Python, opposite function urllib.urlencode












84















How can I convert data after processing urllib.urlencode to dict?
urllib.urldecode does not exist.






python urllib







share|improve this question





























    84















    How can I convert data after processing urllib.urlencode to dict?
    urllib.urldecode does not exist.






    python urllib







    share|improve this question



























      84












      84








      84


      21






      How can I convert data after processing urllib.urlencode to dict?
      urllib.urldecode does not exist.






      python urllib







      share|improve this question
















      How can I convert data after processing urllib.urlencode to dict?
      urllib.urldecode does not exist.






      python urllib




      python urllib






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 29 '16 at 10:44









      guaka

      10.8k74784




      10.8k74784










      asked Aug 22 '10 at 18:59









      ArtyomArtyom

      88821114




      88821114
























          3 Answers
          3






          active

          oldest

          votes


















          116














          As the docs for urlencode say,




          The urlparse module provides the
          functions parse_qs() and parse_qsl()
          which are used to parse query strings
          into Python data structures.




          (In older Python releases, they were in the cgi module). So, for example:



          >>> import urllib
          
>>> import urlparse
          
>>> d = {'a':'b', 'c':'d'}
          
>>> s = urllib.urlencode(d)
          
>>> s
          
'a=b&c=d'
          
>>> d1 = urlparse.parse_qs(s)
          
>>> d1
          
{'a': ['b'], 'c': ['d']}


          The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).



          As an alternative:



          >>> sq = urlparse.parse_qsl(s)
          
>>> sq  
          
[('a', 'b'), ('c', 'd')]
          
>>> dict(sq)
          
{'a': 'b', 'c': 'd'}


          you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).






          share|improve this answer





















          • 1





            Alex Martelli, thank you!

            – Artyom
            Aug 22 '10 at 20:17











          • @Derek, you're welcome!

            – Alex Martelli
            Aug 22 '10 at 20:25






          • 1





            Very thorough answer. Awesome!

            – Hartley Brody
            Dec 21 '11 at 19:53



















          16














          Python 3 code for Alex's solution:



          >>> import urllib.parse
          
>>> d = {'a':'b', 'c':'d'}
          
>>> s = urllib.parse.urlencode(d)
          
>>> s
          
'a=b&c=d'
          
>>> d1 = urllib.parse.parse_qs(s)
          
>>> d1
          
{'a': ['b'], 'c': ['d']}


          The alternative:



          >>> sq = urllib.parse.parse_qsl(s)
          
>>> sq
          
[('a', 'b'), ('c', 'd')]
          
>>> dict(sq)
          
{'a': 'b', 'c': 'd'}


          parse_qsl is reversible:



          >>> urllib.parse.urlencode(sq)
          
'a=b&c=d'





          share|improve this answer































            16














            urllib.unquote_plus() does what you want. It Replace %xx escapes by their single-character equivalent and replaces plus signs with spaces.



            Example: 



            unquote_plus('/%7Ecandidates/?name=john+connolly')


            yields 



            '/~candidates/?name=john connolly'.





            share|improve this answer





















            • 1





              He said, he wanted a dict. So your answer is wrong.

              – balrok
              Jul 23 '14 at 16:45






            • 4





              yay, this is what I was looking for.

              – Joe
              Mar 21 '16 at 12:01











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            3 Answers
            3






            active

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            3 Answers
            3






            active

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            active

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            active

            oldest

            votes









            116














            As the docs for urlencode say,




            The urlparse module provides the
            functions parse_qs() and parse_qsl()
            which are used to parse query strings
            into Python data structures.




            (In older Python releases, they were in the cgi module). So, for example:



            >>> import urllib
            
>>> import urlparse
            
>>> d = {'a':'b', 'c':'d'}
            
>>> s = urllib.urlencode(d)
            
>>> s
            
'a=b&c=d'
            
>>> d1 = urlparse.parse_qs(s)
            
>>> d1
            
{'a': ['b'], 'c': ['d']}


            The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).



            As an alternative:



            >>> sq = urlparse.parse_qsl(s)
            
>>> sq  
            
[('a', 'b'), ('c', 'd')]
            
>>> dict(sq)
            
{'a': 'b', 'c': 'd'}


            you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).






            share|improve this answer





















            • 1





              Alex Martelli, thank you!

              – Artyom
              Aug 22 '10 at 20:17











            • @Derek, you're welcome!

              – Alex Martelli
              Aug 22 '10 at 20:25






            • 1





              Very thorough answer. Awesome!

              – Hartley Brody
              Dec 21 '11 at 19:53
















            116














            As the docs for urlencode say,




            The urlparse module provides the
            functions parse_qs() and parse_qsl()
            which are used to parse query strings
            into Python data structures.




            (In older Python releases, they were in the cgi module). So, for example:



            >>> import urllib
            
>>> import urlparse
            
>>> d = {'a':'b', 'c':'d'}
            
>>> s = urllib.urlencode(d)
            
>>> s
            
'a=b&c=d'
            
>>> d1 = urlparse.parse_qs(s)
            
>>> d1
            
{'a': ['b'], 'c': ['d']}


            The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).



