Remove rest of string after the second to last index of












1















I'm relatively new to JS and pretty confused. I have a URL string: https://a/b/c/d/e/f. These are all dynamic characters and change often (could be https://x/s/g/e/h/d for example) In some cases I want to remove only f, in which I use LastIndexOf. In other cases I want to remove both the last and second to last EG: e/f. How can this be done successfully? I tried using a split but this just replaced the '/' with ',' for some reason.



working f example



var url = https://a/b/c/d/e/f

var new = url.substring(0, url.lastIndexOf('/') +1);


current split e/f example



var new = url.split('/')

console.log(new[new.length -2]);


This prints as: https:,,a,b,c,d,e,f,






javascript reactjs lastindexof







share|improve this question























  • split just splits the string into an array of strings and not replace it

    – Shubham Khatri
    Nov 16 '18 at 11:36






  • 1





    You probably shouldn't use new as a variable name

    – Pete
    Nov 16 '18 at 11:43











  • @Pete thanks, that was just a quick name used for the purpose of the question

    – Uciebila
    Nov 16 '18 at 11:53
















1















I'm relatively new to JS and pretty confused. I have a URL string: https://a/b/c/d/e/f. These are all dynamic characters and change often (could be https://x/s/g/e/h/d for example) In some cases I want to remove only f, in which I use LastIndexOf. In other cases I want to remove both the last and second to last EG: e/f. How can this be done successfully? I tried using a split but this just replaced the '/' with ',' for some reason.



working f example



var url = https://a/b/c/d/e/f

var new = url.substring(0, url.lastIndexOf('/') +1);


current split e/f example



var new = url.split('/')

console.log(new[new.length -2]);


This prints as: https:,,a,b,c,d,e,f,






javascript reactjs lastindexof







share|improve this question























  • split just splits the string into an array of strings and not replace it

    – Shubham Khatri
    Nov 16 '18 at 11:36






  • 1





    You probably shouldn't use new as a variable name

    – Pete
    Nov 16 '18 at 11:43











  • @Pete thanks, that was just a quick name used for the purpose of the question

    – Uciebila
    Nov 16 '18 at 11:53














1












1








1








I'm relatively new to JS and pretty confused. I have a URL string: https://a/b/c/d/e/f. These are all dynamic characters and change often (could be https://x/s/g/e/h/d for example) In some cases I want to remove only f, in which I use LastIndexOf. In other cases I want to remove both the last and second to last EG: e/f. How can this be done successfully? I tried using a split but this just replaced the '/' with ',' for some reason.



working f example



var url = https://a/b/c/d/e/f

var new = url.substring(0, url.lastIndexOf('/') +1);


current split e/f example



var new = url.split('/')

console.log(new[new.length -2]);


This prints as: https:,,a,b,c,d,e,f,






javascript reactjs lastindexof







share|improve this question














I'm relatively new to JS and pretty confused. I have a URL string: https://a/b/c/d/e/f. These are all dynamic characters and change often (could be https://x/s/g/e/h/d for example) In some cases I want to remove only f, in which I use LastIndexOf. In other cases I want to remove both the last and second to last EG: e/f. How can this be done successfully? I tried using a split but this just replaced the '/' with ',' for some reason.



working f example



var url = https://a/b/c/d/e/f

var new = url.substring(0, url.lastIndexOf('/') +1);


current split e/f example



var new = url.split('/')

console.log(new[new.length -2]);


This prints as: https:,,a,b,c,d,e,f,






javascript reactjs lastindexof




javascript reactjs lastindexof






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 16 '18 at 11:33









UciebilaUciebila

5518




5518













  • split just splits the string into an array of strings and not replace it

    – Shubham Khatri
    Nov 16 '18 at 11:36






  • 1





    You probably shouldn't use new as a variable name

    – Pete
    Nov 16 '18 at 11:43











  • @Pete thanks, that was just a quick name used for the purpose of the question

    – Uciebila
    Nov 16 '18 at 11:53



















  • split just splits the string into an array of strings and not replace it

    – Shubham Khatri
    Nov 16 '18 at 11:36






  • 1





    You probably shouldn't use new as a variable name

    – Pete
    Nov 16 '18 at 11:43











  • @Pete thanks, that was just a quick name used for the purpose of the question

    – Uciebila
    Nov 16 '18 at 11:53

















split just splits the string into an array of strings and not replace it

– Shubham Khatri
Nov 16 '18 at 11:36





split just splits the string into an array of strings and not replace it

– Shubham Khatri
Nov 16 '18 at 11:36




1




1





You probably shouldn't use new as a variable name

– Pete
Nov 16 '18 at 11:43





You probably shouldn't use new as a variable name

– Pete
Nov 16 '18 at 11:43













@Pete thanks, that was just a quick name used for the purpose of the question

– Uciebila
Nov 16 '18 at 11:53





@Pete thanks, that was just a quick name used for the purpose of the question

– Uciebila
Nov 16 '18 at 11:53












5 Answers
5






active

oldest

votes


















2














Here you are



const str = 'https://a/b/c/d/e/f'



function removeSegments(url, times) {

    const segments = url.split('/')

    return segments.slice(0, segments.length - times).join('/')

