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  • Detect if 2 HTML fragments have identical hierarchical structure

    - by sergzach
    An example of fragments that have identical hierarchical structure: (1) <div> <span>It's a message</span> </div> (2) <div> <span class='bold'>This is a new text</span> </div> An example of fragments that have different structure: (1) <div> <span><b>It's a message</b></span> </div> (2) <div> <span>This is a new text</span> </div> So, fragments with a similar structure correspond to one hierarchical tree (the same tag names, the same hierarchical structure). How can I detect if 2 elements (html fragments) have the same structure simply with lxml? I have a function that does not work properly for some more difficult case (than the example): def _is_equal( el1, el2 ): # input: 2 elements with possible equal structure and tag names # e.g. root = lxml.html.fromstring( buf ) # el1 = root[ 0 ] # el2 = root[ 1 ] # move from top to bottom, compare elements result = False if el1.tag == el2.tag: # has no children if len( el1 ) == len( el2 ): if len( el1 ) == 0: return True else: # iterate one of them, for example el1 i = 0 for child1 in el1: child2 = el2[ i ] is_equal2 = _is_equal( child1, child2 ) if not is_equal2: return False return True else: return False else: return False The code fails to detect that 2 divs with class='tovar2' have an identical structure: <body> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333193003"> ?????? ?/? </a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333193003,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b> 3 150 </b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"><input value="" type="submit"> </li> </ul> </form> </div> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333124803">?????? ?/?</a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333124803,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b>3 150</b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"> <input value="" type="submit"> </li> </ul> </form> </div> </body>

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  • Django adminsite customize search_fields query

    - by dArignac
    Howdy! In the django admin you can set the search_fields for the ModelAdmin to be able to search over the properties given there. My model class has a property that is not a real model property, means it is not within the database table. The property relates to another database table that is not tied to the current model through relations. But I want to be able to search over it, so I have to somehow customize the query the admin site creates to do the filtering when the search field was filled - is this possible and if, how? I can query the database table of my custom property and it then returns the ids of the model classes fitting the search. This then, as I said, has to flow into the admin site search query. Thanks!

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  • Validation on ManyToManyField before Save in Models.py

    - by Heyl1
    I have the following models: class Application(models.Model): users = models.ManyToManyField(User, through='Permission') folder = models.ForeignKey(Folder) class Folder(models.Model): company = models.ManyToManyField(Compnay) class UserProfile(models.Model): user = models.OneToOneField(User, related_name='profile') company = models.ManyToManyField(Company) What I would like to do is to check whether one of the users of the Application has the same company as the Application (via Folder). If this is the case the Application instance should not be saved. The problem is that the ManyToManyFields aren't updated until after the 'post-save' signal. The only option seems to be the new m2m_changed signal. But I'm not sure how I then roll back the save that has already happened. Another option would be to rewrite the save function (in models.py, because I'm talking about the admin here), but I'm not sure how I could access the manytomanyfield content. Finally I've read something about rewriting the save function in the admin of the model in admin.py, however I still wouldn't know how you would access the manytomanyfield content. I have been searching for this everywhere but nothing I come across seems to work for me. If anything is unclear, please tell me. Thanks for your help! Heleen

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  • Simplifying for-if messes with better structure?

    - by HH
    # Description: you are given a bitwise pattern and a string # you need to find the number of times the pattern matches in the string # any one liner or simple pythonic solution? import random def matchIt(yourString, yourPattern): """find the number of times yourPattern occurs in yourString""" count = 0 matchTimes = 0 # How can you simplify the for-if structures? for coin in yourString: #return to base if count == len(pattern): matchTimes = matchTimes + 1 count = 0 #special case to return to 2, there could be more this type of conditions #so this type of if-conditionals are screaming for a havoc if count == 2 and pattern[count] == 1: count = count - 1 #the work horse #it could be simpler by breaking the intial string of lenght 'l' #to blocks of pattern-length, the number of them is 'l - len(pattern)-1' if coin == pattern[count]: count=count+1 average = len(yourString)/matchTimes return [average, matchTimes] # Generates the list myString =[] for x in range(10000): myString= myString + [int(random.random()*2)] pattern = [1,0,0] result = matchIt(myString, pattern) print("The sample had "+str(result[1])+" matches and its size was "+str(len(myString))+".\n" + "So it took "+str(result[0])+" steps in average.\n" + "RESULT: "+str([a for a in "FAILURE" if result[0] != 8])) # Sample Output # # The sample had 1656 matches and its size was 10000. # So it took 6 steps in average. # RESULT: ['F', 'A', 'I', 'L', 'U', 'R', 'E']

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  • Write xml file with lxml

    - by systempuntoout
    Having a code like this: from lxml import etree root = etree.Element("root") root.set("interesting", "somewhat") child1 = etree.SubElement(root, "test") How do i write root Element object to an xml file using write() method of ElementTree class?