            As an alternative:



            >>> sq = urlparse.parse_qsl(s)
            
>>> sq  
            
[('a', 'b'), ('c', 'd')]
            
>>> dict(sq)
            
{'a': 'b', 'c': 'd'}


            you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).






            share|improve this answer





















            • 1





              Alex Martelli, thank you!

              – Artyom
              Aug 22 '10 at 20:17











            • @Derek, you're welcome!

              – Alex Martelli
              Aug 22 '10 at 20:25






            • 1





              Very thorough answer. Awesome!

              – Hartley Brody
              Dec 21 '11 at 19:53














            116












            116








            116







            As the docs for urlencode say,




            The urlparse module provides the
            functions parse_qs() and parse_qsl()
            which are used to parse query strings
            into Python data structures.




            (In older Python releases, they were in the cgi module). So, for example:



            >>> import urllib
            
>>> import urlparse
            
>>> d = {'a':'b', 'c':'d'}
            
>>> s = urllib.urlencode(d)
            
>>> s
            
'a=b&c=d'
            
>>> d1 = urlparse.parse_qs(s)
            
>>> d1
            
{'a': ['b'], 'c': ['d']}


            The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).



            As an alternative:



            >>> sq = urlparse.parse_qsl(s)
            
>>> sq  
            
[('a', 'b'), ('c', 'd')]
            
>>> dict(sq)
            
{'a': 'b', 'c': 'd'}


            you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).






            share|improve this answer















            As the docs for urlencode say,




            The urlparse module provides the
            functions parse_qs() and parse_qsl()
            which are used to parse query strings
            into Python data structures.




            (In older Python releases, they were in the cgi module). So, for example:



            >>> import urllib
            
>>> import urlparse
            
>>> d = {'a':'b', 'c':'d'}
            
>>> s = urllib.urlencode(d)
            
>>> s
            
'a=b&c=d'
            
>>> d1 = urlparse.parse_qs(s)
            
>>> d1
            
{'a': ['b'], 'c': ['d']}


            The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).



            As an alternative:



            >>> sq = urlparse.parse_qsl(s)
            
>>> sq  
            
[('a', 'b'), ('c', 'd')]
            
>>> dict(sq)
            
{'a': 'b', 'c': 'd'}


            you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Aug 22 '10 at 19:09

























            answered Aug 22 '10 at 19:02









            Alex MartelliAlex Martelli

            629k12810411281




            629k12810411281








            • 1





              Alex Martelli, thank you!

              – Artyom
              Aug 22 '10 at 20:17











            • @Derek, you're welcome!

              – Alex Martelli
              Aug 22 '10 at 20:25






            • 1





              Very thorough answer. Awesome!

              – Hartley Brody
              Dec 21 '11 at 19:53














            • 1





              Alex Martelli, thank you!

              – Artyom
              Aug 22 '10 at 20:17











            • @Derek, you're welcome!

              – Alex Martelli
              Aug 22 '10 at 20:25






            • 1





              Very thorough answer. Awesome!

              – Hartley Brody
              Dec 21 '11 at 19:53








            1




            1





            Alex Martelli, thank you!

            – Artyom
            Aug 22 '10 at 20:17





            Alex Martelli, thank you!

            – Artyom
            Aug 22 '10 at 20:17













            @Derek, you're welcome!

            – Alex Martelli
            Aug 22 '10 at 20:25





            @Derek, you're welcome!

            – Alex Martelli
            Aug 22 '10 at 20:25




            1




            1





            Very thorough answer. Awesome!

            – Hartley Brody
            Dec 21 '11 at 19:53





            Very thorough answer. Awesome!

            – Hartley Brody
            Dec 21 '11 at 19:53













            16














            Python 3 code for Alex's solution:



            >>> import urllib.parse
            
>>> d = {'a':'b', 'c':'d'}
            
>>> s = urllib.parse.urlencode(d)
            
>>> s
            
'a=b&c=d'
            
>>> d1 = urllib.parse.parse_qs(s)
            
>>> d1
            
{'a': ['b'], 'c': ['d']}


            The alternative:



            >>> sq = urllib.parse.parse_qsl(s)
            
>>> sq
            
[('a', 'b'), ('c', 'd')]
            
>>> dict(sq)
            
{'a': 'b', 'c': 'd'}


            parse_qsl is reversible:



            >>> urllib.parse.urlencode(sq)
            
'a=b&c=d'





            share|improve this answer




























              16














              Python 3 code for Alex's solution:



              >>> import urllib.parse
              
>>> d = {'a':'b', 'c':'d'}
              
>>> s = urllib.parse.urlencode(d)
              
>>> s
              
'a=b&c=d'
              
>>> d1 = urllib.parse.parse_qs(s)
              
