How do I render a URLField nicely in a template









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1
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The problem



I have a model for a league like



class League(models.Model):
league_name = models.CharField(max_length=60)
league_url = models.URLField(null=True, blank=True, default='')


where league_url can be a link to an external league page (i.e. not my site).



Now, let's say a user creates a league and sets league_url to "pelicanpark.com/leagues/27". When I render this in a template, league.league_url it shows http://www.pelicanpark.com/leagues/27/ which takes up a lot of real estate.



So, I need to slim it down. My thought is to render something like



<a href="http://www.pelicanpark.com/leagues/27/">pelicanpark.com/...</a>


but this has some obvious complexities. Does anyone have a good solution for this?










share|improve this question





















  • by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
    – Ben
    Mar 14 '15 at 22:12










  • Good question. I mean pelicanpark.com/...
    – Ben
    Mar 14 '15 at 22:21










  • Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
    – Ben
    Mar 14 '15 at 22:31










  • @Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
    – Ben
    Mar 15 '15 at 0:23














up vote
1
down vote

favorite












The problem



I have a model for a league like



class League(models.Model):
league_name = models.CharField(max_length=60)
league_url = models.URLField(null=True, blank=True, default='')


where league_url can be a link to an external league page (i.e. not my site).



Now, let's say a user creates a league and sets league_url to "pelicanpark.com/leagues/27". When I render this in a template, league.league_url it shows http://www.pelicanpark.com/leagues/27/ which takes up a lot of real estate.



So, I need to slim it down. My thought is to render something like



<a href="http://www.pelicanpark.com/leagues/27/">pelicanpark.com/...</a>


but this has some obvious complexities. Does anyone have a good solution for this?










share|improve this question





















  • by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
    – Ben
    Mar 14 '15 at 22:12










  • Good question. I mean pelicanpark.com/...
    – Ben
    Mar 14 '15 at 22:21










  • Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
    – Ben
    Mar 14 '15 at 22:31










  • @Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
    – Ben
    Mar 15 '15 at 0:23












up vote
1
down vote

favorite









up vote
1
down vote

favorite











The problem



I have a model for a league like



class League(models.Model):
league_name = models.CharField(max_length=60)
league_url = models.URLField(null=True, blank=True, default='')


where league_url can be a link to an external league page (i.e. not my site).



Now, let's say a user creates a league and sets league_url to "pelicanpark.com/leagues/27". When I render this in a template, league.league_url it shows http://www.pelicanpark.com/leagues/27/ which takes up a lot of real estate.



So, I need to slim it down. My thought is to render something like



<a href="http://www.pelicanpark.com/leagues/27/">pelicanpark.com/...</a>


but this has some obvious complexities. Does anyone have a good solution for this?










share|improve this question













The problem



I have a model for a league like



class League(models.Model):
league_name = models.CharField(max_length=60)
league_url = models.URLField(null=True, blank=True, default='')


where league_url can be a link to an external league page (i.e. not my site).



Now, let's say a user creates a league and sets league_url to "pelicanpark.com/leagues/27". When I render this in a template, league.league_url it shows http://www.pelicanpark.com/leagues/27/ which takes up a lot of real estate.



So, I need to slim it down. My thought is to render something like



<a href="http://www.pelicanpark.com/leagues/27/">pelicanpark.com/...</a>


but this has some obvious complexities. Does anyone have a good solution for this?







django django-templates






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 14 '15 at 20:34









Ben

5,8961259117




5,8961259117











  • by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
    – Ben
    Mar 14 '15 at 22:12










  • Good question. I mean pelicanpark.com/...
    – Ben
    Mar 14 '15 at 22:21










  • Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
    – Ben
    Mar 14 '15 at 22:31










  • @Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
    – Ben
    Mar 15 '15 at 0:23
















  • by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
    – Ben
    Mar 14 '15 at 22:12










  • Good question. I mean pelicanpark.com/...
    – Ben
    Mar 14 '15 at 22:21










  • Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
    – Ben
    Mar 14 '15 at 22:31










  • @Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
    – Ben
    Mar 15 '15 at 0:23















by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
– Ben
Mar 14 '15 at 22:12




by ... do you mean you want pelicanpark.com/... or pelicanpark.com/leagues/27/?
– Ben
Mar 14 '15 at 22:12












Good question. I mean pelicanpark.com/...
– Ben
Mar 14 '15 at 22:21




Good question. I mean pelicanpark.com/...
– Ben
Mar 14 '15 at 22:21












Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
– Ben
Mar 14 '15 at 22:31




Can you expand more on the complications by doing the following in the template? <a href=" league.league_url ">pelicanpark.com/...</a> wherein league is the user's league instance passed to the template. That seem like it would be the easiest?
– Ben
Mar 14 '15 at 22:31












@Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
– Ben
Mar 15 '15 at 0:23




@Ben pelicanpark.com is just an example url. In reality the league_url could be anything. E.g. if the url was "myleaguesite.com/leagues/123" I'd want to display <a href=" league.league_url ">myleaguesite.com/...</a>. In other words, the "domain.com/..." should be rendered dynamically depending on the actual league_url.
– Ben
Mar 15 '15 at 0:23












3 Answers
3






active

oldest

votes

















up vote
1
down vote













There's probably some fancy way of using regex for this kind of thing, but that's above my head. Luckly it looks like there's urlparse for python. To implement it in your example:



from urlparse import urlparse

class League(models.Model):
league_name = models.CharField(max_length=60)
league_url = models.URLField(null=True, blank=True, default='')

def url_text(self):
parsed_url = urlparse(self.league_url)
return parsed_url.hostname.replace("www.", "") + "/..."