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  • threading.local equivalent for twisted.web?

    - by defnull
    In asynchronous environments, threading.local is not guaranteed to be context-local anymore, because several contexts may coexist within a single thread. Most asynchronous frameworks (gevent, eventlet) provide a get_current_context() functionality to identify the current context. Some offer a way to monkey-patch threading.local so it is local to 'greenthreads' or other framework-specific contexts. I cannot find such a functionality in the twisted documentation. How do I do this?

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  • Django: Serving a Download in a Generic View

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

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  • Test assertions for tuples with floats

    - by Space_C0wb0y
    I have a function that returns a tuple that, among others, contains a float value. Usually I use assertAlmostEquals to compare those, but this does not work with tuples. Also, the tuple contains other data-types as well. Currently I am asserting every element of the tuple individually, but that gets too much for a list of such tuples. Is there any good way to write assertions for such cases?

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  • Check result of AX_PYTHON_MODULE in configure.ac

    - by tmatth
    In using the m4_ax_python_module.m4 macro in configure.ac (AX_PYTHON_MODULE), one can know at configure time if a given module is installed. It takes two arguments, the module name, and second argument which if not empty, will lead to an exit, useful when the module is a must-have. In the case where you don't want a fatal exit, how do you test in configure.ac which modules were found or not? They output "yes" or "no" when configure is run, but that's all I've found so far. Basically If I have these lines in configure.ac: AX_PYTHON_MODULE(json,[]) AX_PYTHON_MODULE(simplejson,[]) How do I test which of the two modules were found? See http://www.gnu.org/software/autoconf-archive/ax_python_module.html#ax_python_module for documentation about this macro.

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  • How to update Geo-Location in fireeagle

    - by Ganesh
    Hi Every One, I am developing an application on fireeagle, there i need to update the users exact location, with out asking any information from the user (i.e) lat, long e.t.c., If it is not possible using yahoo fireeagle, please let me know if there exists any other api's other than yahoo fireeagle. If they can get the exact location of web user in 'Lat' and 'Long', either from 'Pc' or from 'Mobile' browser. Thanks in advance.

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  • Django: How do I get logging working?

    - by swisstony
    I've added the following to my settings.py file: import logging ... logging.basicConfig(level=logging.DEBUG, format='%(asctime)s %(levelname)s %(message)s', filename=os.path.join(rootdir, 'django.log'), filemode='a+') And in views.py, I've added: import logging log = logging.getLogger(__name__) ... log.info("testing 123!") Unfortunately, no log file is being created. Any ideas what I am doing wrong? And also is their a better method I should be using for logging? I am doing this on Webfaction.

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  • How do I restrict foreign keys choices to related objects only in django

    - by Jeff Mc
    I have a two way foreign relation similar to the following class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True) class Child(models.Model): name = models.CharField(max_length=255) myparent = models.ForeignKey(Parent) How do I restrict the choices for Parent.favoritechild to only children whose parent is itself? I tried class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True, limit_choices_to = {"myparent": "self"}) but that causes the admin interface to not list any children.

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  • Is there a replacement for Paste.Template?

    - by Jorge Vargas
    I have grown tired of all the little issues with paste template, it's horrible to maintain the templates, it has no way of updating an old project and it's very hard to test. I'm wondering if someone knows of an alternative for quickstart generators as they have proven to be useful.