}



console.log(removeSegments(str, 1)) // 'https://a/b/c/d/e'

console.log(removeSegments(str, 2)) // 'https://a/b/c/d'





share|improve this answer
























  • This was the easiest and cleanest way for me to replicate this into my code, thank you

    – Uciebila
    Nov 16 '18 at 11:52











  • Glad it helped you!

    – Nurbol Alpysbayev
    Nov 16 '18 at 11:54



















1














Here is one option, using regex replacement. To remove the final path only:



url = "https://a/b/c/d/e/f";

url = url.replace(//[^/]+$/mg, "");


To remove the final two paths:



url = "https://a/b/c/d/e/f";

url = url.replace(//[^/]+/[^/]+$/mg, "");





share|improve this answer
























  • Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

    – Uciebila
    Nov 16 '18 at 11:45











  • @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

    – Tim Biegeleisen
    Nov 16 '18 at 13:36











  • From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

    – Uciebila
    Nov 16 '18 at 13:57











  • Strange. It is too bad that your framework is limiting what you can and cannot do.

    – Tim Biegeleisen
    Nov 16 '18 at 13:58





















0














You can easily do that using the URL API.



You need to split the pathname by / and then use Array#slice to take only the parts you want. Then join those again by /.



Here is an example:






var url = new URL('https://a/b/c/d/e/f');



console.log(

  url.origin +

  url.pathname.split('/').slice(0, -2).join('/')

);








share|improve this answer































    0














    split() creates an array of substrings that is passed in the split function.
    So when you write



    url.split('/') it will create an array of substrings with / as delimeter.



    The answer could be :



    get the index of the element till when you want to remove your string by using :



    var index = indexOf(x)


    then pass it in the function.



    var str1 = str.substr(0, index)


    str1 will be your answer






    share|improve this answer































      0














      You can use:



      var num;
      
for ( var i=0; i< num; i++) {
      
    url= url.substring(0, url.lastIndexOf('/'));
      
}





      share|improve this answer

























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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        Here you are



        const str = 'https://a/b/c/d/e/f'
        

        
function removeSegments(url, times) {
        
    const segments = url.split('/')
        
    return segments.slice(0, segments.length - times).join('/')
        
}
        

        
console.log(removeSegments(str, 1)) // 'https://a/b/c/d/e'
        
console.log(removeSegments(str, 2)) // 'https://a/b/c/d'





        share|improve this answer
























        • This was the easiest and cleanest way for me to replicate this into my code, thank you

          – Uciebila
          Nov 16 '18 at 11:52











        • Glad it helped you!

          – Nurbol Alpysbayev
          Nov 16 '18 at 11:54
















        2














        Here you are



        const str = 'https://a/b/c/d/e/f'
        

        
function removeSegments(url, times) {
        
    const segments = url.split('/')
        
    return segments.slice(0, segments.length - times).join('/')
        
}
        

        
console.log(removeSegments(str, 1)) // 'https://a/b/c/d/e'
        
console.log(removeSegments(str, 2)) // 'https://a/b/c/d'





        share|improve this answer
























        • This was the easiest and cleanest way for me to replicate this into my code, thank you

          – Uciebila
          Nov 16 '18 at 11:52











        • Glad it helped you!

          – Nurbol Alpysbayev
          Nov 16 '18 at 11:54














        2












        2








        2







        Here you are



        const str = 'https://a/b/c/d/e/f'
        

        
function removeSegments(url, times) {
        
    const segments = url.split('/')
        
    return segments.slice(0, segments.length - times).join('/')
        
}
        

        
console.log(removeSegments(str, 1)) // 'https://a/b/c/d/e'
        
console.log(removeSegments(str, 2)) // 'https://a/b/c/d'





        share|improve this answer













        Here you are



        const str = 'https://a/b/c/d/e/f'
        

        
function removeSegments(url, times) {
        
    const segments = url.split('/')
        
    return segments.slice(0, segments.length - times).join('/')
        
}
        

        
console.log(removeSegments(str, 1)) // 'https://a/b/c/d/e'
        
console.log(removeSegments(str, 2)) // 'https://a/b/c/d'






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 16 '18 at 11:39









        Nurbol AlpysbayevNurbol Alpysbayev

        4,0191327




        4,0191327













        • This was the easiest and cleanest way for me to replicate this into my code, thank you

          – Uciebila
          Nov 16 '18 at 11:52











        • Glad it helped you!

