Regex: Find strings without character












0















i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:



rt bl/ws se gn/wd wk bl/rt



/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i


and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:



rt/rt bl/ws se/se gn/wd wk/wk bl/rt


I tried it some time but without success. Could you help me with that ?



Update 1:



I've started with regex that matches words not containing 'at'



(b((?!(at))w)+b)


Et the and I want to replace matched elements with python like



re.sub(r'(b((?!(at))w)+b)', r'1/1', text)


but first have to find right elements ...






javascript regex regex-negation







share|improve this question




















  • 1





    Show us what you tried, it will help identify your problem.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:11











  • Also, it would be great if you mentioned the programming language you are using the regex in.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:15











  • Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

    – Socowi
    Nov 20 '18 at 22:16











  • @Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

    – walko1234
    Nov 20 '18 at 22:53











  • Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

    – Wiktor Stribiżew
    Nov 21 '18 at 0:40


















0















i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:



rt bl/ws se gn/wd wk bl/rt



/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i


and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:



rt/rt bl/ws se/se gn/wd wk/wk bl/rt


I tried it some time but without success. Could you help me with that ?



Update 1:



I've started with regex that matches words not containing 'at'



(b((?!(at))w)+b)


Et the and I want to replace matched elements with python like



re.sub(r'(b((?!(at))w)+b)', r'1/1', text)


but first have to find right elements ...






javascript regex regex-negation







share|improve this question




















  • 1





    Show us what you tried, it will help identify your problem.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:11











  • Also, it would be great if you mentioned the programming language you are using the regex in.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:15











  • Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

    – Socowi
    Nov 20 '18 at 22:16











  • @Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

    – walko1234
    Nov 20 '18 at 22:53











  • Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

    – Wiktor Stribiżew
    Nov 21 '18 at 0:40
















0












0








0


0






i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:



rt bl/ws se gn/wd wk bl/rt



/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i


and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:



rt/rt bl/ws se/se gn/wd wk/wk bl/rt


I tried it some time but without success. Could you help me with that ?



Update 1:



I've started with regex that matches words not containing 'at'



(b((?!(at))w)+b)


Et the and I want to replace matched elements with python like



re.sub(r'(b((?!(at))w)+b)', r'1/1', text)


but first have to find right elements ...






javascript regex regex-negation







share|improve this question
















i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:



rt bl/ws se gn/wd wk bl/rt



/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i


and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:



rt/rt bl/ws se/se gn/wd wk/wk bl/rt


I tried it some time but without success. Could you help me with that ?



Update 1:



I've started with regex that matches words not containing 'at'



(b((?!(at))w)+b)


Et the and I want to replace matched elements with python like



re.sub(r'(b((?!(at))w)+b)', r'1/1', text)


but first have to find right elements ...






javascript regex regex-negation




javascript regex regex-negation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 '18 at 7:57









JohnyL

3,7231924




3,7231924










asked Nov 20 '18 at 22:09









walko1234walko1234

466




466








  • 1





    Show us what you tried, it will help identify your problem.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:11











  • Also, it would be great if you mentioned the programming language you are using the regex in.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:15











  • Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

    – Socowi
    Nov 20 '18 at 22:16











  • @Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

    – walko1234
    Nov 20 '18 at 22:53











  • Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

    – Wiktor Stribiżew
    Nov 21 '18 at 0:40
















  • 1





    Show us what you tried, it will help identify your problem.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:11











  • Also, it would be great if you mentioned the programming language you are using the regex in.

    – Wiktor Stribiżew
    Nov 20 '18 at 22:15











  • Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

    – Socowi
    Nov 20 '18 at 22:16











  • @Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

    – walko1234
    Nov 20 '18 at 22:53











  • Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

    – Wiktor Stribiżew
    Nov 21 '18 at 0:40










1




1





Show us what you tried, it will help identify your problem.

– Wiktor Stribiżew
Nov 20 '18 at 22:11





Show us what you tried, it will help identify your problem.

– Wiktor Stribiżew
Nov 20 '18 at 22:11













Also, it would be great if you mentioned the programming language you are using the regex in.

– Wiktor Stribiżew
Nov 20 '18 at 22:15





Also, it would be great if you mentioned the programming language you are using the regex in.