Then in template:



<a href=" league.league_url "> league.url_text </a>


I would refrain from rendering the html in the backend. I've always been told there's security risks in doing that.






share|improve this answer





























    up vote
    0
    down vote













    You can create a field with MethodSerializer



    class League(models.Model):
    league_name = models.CharField(max_length=60)
    nice_league_url = serializers.SerializerMethodField()


    def get_league_url(self,obj):
    final_url = "<a href="+obj.league_url+">pelicanpark.com/...</a>"
    return final_url





    share|improve this answer





























      up vote
      0
      down vote













      The above answers are spot on. I just ran into the same problem and found some changes. For Python3 and higher, use:



      from urllib.parse import urlparse


      instead.






      share|improve this answer






















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote













        There's probably some fancy way of using regex for this kind of thing, but that's above my head. Luckly it looks like there's urlparse for python. To implement it in your example:



        from urlparse import urlparse

        class League(models.Model):
        league_name = models.CharField(max_length=60)
        league_url = models.URLField(null=True, blank=True, default='')

        def url_text(self):
        parsed_url = urlparse(self.league_url)
        return parsed_url.hostname.replace("www.", "") + "/..."


        Then in template:



        <a href=" league.league_url "> league.url_text </a>


        I would refrain from rendering the html in the backend. I've always been told there's security risks in doing that.






        share|improve this answer


























          up vote
          1
          down vote













          There's probably some fancy way of using regex for this kind of thing, but that's above my head. Luckly it looks like there's urlparse for python. To implement it in your example:



          from urlparse import urlparse

          class League(models.Model):
          league_name = models.CharField(max_length=60)
          league_url = models.URLField(null=True, blank=True, default='')

          def url_text(self):
          parsed_url = urlparse(self.league_url)
          return parsed_url.hostname.replace("www.", "") + "/..."


          Then in template:



          <a href=" league.league_url "> league.url_text </a>


          I would refrain from rendering the html in the backend. I've always been told there's security risks in doing that.






          share|improve this answer
























            up vote
            1
            down vote










            up vote
            1
            down vote









            There's probably some fancy way of using regex for this kind of thing, but that's above my head. Luckly it looks like there's urlparse for python. To implement it in your example:



            from urlparse import urlparse

            class League(models.Model):
            league_name = models.CharField(max_length=60)
            league_url = models.URLField(null=True, blank=True, default='')

            def url_text(self):
            parsed_url = urlparse(self.league_url)
            return parsed_url.hostname.replace("www.", "") + "/..."


            Then in template:



            <a href=" league.league_url "> league.url_text </a>


            I would refrain from rendering the html in the backend. I've always been told there's security risks in doing that.






            share|improve this answer














            There's probably some fancy way of using regex for this kind of thing, but that's above my head. Luckly it looks like there's urlparse for python. To implement it in your example:



            from urlparse import urlparse

            class League(models.Model):
            league_name = models.CharField(max_length=60)
            league_url = models.URLField(null=True, blank=True, default='')

            def url_text(self):
            parsed_url = urlparse(self.league_url)
            return parsed_url.hostname.replace("www.", "") + "/..."


            Then in template:



            <a href=" league.league_url "> league.url_text </a>


            I would refrain from rendering the html in the backend. I've always been told there's security risks in doing that.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 17 '15 at 18:26

























            answered Mar 15 '15 at 0:49









            Ben

            3,37822144




            3,37822144






















                up vote
                0
                down vote













                You can create a field with MethodSerializer



                class League(models.Model):
                league_name = models.CharField(max_length=60)
                nice_league_url = serializers.SerializerMethodField()


                def get_league_url(self,obj):
                final_url = "<a href="+obj.league_url+">pelicanpark.com/...</a>"
                return final_url





                share|improve this answer


























                  up vote
                  0
                  down vote













                  You can create a field with MethodSerializer



                  class League(models.Model):
                  league_name = models.CharField(max_length=60)
                  nice_league_url = serializers.SerializerMethodField()


                  def get_league_url(self,obj):
                  final_url = "<a href="+obj.league_url+">pelicanpark.com/...</a>"
                  return final_url





                  share|improve this answer
























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    You can create a field with MethodSerializer



                    class League(models.Model):
                    league_name = models.CharField(max_length=60)
                    nice_league_url = serializers.SerializerMethodField()


                    def get_league_url(self,obj):
                    final_url = "<a href="+obj.league_url+">pelicanpark.com/...</a>"
                    return final_url





                    share|improve this answer














                    You can create a field with MethodSerializer



                    class League(models.Model):
                    league_name = models.CharField(max_length=60)
                    nice_league_url = serializers.SerializerMethodField()


                    def get_league_url(self,obj):
                    final_url = "<a href="+obj.league_url+">pelicanpark.com/...</a>"
                    return final_url






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Mar 14 '15 at 20:45

























                    answered Mar 14 '15 at 20:38









                    levi

                    14k24553




                    14k24553




















                        up vote
                        0
                        down vote













                        The above answers are spot on. I just ran into the same problem and found some changes. For Python3 and higher, use:



                        from urllib.parse import urlparse


                        instead.






                        share|improve this answer


























                          up vote
                          0
                          down vote













                          The above answers are spot on. I just ran into the same problem and found some changes. For Python3 and higher, use:



                          from urllib.parse import urlparse


                          instead.






                          share|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            The above answers are spot on. I just ran into the same problem and found some changes. For Python3 and higher, use:



                            from urllib.parse import urlparse


                            instead.






                            share|improve this answer














                            The above answers are spot on. I just ran into the same problem and found some changes. For Python3 and higher, use:



                            from urllib.parse import urlparse


                            instead.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 11 at 2:01









                            Stephen Rauch

                            27.6k153256




                            27.6k153256










                            answered Nov 11 at 1:43









                            Shrijana Ghimire

                            212




                            212



























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