience required for the job. This should not be too difficult as job posts are usually very explicit about this ("must have 5 years experience in..."), but there may be some issues with overly simple solutions. In my case, I'm looking for entry-level positions. Often they don't say "entry-level", but inclusion of the words probably means the job should be saved. Next, I can safely exclude a job the says it requires "5 years" of experience in whatever, so a regex like /\d\syears/ seems reasonable to exclude jobs. But then, I realized some jobs say they'll take 0-2 years of experience, matches the exclusion regex but is clearly a job I want to take a look at. Hmmm, I can handle that with another regex. But some say "less than 2 years" or "fewer than 2 years". Can handle that too, but it makes me wonder what other patterns I'm not thinking of, and possibly excluding many jobs. That's what brings me here, to find a better way to do this than regexes, if there is one. I'd like to minimize the false negative rate and save all the jobs that seem like they might not require many years of experience. Does excluding anything that matches /[3-9]\syears|1\d\syears/ seem reasonable? Or is there a better way? Training a bayesian filter maybe?

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  • PyGTK: Radiobuttons are still displayed after removal

    - by canavanin
    Hi everyone! I am using PyGTK and the gtk.assistant. On one page I would like to display two radiobuttons in case the user selected a certain option on a previous page. The labels of the buttons - and whether the buttons are to be present at all - are to depend entirely on that earlier selection. Furthermore, if the user goes back and changes that selection, the page containing the radiobuttons is to be updated accordingly. I have got as far as having the radiobuttons displayed when necessary, and with the correct labels. The trouble is that if I go back and change the determining selection, or if I move one page further than the 'radiobutton page' and then move back, the buttons are not only not removed (in case that would have been required), their number has also doubled. To show you what I'm doing, here's part of my code (I've left out bits that do unrelated things, that's why the function name doesn't fit). The function is called when the "prepare" signal is emitted prior to construction of the 'radiobutten page'. def make_class_skills_treestore(self): print self.trained_by_default_hbox.get_children() # PRINT 1 for child in self.trained_by_default_hbox.get_children(): if type(child) == gtk.RadioButton: self.trained_by_default_hbox.remove(child) #child.destroy() # <-- removed the labels, but not the buttons print self.trained_by_default_hbox.get_children() # PRINT 2 class_skills = self.data.data['classes'][selected_class].class_skills.values() default_trained_count = (class_skills.count([True, True]) , class_skills.count([True, False])) num_default_trained_skills = default_trained_count[1] / 2 # you have to pick one of a pair --> don't # count each as trained by default for i in range(default_trained_count[0]): # those are trained by default --> no choice num_default_trained_skills +=1 selected_class = self.get_classes_key_from_class_selection() if default_trained_count[1]: for skill in self.data.data['classes'][selected_class].class_skills.keys(): if self.data.data['classes'][selected_class].class_skills[skill] == [ True, False ] and not self.default_radio: self.default_radio.append(gtk.RadioButton(group=None, label=skill)) elif self.data.data['classes'][selected_class].class_skills[skill] == [ True, False ] and self.default_radio: self.default_radio.append(gtk.RadioButton(group=self.default_radio[0], label=skill)) if self.default_radio: for radio in self.default_radio: self.trained_by_default_hbox.add(radio) self.trained_by_default_hbox.show_all() self.trained_by_default_hbox and self.trained_by_default_label, as well as self.default_radio stem from the above function's class. I have two print statements (PRINT 1 and PRINT 2) in there for debugging. Here's what they give me: PRINT 1: [<gtk.Label object at 0x8fc4c84 (GtkLabel at 0x90a2f20)>, <gtk.RadioButton object at 0x8fc4d4c (GtkRadioButton at 0x90e4018)>, <gtk.RadioButton object at 0x8fc4cac (GtkRadioButton at 0x90ceec0)>] PRINT 2: [<gtk.Label object at 0x8fc4c84 (GtkLabel at 0x90a2f20)>] So the buttons have indeed been removed, yet they still show up on the page. I know the code requires some refactoring, but first I'd like to get it to work at all... If someone could help me out that would be great! Thanks a lot in advance for your replies - any kind of help is highly appreciated.

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  • removing elements incrementally from a list

    - by Javier
    Dear all, I've a list of float numbers and I would like to delete incrementally a set of elements in a given range of indexes, sth. like: for j in range(beginIndex, endIndex+1): print ("remove [%d] => val: %g" % (j, myList[j])) del myList[j] However, since I'm iterating over the same list, the indexes (range) are not valid any more for the new list. Does anybody has some suggestions on how to delete the elements properly? Best wishes

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  • How do you get SQLAlchemy to override MySQL "on update CURRENT_TIMESTAMP"

    - by nocola
    I've inherited an older database that was setup with a "on update CURRENT_TIMESTAMP" put on a field that should only describe an item's creation. With PHP I have been using "timestamp=timestamp" on UPDATE clauses, but in SQLAlchemy I can't seem to force the system to use the set timestamp. Do I have no choice and need to update the MySQL table (millions of rows)? foo = session.query(f).get(int(1)) ts = foo.timestamp setattr(foo, 'timestamp', ts) setattr(foo, 'bar', bar) www_model.www_Session.commit() I have also tried: foo = session.query(f).get(int(1)) setattr(foo, 'timestamp', foo.timestamp) setattr(foo, 'bar', bar) www_model.www_Session.commit()

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  • Is there something similar to 'rake routes' in django?