          – Nurbol Alpysbayev
          Nov 16 '18 at 11:54



















        • This was the easiest and cleanest way for me to replicate this into my code, thank you

          – Uciebila
          Nov 16 '18 at 11:52











        • Glad it helped you!

          – Nurbol Alpysbayev
          Nov 16 '18 at 11:54

















        This was the easiest and cleanest way for me to replicate this into my code, thank you

        – Uciebila
        Nov 16 '18 at 11:52





        This was the easiest and cleanest way for me to replicate this into my code, thank you

        – Uciebila
        Nov 16 '18 at 11:52













        Glad it helped you!

        – Nurbol Alpysbayev
        Nov 16 '18 at 11:54





        Glad it helped you!

        – Nurbol Alpysbayev
        Nov 16 '18 at 11:54













        1














        Here is one option, using regex replacement. To remove the final path only:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+$/mg, "");


        To remove the final two paths:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+/[^/]+$/mg, "");





        share|improve this answer
























        • Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

          – Uciebila
          Nov 16 '18 at 11:45











        • @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

          – Tim Biegeleisen
          Nov 16 '18 at 13:36











        • From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

          – Uciebila
          Nov 16 '18 at 13:57











        • Strange. It is too bad that your framework is limiting what you can and cannot do.

          – Tim Biegeleisen
          Nov 16 '18 at 13:58


















        1














        Here is one option, using regex replacement. To remove the final path only:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+$/mg, "");


        To remove the final two paths:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+/[^/]+$/mg, "");





        share|improve this answer
























        • Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

          – Uciebila
          Nov 16 '18 at 11:45











        • @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

          – Tim Biegeleisen
          Nov 16 '18 at 13:36











        • From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

          – Uciebila
          Nov 16 '18 at 13:57











        • Strange. It is too bad that your framework is limiting what you can and cannot do.

          – Tim Biegeleisen
          Nov 16 '18 at 13:58
















        1












        1








        1







        Here is one option, using regex replacement. To remove the final path only:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+$/mg, "");


        To remove the final two paths:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+/[^/]+$/mg, "");





        share|improve this answer













        Here is one option, using regex replacement. To remove the final path only:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+$/mg, "");


        To remove the final two paths:



        url = "https://a/b/c/d/e/f";
        
url = url.replace(//[^/]+/[^/]+$/mg, "");






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 16 '18 at 11:38









        Tim BiegeleisenTim Biegeleisen

        222k1390143




        222k1390143













        • Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

          – Uciebila
          Nov 16 '18 at 11:45











        • @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

          – Tim Biegeleisen
          Nov 16 '18 at 13:36











        • From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

          – Uciebila
          Nov 16 '18 at 13:57











        • Strange. It is too bad that your framework is limiting what you can and cannot do.

          – Tim Biegeleisen
          Nov 16 '18 at 13:58





















        • Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

          – Uciebila
          Nov 16 '18 at 11:45











        • @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

          – Tim Biegeleisen
          Nov 16 '18 at 13:36











        • From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

          – Uciebila
          Nov 16 '18 at 13:57











        • Strange. It is too bad that your framework is limiting what you can and cannot do.

          – Tim Biegeleisen
          Nov 16 '18 at 13:58



















        Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

        – Uciebila
        Nov 16 '18 at 11:45





        Thanks for this, when I add the reg ex in, on replace(/**** it gives an error reading: unterminated Regular expression literal.

        – Uciebila
        Nov 16 '18 at 11:45













        @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

        – Tim Biegeleisen
        Nov 16 '18 at 13:36





        @Uciebila Your comment makes no sense, and I can only comment that my answer solved your problem in fewer lines of code than the answer you accepted.

        – Tim Biegeleisen
        Nov 16 '18 at 13:36













        From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

        – Uciebila
        Nov 16 '18 at 13:57





        From some research I found thatbecause my code is within a react app, it is a .js and not a .jsx. Which meant that using the above code threw an error that wouldnt be fixed until I changed my file extension, which would break the react app. With the way I added the accepted answer into my code it only needed two lines

        – Uciebila
        Nov 16 '18 at 13:57













        Strange. It is too bad that your framework is limiting what you can and cannot do.

