Custom regex to format number












2














i have this number
1234567890



this is how i want to display it



1234 567 890



im trying this



console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));





javascript angularjs regex







share|improve this question






















  • which angular version are you using?
    – Abhishek Mani
    Nov 23 '18 at 6:33










  • Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
    – CertainPerformance
    Nov 23 '18 at 6:33
















2














i have this number
1234567890



this is how i want to display it



1234 567 890



im trying this



console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));





javascript angularjs regex







share|improve this question






















  • which angular version are you using?
    – Abhishek Mani
    Nov 23 '18 at 6:33










  • Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
    – CertainPerformance
    Nov 23 '18 at 6:33














2












2








2







i have this number
1234567890



this is how i want to display it



1234 567 890



im trying this



console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));





javascript angularjs regex







share|improve this question













i have this number
1234567890



this is how i want to display it



1234 567 890



im trying this



console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));





javascript angularjs regex




javascript angularjs regex






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 6:32









Manav

76111




76111












  • which angular version are you using?
    – Abhishek Mani
    Nov 23 '18 at 6:33










  • Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
    – CertainPerformance
    Nov 23 '18 at 6:33


















  • which angular version are you using?
    – Abhishek Mani
    Nov 23 '18 at 6:33










  • Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
    – CertainPerformance
    Nov 23 '18 at 6:33
















which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33




which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33












Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33




Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33












4 Answers
4






active

oldest

votes


















2














One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):






const format = str => str.replace(

  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,

  //       ^^^ change these to change the number of unbroken leading digits

  '$1$2 '

);



console.log(format('1234567899'));

console.log(format('01234567899'));

console.log(format('101234567899'));





The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.



The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.






share|improve this answer





























    0















    • Divide your string into 2 parts.



        1. String contains space after 4 digits



        1. String contains space after 3 digits.




    • Add space after n digits


    Try with this






    var str = "1234567890";
    
var temp = str.substring(0,4); // get first 4 digits
    
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
    
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
    

    
//Concatenate both strings
    
var result = temp + ' ' + z;
    

    
//Display result 
    
console.log(temp);
    
console.log(z);
    
console.log(result);








    share|improve this answer































      0














      Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.



      Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1



      Here is some sample javascript code,






      console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
      
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
      
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
      
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));








      share|improve this answer































        0














        You can do with this way.



        let number = '1234567890';
        
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
        
console.warn(result);





        share|improve this answer























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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):






          const format = str => str.replace(
          
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
          
  //       ^^^ change these to change the number of unbroken leading digits
          
  '$1$2 '
          
);
          

          
console.log(format('1234567899'));
          
console.log(format('01234567899'));
          
console.log(format('101234567899'));





          The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.



          The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.






          share|improve this answer


























            2














            One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):






            const format = str => str.replace(
            
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
            
  //       ^^^ change these to change the number of unbroken leading digits
            
  '$1$2 '
            
);
            

            
console.log(format('1234567899'));
            
console.log(format('01234567899'));
            
console.log(format('101234567899'));





            The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.



            The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.






            share|improve this answer
























              2












              2








              2






              One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):






              const format = str => str.replace(
              
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
              
  //       ^^^ change these to change the number of unbroken leading digits
              
  '$1$2 '
              
);
              

              
console.log(format('1234567899'));
              
console.log(format('01234567899'));
              
console.log(format('101234567899'));





              The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.



              The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.






              share|improve this answer












              One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):






              const format = str => str.replace(
              
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
              
  //       ^^^ change these to change the number of unbroken leading digits
              
  '$1$2 '
              
);
              

              
console.log(format('1234567899'));
              
console.log(format('01234567899'));
              
console.log(format('101234567899'));





              The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.



              The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.






              const format = str => str.replace(
              
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
              
  //       ^^^ change these to change the number of unbroken leading digits
              
  '$1$2 '
              
);
              

              
console.log(format('1234567899'));
              
console.log(format('01234567899'));
              
console.log(format('101234567899'));





              const format = str => str.replace(
              
  /(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
              
  //       ^^^ change these to change the number of unbroken leading digits
              
  '$1$2 '
              
);
              

              
console.log(format('1234567899'));
              
console.log(format('01234567899'));
              
console.log(format('101234567899'));






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 23 '18 at 6:53









              CertainPerformance

              76.5k143862




              76.5k143862

























                  0















                  • Divide your string into 2 parts.



                      1. String contains space after 4 digits



                      1. String contains space after 3 digits.




                  • Add space after n digits


                  Try with this






                  var str = "1234567890";
                  
var temp = str.substring(0,4); // get first 4 digits
                  
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                  
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                  

                  
//Concatenate both strings
                  
var result = temp + ' ' + z;
                  

                  
//Display result 
                  
console.log(temp);
                  
console.log(z);
                  
console.log(result);








                  share|improve this answer




























                    0















                    • Divide your string into 2 parts.



