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









ManavManav

72111




72111













  • 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









              CertainPerformanceCertainPerformance

              85.7k154371




              85.7k154371

























                  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 TelkikarPrasad Telkikar

                      1,951419




                      1,951419























                          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 RajwanshiPushpesh Kumar Rajwanshi

                              7,6202927




                              7,6202927























                                  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









                                      trontron

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                                          Costa Masnaga

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                                          .mw-parser-output .avviso .mbox-text-div>div,.mw-parser-output .avviso .mbox-text-full-div>div{font-size:90%}.mw-parser-output .avviso .mbox-image div{width:52px}.mw-parser-output .avviso .mbox-text-full-div .hide-when-compact{display:block} Questa voce o sezione sull'argomento Lombardia non cita le fonti necessarie o quelle presenti sono insufficienti . Puoi migliorare questa voce aggiungendo citazioni da fonti attendibili secondo le linee guida sull'uso delle fonti. Segui i suggerimenti del progetto di riferimento. Costa Masnaga comune Localizzazione Stato   Italia Regione Lombardia Provincia Lecco Amministrazione Sindaco Sabina Panzeri (Al centro il cittadino) dall'08/06/2009 Territorio Coordinate 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Coordinate: 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Altitudine 318 m s.l.m
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                                          Fotorealismo

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                                          Guerrino Boatto, Venice's St. Barnaba Church and canal Il fotorealismo è un genere di pittura basato sull'uso di una o più fotografie dalle quali l'artista prende le informazioni necessarie da utilizzare poi nel processo della creazione dell'opera, in genere ad olio o ad acrilico, al fine di avere un aspetto finale il più simile possibile alla fotografia. Indice 1 Storia 2 Lista dei fotorealisti 3 Note 4 Voci correlate 5 Altri progetti Storia | Il fotorealismo è un movimento che nasce negli Stati Uniti d'America alla fine degli anni sessanta [1] sulla scia della Pop Art [2] [3] [4] , e insieme condividono il carattere reazionario verso la travolgente ascesa dei media, che alla metà del ventesimo secolo si trasforma in un fenomeno di massa talmente imponente da minacciare il valore dell'immagine nell'arte. [5] [6] Ne consegue così che, colto il gesto d'inizio che fu della Pop Art, i fotorealisti aprono ad un
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                                          Sidney Franklin

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                                          Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-
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