Need regex for a string having characters followed by even number of digits












-1














Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?



Valid:



a11a1121

bbbb11a1121


Invalid:



a11a1


I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.



Can you please help in this regard?






java regex







share|improve this question




















  • 2




    Possible duplicate of Learning Regular Expressions
    – Biffen
    Nov 12 at 11:52
















-1














Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?



Valid:



a11a1121

bbbb11a1121


Invalid:



a11a1


I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.



Can you please help in this regard?






java regex







share|improve this question




















  • 2




    Possible duplicate of Learning Regular Expressions
    – Biffen
    Nov 12 at 11:52














-1












-1








-1







Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?



Valid:



a11a1121

bbbb11a1121


Invalid:



a11a1


I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.



Can you please help in this regard?






java regex







share|improve this question















Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?



Valid:



a11a1121

bbbb11a1121


Invalid:



a11a1


I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.



Can you please help in this regard?






java regex




java regex






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 11:53









Biffen

4,16352129




4,16352129










asked Nov 12 at 11:47









Icche Guri

96211130




96211130








  • 2




    Possible duplicate of Learning Regular Expressions
    – Biffen
    Nov 12 at 11:52














  • 2




    Possible duplicate of Learning Regular Expressions
    – Biffen
    Nov 12 at 11:52








2




2




Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52




Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52












2 Answers
2






active

oldest

votes


















1














The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]



You can break down your requirement to groups and then handle it accordingly.



Like you require at least one character. so you start with ^([a-zA-Z]+)$



Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$



Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$



You can use online tools like regex101 for these purpose. The provided regex in action here






share|improve this answer





























    1














    You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$



    Explanation :





    • [a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one


    • ([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use ([0-9]{2})+ instead.






    share|improve this answer























      Your Answer






      StackExchange.ifUsing("editor", function () {
      StackExchange.using("externalEditor", function () {
      StackExchange.using("snippets", function () {
      StackExchange.snippets.init();
      });
      });
      }, "code-snippets");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "1"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53261522%2fneed-regex-for-a-string-having-characters-followed-by-even-number-of-digits%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]



      You can break down your requirement to groups and then handle it accordingly.



      Like you require at least one character. so you start with ^([a-zA-Z]+)$



      Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$



      Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$



      You can use online tools like regex101 for these purpose. The provided regex in action here






      share|improve this answer


























        1














        The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]



        You can break down your requirement to groups and then handle it accordingly.



        Like you require at least one character. so you start with ^([a-zA-Z]+)$



        Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$



        Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$



        You can use online tools like regex101 for these purpose. The provided regex in action here






        share|improve this answer
























          1












          1








          1






          The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]



          You can break down your requirement to groups and then handle it accordingly.



          Like you require at least one character. so you start with ^([a-zA-Z]+)$



          Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$



          Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$



          You can use online tools like regex101 for these purpose. The provided regex in action here






          share|improve this answer












          The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]



          You can break down your requirement to groups and then handle it accordingly.



          Like you require at least one character. so you start with ^([a-zA-Z]+)$



          Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$



          Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$



          You can use online tools like regex101 for these purpose. The provided regex in action here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 12 at 12:01









          suvartheec

          1,121917




          1,121917

























              1














              You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$



              Explanation :





              • [a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one


              • ([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use ([0-9]{2})+ instead.






              share|improve this answer




























                1














                You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$



                Explanation :





                • [a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one


                • ([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use ([0-9]{2})+ instead.






                share|improve this answer


























                  1












                  1








                  1






                  You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$



                  Explanation :





                  • [a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one


                  • ([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use ([0-9]{2})+ instead.






                  share|improve this answer














                  You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$



                  Explanation :





                  • [a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one


                  • ([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use ([0-9]{2})+ instead.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 12 at 12:29

























                  answered Nov 12 at 11:54









                  Amessihel

                  1,9631723




                  1,9631723






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Stack Overflow!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53261522%2fneed-regex-for-a-string-having-characters-followed-by-even-number-of-digits%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Bressuire

                      Image
                      Subprefecture and commune in Nouvelle-Aquitaine, France Bressuire Subprefecture and commune Chateau de Bressuire and the Eglise Notre-Dame Coat of arms Location of Bressuire Bressuire Show map of France Bressuire Show map of Nouvelle-Aquitaine Coordinates: 46°50′27″N 0°29′14″W  /  46.8408°N 0.4872°W  / 46.8408; -0.4872 Coordinates: 46°50′27″N 0°29′14″W  /  46.8408°N 0.4872°W  / 46.8408; -0.4872 Country France Region Nouvelle-Aquitaine Department Deux-Sèvres Arrondissement Bressuire Canton Bressuire Government  • Mayor .mw-parser-output .nobold{font-weight:normal} (2014–20) Jean Michel Bernier Area 1 180.59 km 2 (69.73 sq mi) Population (2014) 2 19,300  • Density 110/km 2 (280/sq mi) Time zone UTC+01:00 (CET)  • Summer (DST) UTC+02:00 (CEST) INSEE/Postal code 79049 /79300 Elevation 98–236 m (322–774 ft) (avg. 173 m or 568 ft) 1 French Land Register data, which exclude...
                      Read more

                      Vorschmack

                      Image
                      Vorschmack Ukrainian Jewish-style vorschmack served on rye bread Course Hors d'oeuvre Region or state Eastern Europe Associated national cuisine Ashkenazi Jewish, Finnish, German, Ukrainian, Polish, Russian Main ingredients Ground meat and/or fish Cookbook: Vorschmack   Media: Vorschmack Vorschmack or forshmak (Yiddish: פֿאָרשמאַק ‎, from archaic German Vorschmack , "foretaste" [1] or "appetizer" [2] ) is an originally East European dish made of salty minced fish or meat. Different variants of this dish are especially common in Ashkenazi Jewish and Finnish cuisine. Some varieties are also known in Russian and Polish cuisine. Contents 1 In Jewish cuisine 2 In Russian cuisine 3 In Polish cuisine 4 In Finnish cuisine 5 See also 6 References In Jewish cuisine According to Gil Marks, the German name points to the possible Germanic origin of this dish. [1] William Pokhlyobkin descr...
                      Read more

                      Quarantine

                      Image
                      For other uses, see Quarantine (disambiguation). Signal flag "Lima" called the "Yellow Jack" which when flown in harbor means ship is under quarantine. A simple yellow flag (also called the "Yellow Jack") had historically been used to signal quarantine (it stands for Q among signal flags), but now indicates the opposite, as a signal of a ship free of disease that requests boarding and inspection. A quarantine is used to separate and restrict the movement of people; it is 'a restraint upon the activities or communication of persons or the transport of goods designed to prevent the spread of disease or pests', for a certain period of time. [1] This is often used in connection to disease and illness, such as those who may possibly have been exposed to a communicable disease. [2] The term is often erroneously used to mean medical isolation, which is "to separate ill persons who have a communicable disease from those who are healthy...
                      Read more