        – Tim Biegeleisen
        Nov 16 '18 at 13:58







        Strange. It is too bad that your framework is limiting what you can and cannot do.

        – Tim Biegeleisen
        Nov 16 '18 at 13:58













        0














        You can easily do that using the URL API.



        You need to split the pathname by / and then use Array#slice to take only the parts you want. Then join those again by /.



        Here is an example:






        var url = new URL('https://a/b/c/d/e/f');
        

        
console.log(
        
  url.origin +
        
  url.pathname.split('/').slice(0, -2).join('/')
        
);








        share|improve this answer




























          0














          You can easily do that using the URL API.



          You need to split the pathname by / and then use Array#slice to take only the parts you want. Then join those again by /.



          Here is an example:






          var url = new URL('https://a/b/c/d/e/f');
          

          
console.log(
          
  url.origin +
          
  url.pathname.split('/').slice(0, -2).join('/')
          
);








          share|improve this answer


























            0












            0








            0







            You can easily do that using the URL API.



            You need to split the pathname by / and then use Array#slice to take only the parts you want. Then join those again by /.



            Here is an example:






            var url = new URL('https://a/b/c/d/e/f');
            

            
console.log(
            
  url.origin +
            
  url.pathname.split('/').slice(0, -2).join('/')
            
);








            share|improve this answer













            You can easily do that using the URL API.



            You need to split the pathname by / and then use Array#slice to take only the parts you want. Then join those again by /.



            Here is an example:






            var url = new URL('https://a/b/c/d/e/f');
            

            
console.log(
            
  url.origin +
            
  url.pathname.split('/').slice(0, -2).join('/')
            
);








            var url = new URL('https://a/b/c/d/e/f');
            

            
console.log(
            
  url.origin +
            
  url.pathname.split('/').slice(0, -2).join('/')
            
);





            var url = new URL('https://a/b/c/d/e/f');
            

            
console.log(
            
  url.origin +
            
  url.pathname.split('/').slice(0, -2).join('/')
            
);






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 16 '18 at 11:40









            31piy31piy

            17.2k52342




            17.2k52342























                0














                split() creates an array of substrings that is passed in the split function.
                So when you write



                url.split('/') it will create an array of substrings with / as delimeter.



                The answer could be :



                get the index of the element till when you want to remove your string by using :



                var index = indexOf(x)


                then pass it in the function.



                var str1 = str.substr(0, index)


                str1 will be your answer






                share|improve this answer




























                  0














                  split() creates an array of substrings that is passed in the split function.
                  So when you write



                  url.split('/') it will create an array of substrings with / as delimeter.



                  The answer could be :



                  get the index of the element till when you want to remove your string by using :



                  var index = indexOf(x)


                  then pass it in the function.



                  var str1 = str.substr(0, index)


                  str1 will be your answer






                  share|improve this answer


























                    0












                    0








                    0







                    split() creates an array of substrings that is passed in the split function.
                    So when you write



                    url.split('/') it will create an array of substrings with / as delimeter.



                    The answer could be :



                    get the index of the element till when you want to remove your string by using :



                    var index = indexOf(x)


                    then pass it in the function.



                    var str1 = str.substr(0, index)


                    str1 will be your answer






                    share|improve this answer













                    split() creates an array of substrings that is passed in the split function.
                    So when you write



                    url.split('/') it will create an array of substrings with / as delimeter.



                    The answer could be :



                    get the index of the element till when you want to remove your string by using :



                    var index = indexOf(x)


                    then pass it in the function.



                    var str1 = str.substr(0, index)


                    str1 will be your answer







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 16 '18 at 11:43









                    vertikavertika

                    870419




                    870419























                        0














                        You can use:



                        var num;
                        
for ( var i=0; i< num; i++) {
                        
    url= url.substring(0, url.lastIndexOf('/'));
                        
}





                        share|improve this answer






























                          0














                          You can use:



                          var num;
                          
for ( var i=0; i< num; i++) {
                          
    url= url.substring(0, url.lastIndexOf('/'));
                          
}





                          share|improve this answer




























                            0












                            0








                            0







                            You can use:



                            var num;
                            
for ( var i=0; i< num; i++) {
                            
    url= url.substring(0, url.lastIndexOf('/'));
                            
}





                            share|improve this answer















                            You can use:



                            var num;
                            
for ( var i=0; i< num; i++) {
                            
    url= url.substring(0, url.lastIndexOf('/'));
                            
}






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 16 '18 at 11:50









                            SamTebbs33

                            3,86031632




                            3,86031632










                            answered Nov 16 '18 at 11:47









                            simonasimona

                            12




                            12






























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