                        1. String contains space after 4 digits



                        1. String contains space after 3 digits.




                    • Add space after n digits


                    Try with this






                    var str = "1234567890";
                    
var temp = str.substring(0,4); // get first 4 digits
                    
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                    
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                    

                    
//Concatenate both strings
                    
var result = temp + ' ' + z;
                    

                    
//Display result 
                    
console.log(temp);
                    
console.log(z);
                    
console.log(result);








                    share|improve this answer


























                      0












                      0








                      0







                      • Divide your string into 2 parts.



                          1. String contains space after 4 digits



                          1. String contains space after 3 digits.




                      • Add space after n digits


                      Try with this






                      var str = "1234567890";
                      
var temp = str.substring(0,4); // get first 4 digits
                      
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                      
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                      

                      
//Concatenate both strings
                      
var result = temp + ' ' + z;
                      

                      
//Display result 
                      
console.log(temp);
                      
console.log(z);
                      
console.log(result);








                      share|improve this answer















                      • Divide your string into 2 parts.



                          1. String contains space after 4 digits



                          1. String contains space after 3 digits.




                      • Add space after n digits


                      Try with this






                      var str = "1234567890";
                      
var temp = str.substring(0,4); // get first 4 digits
                      
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                      
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                      

                      
//Concatenate both strings
                      
var result = temp + ' ' + z;
                      

                      
//Display result 
                      
console.log(temp);
                      
console.log(z);
                      
console.log(result);








                      var str = "1234567890";
                      
var temp = str.substring(0,4); // get first 4 digits
                      
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                      
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                      

                      
//Concatenate both strings
                      
var result = temp + ' ' + z;
                      

                      
//Display result 
                      
console.log(temp);
                      
console.log(z);
                      
console.log(result);





                      var str = "1234567890";
                      
var temp = str.substring(0,4); // get first 4 digits
                      
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
                      
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
                      

                      
//Concatenate both strings
                      
var result = temp + ' ' + z;
                      

                      
//Display result 
                      
console.log(temp);
                      
console.log(z);
                      
console.log(result);






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Nov 23 '18 at 7:00

























                      answered Nov 23 '18 at 6:54









                      Prasad Telkikar

                      1,862419




                      1,862419























                          0














                          Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.



                          Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1



                          Here is some sample javascript code,






                          console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                          
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                          
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                          
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));








                          share|improve this answer




























                            0














                            Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.



                            Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1



                            Here is some sample javascript code,






                            console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                            
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                            
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                            
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));








                            share|improve this answer


























                              0












                              0








                              0






                              Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.



                              Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1



                              Here is some sample javascript code,






                              console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));








                              share|improve this answer














                              Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.



                              Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1



                              Here is some sample javascript code,






                              console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));








                              console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));





                              console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
                              
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));






                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited Nov 23 '18 at 8:06

























                              answered Nov 23 '18 at 7:38









                              Pushpesh Kumar Rajwanshi

                              5,3242827




                              5,3242827























                                  0














                                  You can do with this way.



                                  let number = '1234567890';
                                  
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
                                  
console.warn(result);





                                  share|improve this answer




























                                    0














                                    You can do with this way.



                                    let number = '1234567890';
                                    
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
                                    
console.warn(result);





                                    share|improve this answer


























                                      0












                                      0








                                      0






                                      You can do with this way.



                                      let number = '1234567890';
                                      
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
                                      
console.warn(result);





                                      share|improve this answer














                                      You can do with this way.



                                      let number = '1234567890';
                                      
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
                                      
console.warn(result);






                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited Nov 23 '18 at 11:13

























                                      answered Nov 23 '18 at 6:44









                                      tron

                                      4318




                                      4318






























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                                          How to ignore python UserWarning in pytest?

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                                          up vote 1 down vote favorite I am using openpyxl for parsing .xlsm files, and pytest for testing. When I open a file, I get the: OpenPyxl -> UserWarning: Data Validation extension is not supported and will be removed That is not really a problem bcs. program works and I can't change the .xlsm file to fix that. But... When I run pytest with something like: def test_wrong_file_format(): assert check_excel(open_excel('file.xlsm')) == True I will get the warning i mentioned altought check_excel(open_excel('file.xlsm')) returns True and the test should suceed... Is there a nice way to tell the pytest that "It's not a bug it's a feature" and tests should pass even when they get this warning? Is there other way than using something like: with pytest.warns(UserWarning): ...
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                                          Alexandru Averescu

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                                          Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center ...
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