– Wiktor Stribiżew
Nov 20 '18 at 22:15













Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

– Socowi
Nov 20 '18 at 22:16





Your regex is very complicated. You could simplify it. For instance, (s+)? is the same as s*. Also have a look at w and b.

– Socowi
Nov 20 '18 at 22:16













@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

– walko1234
Nov 20 '18 at 22:53





@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?

– walko1234
Nov 20 '18 at 22:53













Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

– Wiktor Stribiżew
Nov 21 '18 at 0:40







Then you may use re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)

– Wiktor Stribiżew
Nov 21 '18 at 0:40














2 Answers
2






active

oldest

votes


















0














in python you can do something like:



re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')



'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'





share|improve this answer































    0














    If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:



    (?<!/)([A-Za-z]{2}(?!/))



    (?<!/) - makes sure there isn't a / behind the match
    ([A-Za-z]{2} looks for two alphabetic characters
    (?!/)) makes sure there isn't a trailing /



    In Python you would use it like this:



    print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))


    Output:



    rt/rt bl/ws se/se gn/wd wk/wk bl/rt


    Demo on rextester



    In PHP you would use it like so:



    $str = 'rt bl/ws se gn/wd wk bl/rt';
    
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);


    Output:



    rt/rt bl/ws se/se gn/wd wk/wk bl/rt


    Demo on 3v4l.org






    share|improve this answer

























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      2 Answers
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      0














      in python you can do something like:



      re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
      

      
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'





      share|improve this answer




























        0














        in python you can do something like:



        re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
        

        
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'





        share|improve this answer


























          0












          0








          0







          in python you can do something like:



          re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
          

          
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'





          share|improve this answer













          in python you can do something like:



          re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
          

          
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 20 '18 at 23:17









          OnyambuOnyambu

          16k1522




          16k1522

























              0














              If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:



              (?<!/)([A-Za-z]{2}(?!/))



              (?<!/) - makes sure there isn't a / behind the match
              ([A-Za-z]{2} looks for two alphabetic characters
              (?!/)) makes sure there isn't a trailing /



              In Python you would use it like this:



              print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))


              Output:



              rt/rt bl/ws se/se gn/wd wk/wk bl/rt


              Demo on rextester



              In PHP you would use it like so:



              $str = 'rt bl/ws se gn/wd wk bl/rt';
              
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);


              Output:



              rt/rt bl/ws se/se gn/wd wk/wk bl/rt


              Demo on 3v4l.org






              share|improve this answer






























                0














                If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:



                (?<!/)([A-Za-z]{2}(?!/))



                (?<!/) - makes sure there isn't a / behind the match
                ([A-Za-z]{2} looks for two alphabetic characters
                (?!/)) makes sure there isn't a trailing /



                In Python you would use it like this:



                print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))


                Output:



                rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                Demo on rextester



                In PHP you would use it like so:



                $str = 'rt bl/ws se gn/wd wk bl/rt';
                
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);


                Output:



                rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                Demo on 3v4l.org






                share|improve this answer




























                  0












                  0








                  0







                  If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:



                  (?<!/)([A-Za-z]{2}(?!/))



                  (?<!/) - makes sure there isn't a / behind the match
                  ([A-Za-z]{2} looks for two alphabetic characters
                  (?!/)) makes sure there isn't a trailing /



                  In Python you would use it like this:



                  print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))


                  Output:



                  rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                  Demo on rextester



                  In PHP you would use it like so:



                  $str = 'rt bl/ws se gn/wd wk bl/rt';
                  
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);


                  Output:



                  rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                  Demo on 3v4l.org






                  share|improve this answer















                  If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:



                  (?<!/)([A-Za-z]{2}(?!/))



                  (?<!/) - makes sure there isn't a / behind the match
                  ([A-Za-z]{2} looks for two alphabetic characters
                  (?!/)) makes sure there isn't a trailing /



                  In Python you would use it like this:



                  print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))


                  Output:



                  rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                  Demo on rextester



                  In PHP you would use it like so:



                  $str = 'rt bl/ws se gn/wd wk bl/rt';
                  
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);


                  Output:



                  rt/rt bl/ws se/se gn/wd wk/wk bl/rt


                  Demo on 3v4l.org







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 22 '18 at 7:09

























                  answered Nov 20 '18 at 23:03









                  NickNick

                  33.3k132042




                  33.3k132042






























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