Replacing each letter wich another letter












0















I am wondering how to replace each letter of a string with another letter. For example A -> D.



I have tried using .replace() (repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH I end up with AD -> HH.



Im sure there is a better and simpler way of doing this.



Thank you in advance.










share|improve this question



























    0















    I am wondering how to replace each letter of a string with another letter. For example A -> D.



    I have tried using .replace() (repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH I end up with AD -> HH.



    Im sure there is a better and simpler way of doing this.



    Thank you in advance.










    share|improve this question

























      0












      0








      0








      I am wondering how to replace each letter of a string with another letter. For example A -> D.



      I have tried using .replace() (repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH I end up with AD -> HH.



      Im sure there is a better and simpler way of doing this.



      Thank you in advance.










      share|improve this question














      I am wondering how to replace each letter of a string with another letter. For example A -> D.



      I have tried using .replace() (repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH I end up with AD -> HH.



      Im sure there is a better and simpler way of doing this.



      Thank you in advance.







      javascript node.js






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 20 '18 at 18:51









      Artur DudekArtur Dudek

      1




      1
























          1 Answer
          1






          active

          oldest

          votes


















          2














          Build up a dictionary:



           const dict = { A: "D", D: "A", /*...*/ };


          Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:



           const output = input.split("").map(char => dict[char]).join("");





          share|improve this answer



















          • 2





            You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

            – Elias Toivanen
            Nov 20 '18 at 19:09











          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%2f53399653%2freplacing-each-letter-wich-another-letter%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Build up a dictionary:



           const dict = { A: "D", D: "A", /*...*/ };


          Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:



           const output = input.split("").map(char => dict[char]).join("");





          share|improve this answer



















          • 2





            You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

            – Elias Toivanen
            Nov 20 '18 at 19:09
















          2














          Build up a dictionary:



           const dict = { A: "D", D: "A", /*...*/ };


          Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:



           const output = input.split("").map(char => dict[char]).join("");





          share|improve this answer



















          • 2





            You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

            – Elias Toivanen
            Nov 20 '18 at 19:09














          2












          2








          2







          Build up a dictionary:



           const dict = { A: "D", D: "A", /*...*/ };


          Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:



           const output = input.split("").map(char => dict[char]).join("");





          share|improve this answer













          Build up a dictionary:



           const dict = { A: "D", D: "A", /*...*/ };


          Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:



           const output = input.split("").map(char => dict[char]).join("");






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 20 '18 at 18:54









          Jonas WilmsJonas Wilms

          56k42851




          56k42851








          • 2





            You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

            – Elias Toivanen
            Nov 20 '18 at 19:09














          • 2





            You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

            – Elias Toivanen
            Nov 20 '18 at 19:09








          2




          2





          You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

          – Elias Toivanen
          Nov 20 '18 at 19:09





          You may also use the replace method directly like so: input.replace(/./g, c => dict[c] || c).

          – Elias Toivanen
          Nov 20 '18 at 19:09


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53399653%2freplacing-each-letter-wich-another-letter%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

          "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

          Alcedinidae

          RAC Tourist Trophy