>>> d1
              
{'a': ['b'], 'c': ['d']}


              The alternative:



              >>> sq = urllib.parse.parse_qsl(s)
              
>>> sq
              
[('a', 'b'), ('c', 'd')]
              
>>> dict(sq)
              
{'a': 'b', 'c': 'd'}


              parse_qsl is reversible:



              >>> urllib.parse.urlencode(sq)
              
'a=b&c=d'





              share|improve this answer


























                16












                16








                16







                Python 3 code for Alex's solution:



                >>> import urllib.parse
                
>>> d = {'a':'b', 'c':'d'}
                
>>> s = urllib.parse.urlencode(d)
                
>>> s
                
'a=b&c=d'
                
>>> d1 = urllib.parse.parse_qs(s)
                
>>> d1
                
{'a': ['b'], 'c': ['d']}


                The alternative:



                >>> sq = urllib.parse.parse_qsl(s)
                
>>> sq
                
[('a', 'b'), ('c', 'd')]
                
>>> dict(sq)
                
{'a': 'b', 'c': 'd'}


                parse_qsl is reversible:



                >>> urllib.parse.urlencode(sq)
                
'a=b&c=d'





                share|improve this answer













                Python 3 code for Alex's solution:



                >>> import urllib.parse
                
>>> d = {'a':'b', 'c':'d'}
                
>>> s = urllib.parse.urlencode(d)
                
>>> s
                
'a=b&c=d'
                
>>> d1 = urllib.parse.parse_qs(s)
                
>>> d1
                
{'a': ['b'], 'c': ['d']}


                The alternative:



                >>> sq = urllib.parse.parse_qsl(s)
                
>>> sq
                
[('a', 'b'), ('c', 'd')]
                
>>> dict(sq)
                
{'a': 'b', 'c': 'd'}


                parse_qsl is reversible:



                >>> urllib.parse.urlencode(sq)
                
'a=b&c=d'






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Apr 17 '12 at 0:02









                phobiephobie

                1,61511519




                1,61511519























                    16














                    urllib.unquote_plus() does what you want. It Replace %xx escapes by their single-character equivalent and replaces plus signs with spaces.



                    Example: 



                    unquote_plus('/%7Ecandidates/?name=john+connolly')


                    yields 



                    '/~candidates/?name=john connolly'.





                    share|improve this answer





















                    • 1





                      He said, he wanted a dict. So your answer is wrong.

                      – balrok
                      Jul 23 '14 at 16:45






                    • 4





                      yay, this is what I was looking for.

                      – Joe
                      Mar 21 '16 at 12:01
















                    16














                    urllib.unquote_plus() does what you want. It Replace %xx escapes by their single-character equivalent and replaces plus signs with spaces.



                    Example: 



                    unquote_plus('/%7Ecandidates/?name=john+connolly')


                    yields 



                    '/~candidates/?name=john connolly'.





                    share|improve this answer





















                    • 1





                      He said, he wanted a dict. So your answer is wrong.

                      – balrok
                      Jul 23 '14 at 16:45






                    • 4





                      yay, this is what I was looking for.

                      – Joe
                      Mar 21 '16 at 12:01














                    16












                    16








                    16







                    urllib.unquote_plus() does what you want. It Replace %xx escapes by their single-character equivalent and replaces plus signs with spaces.



                    Example: 



                    unquote_plus('/%7Ecandidates/?name=john+connolly')


                    yields 



                    '/~candidates/?name=john connolly'.





                    share|improve this answer















                    urllib.unquote_plus() does what you want. It Replace %xx escapes by their single-character equivalent and replaces plus signs with spaces.



                    Example: 



                    unquote_plus('/%7Ecandidates/?name=john+connolly')


                    yields 



                    '/~candidates/?name=john connolly'.






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 15 '18 at 8:11









                    Jean-François Corbett

                    28.8k22110161




                    28.8k22110161










                    answered Feb 26 '14 at 15:36









                    Andrew FarrellAndrew Farrell

                    1,3131218




                    1,3131218








                    • 1





                      He said, he wanted a dict. So your answer is wrong.

                      – balrok
                      Jul 23 '14 at 16:45






                    • 4





                      yay, this is what I was looking for.

                      – Joe
                      Mar 21 '16 at 12:01














                    • 1





                      He said, he wanted a dict. So your answer is wrong.

                      – balrok
                      Jul 23 '14 at 16:45






                    • 4





                      yay, this is what I was looking for.

                      – Joe
                      Mar 21 '16 at 12:01








                    1




                    1





                    He said, he wanted a dict. So your answer is wrong.

                    – balrok
                    Jul 23 '14 at 16:45





                    He said, he wanted a dict. So your answer is wrong.

                    – balrok
                    Jul 23 '14 at 16:45




                    4




                    4





                    yay, this is what I was looking for.

                    – Joe
                    Mar 21 '16 at 12:01





                    yay, this is what I was looking for.

                    – Joe
                    Mar 21 '16 at 12:01


















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