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  • Paramiko ssh output stops at --more--

    - by Anesh
    The output stops printing at --more-- any idea how to get the end of the output >>> import paramiko >>> ssh = paramiko.SSHClient() >>> ssh.set_missing_host_key_policy(paramiko.AutoAddPolicy()) >>> conn=ssh.connect("ipaddress",username="user", password="pass") >>> channel = ssh.invoke_shell() >>> channel.send("en\n") 3 >>> channel.send("password\n") 9 >>> channel.send("show security local-user-list\n") 30 >>> results = '' >>> channel.send("\n") 1 >>> results += channel.recv(5000) >>> print results bluecoat>en Password: bluecoat#show security local-user-list Default List: local_user_database Append users loaded from file to default list: false local_user_database Lockout parameters: Max failed attempts: 60 Lockout duration: 3600 Reset interval: 7200 Users: Groups: admin_local Lockout parameters: Max failed attempts: 60 Lockout duration: 3600 Reset interval: 7200 Users: <username> Hashed Password: Enabled: true Groups: <username> Hashed Password: Enabled: true **--More--** As you can see above the output stops printing at --more-- any idea how to get the output to print till the end.

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  • How to cutomize a modelform widget in django 1.1?

    - by muudscope
    I'm trying to modify a django form to use a textarea instead of a normal input for the "address" field in my house form. The docs seem to imply this changed from django 1.1 (which I'm using) to 1.2. But neither approach is working for me. Here's what I've tried: class HouseForm(forms.ModelForm): address = forms.Textarea() # Should work with django 1.1, but doesn't class Meta: model = House #widgets = { 'address': forms.Textarea() } # 1.2 style - doesn't work either.

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  • A good data model for finding a user's favorite stories

    - by wings
    Original Design Here's how I originally had my Models set up: class UserData(db.Model): user = db.UserProperty() favorites = db.ListProperty(db.Key) # list of story keys # ... class Story(db.Model): title = db.StringProperty() # ... On every page that displayed a story I would query UserData for the current user: user_data = UserData.all().filter('user =' users.get_current_user()).get() story_is_favorited = (story in user_data.favorites) New Design After watching this talk: Google I/O 2009 - Scalable, Complex Apps on App Engine, I wondered if I could set things up more efficiently. class FavoriteIndex(db.Model): favorited_by = db.StringListProperty() The Story Model is the same, but I got rid of the UserData Model. Each instance of the new FavoriteIndex Model has a Story instance as a parent. And each FavoriteIndex stores a list of user id's in it's favorited_by property. If I want to find all of the stories that have been favorited by a certain user: index_keys = FavoriteIndex.all(keys_only=True).filter('favorited_by =', users.get_current_user().user_id()) story_keys = [k.parent() for k in index_keys] stories = db.get(story_keys) This approach avoids the serialization/deserialization that's otherwise associated with the ListProperty. Efficiency vs Simplicity I'm not sure how efficient the new design is, especially after a user decides to favorite 300 stories, but here's why I like it: A favorited story is associated with a user, not with her user data On a page where I display a story, it's pretty easy to ask the story if it's been favorited (without calling up a separate entity filled with user data). fav_index = FavoriteIndex.all().ancestor(story).get() fav_of_current_user = users.get_current_user().user_id() in fav_index.favorited_by It's also easy to get a list of all the users who have favorited a story (using the method in #2) Is there an easier way? Please help. How is this kind of thing normally done?

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  • OpenCV performance in different languages

    - by h0b0
    I'm doing some prototyping with OpenCV for a hobby project involving processing of real time camera data. I wonder if it is worth the effort to reimplement this in C or C++ when I have it all figured out or if no significant performance boost can be expected. The program basically chains OpenCV functions, so the main part of the work should be done in native code anyway.

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  • Google App Engine django model form does not pick up BlobProperty

    - by Wes
    I have the following model: class Image(db.Model): auction = db.ReferenceProperty(Auction) image = db.BlobProperty() thumb = db.BlobProperty() caption = db.StringProperty() item_to_tag = db.StringProperty() And the following form: class ImageForm(djangoforms.ModelForm): class Meta: model = Image When I call ImageForm(), only the non-Blob fields are created, like this: <tr><th><label for="id_auction">Auction:</label></th><td><select name="auction" id="id_auction"> <option value="" selected="selected">---------</option> <option value="ahRoYXJ0bWFuYXVjdGlvbmVlcmluZ3INCxIHQXVjdGlvbhgKDA">2010-06-19 11:00:00</option> </select></td></tr> <tr><th><label for="id_caption">Caption:</label></th><td><input type="text" name="caption" id="id_caption" /></td></tr> <tr><th><label for="id_item_to_tag">Item to tag:</label></th><td><input type="text" name="item_to_tag" id="id_item_to_tag" /></td></tr> I want the Blob fields to be included in the form as well (as file inputs). What am I doing wrong?

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