    - by The MYYN
    In rails, on can show the active routes with rake (http://guides.rubyonrails.org/routing.html): $ rake routes users GET /users {:controller=>"users", :action=>"index"} formatted_users GET /users.:format {:controller=>"users", :action=>"index"} POST /users {:controller=>"users", :action=>"create"} POST /users.:format {:controller=>"users", :action=>"create"} Is there a similar tool/command for django showing the e.g. the URL pattern, the name of the pattern (if any) and the associated function in the views?

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  • Query distinct list of choices for Django form with App Engine Datastore

    - by Brian
    I've been trying to figure this out for hours across a couple of days, and can not get it to work. I've been everywhere. I'll continue trying to figure it out, but was hoping for a quicker solution. I'm using App Engine datastore + Django. Using a query in a view and custom forms, I was able to get a list to the form but then I was not able to post. I have been trying to figure out how to dynamically add the choices as part of the Django form... I've tried various ways with no success. Help! Below are the two models. I'd like to get a distinct list of address_id to show in the location field in InfoForm. This fields could (and maybe should) be named the same, but I thought it'd be easier if they were named different. class Info(db.Model): user = db.UserProperty() location = db.StringProperty() info = db.StringProperty() created = db.DateTimeProperty(auto_now_add=True) modified = db.DateTimeProperty(auto_now=True) class Locations(db.Model): user = db.UserProperty() address_id = db.StringProperty() address = db.StringProperty() class InfoForm(djangoforms.ModelForm): info = forms.ChoiceField(choices=INFO_CHOICES) location = forms.ChoiceField() class Meta: model = Info exclude = ['user','created','modified']

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  • Django: how to create sites dynamically?

    - by Leandro Ardissone
    Hi, I need to create an application for the company where I can create sites dynamically. For example, I need an admin interface (Django's admin is enough) where I can setup a new site and add some settings to it. Each site must hold a domain (domains can be manually added to apache conf, but if Django can handle it too would be awesome). Each site must be independent of the others, I mean, I shouldn't be able to see the data content of other sites but I can share same applications/models. I've seen the Django's Sites framework, but I'm not sure if it's possible to implement that way. Should I use Sites framework or create a new app that can handle sites better? What do you think?

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  • Cant get the child dir in django hosting (alwaysdata.com) .

    - by zjm1126
    this is my file : mysite templates homepage.html accounts a.html login_view.html i can get the homepage.html and accounts\a.html on 127.0.0.1:8000 but in http://zjm1126.alwaysdata.net , i can only get the homepage.html ,and cant get the account\a.html , this is my code : return render_to_response('accounts/login_view.html') and the accounts/login_view.html is : {% include "accounts\a.html" %} what can i do , thanks ,

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  • default model field attribute in Django

    - by Rosarch
    I have a Django model: @staticmethod def getdefault(): print "getdefault called" return cPickle.dumps(set()) _applies_to = models.TextField(db_index=True, default=getdefault) For some reason, getdefault() is never called, even as I construct instances of this model and save them to the database. This seems to contradict the Django documentation: Field.default The default value for the field. This can be a value or a callable object. If callable it will be called every time a new object is created. Am I doing something wrong? Update: Originally, I had this, but then I switched to the above version to debug: _applies_to = models.TextField(db_index=True, default=cPickle.dumps(set())) I'm not sure why that wouldn't work.

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  • wxPython ,Change the background colour of a StyledTextCtrl

    - by user1357159
    I tried (but nothing happens) self.txt.SetBackgroundColour ((255,0,0)) As said in the title I'm trying to change the background colour StyledTextCtrl. Does anyone know a method that could be used? I've checked the API docs but I couldn't seem to find one, http://wxpython.org/docs/api/wx.stc.StyledTextCtrl-class.html (by background colour, I mean the whole writing area, of course) Does anyone know a way I could do this? EDIT: The background doesn't change in the following code import wx import wx.stc app = wx.App(redirect=True) top = wx.Frame(None, title="StyledTXTCtrl", size=(300,200)) txt=wx.stc.StyledTextCtrl(top) txt.SetBackgroundColour((255,255,0)) txt.Refresh() top.Show() app.MainLoop()

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