How to avoid printing a newline when seq completes?












1














To create a column header, looking like:



1234567890123456789


I (am trying to) use seq and echo:



seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9


However, seq outputs a newline after each run. How can I avoid that?






seq







share|improve this question
























  • Fair enough. Sorry about that, it seemed more natural to me.
    – terdon♦
    32 mins ago
















1














To create a column header, looking like:



1234567890123456789


I (am trying to) use seq and echo:



seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9


However, seq outputs a newline after each run. How can I avoid that?






seq







share|improve this question
























  • Fair enough. Sorry about that, it seemed more natural to me.
    – terdon♦
    32 mins ago














1












1








1







To create a column header, looking like:



1234567890123456789


I (am trying to) use seq and echo:



seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9


However, seq outputs a newline after each run. How can I avoid that?






seq







share|improve this question















To create a column header, looking like:



1234567890123456789


I (am trying to) use seq and echo:



seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9


However, seq outputs a newline after each run. How can I avoid that?






seq




seq






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 36 mins ago

























asked 59 mins ago









GreenMatt

20617




20617












  • Fair enough. Sorry about that, it seemed more natural to me.
    – terdon♦
    32 mins ago


















  • Fair enough. Sorry about that, it seemed more natural to me.
    – terdon♦
    32 mins ago
















Fair enough. Sorry about that, it seemed more natural to me.
– terdon♦
32 mins ago




Fair enough. Sorry about that, it seemed more natural to me.
– terdon♦
32 mins ago










4 Answers
4






active

oldest

votes


















3














Assuming you just want to print 1234567890123456789, you can do it with:



$ printf "%s" $(seq 1 9) $(seq 0 9)

1234567890123456789$


That won't have a trailing newline at all though, so maybe you prefer:



$ printf "%s" $(seq 1 9) $(seq 0 9) $'n'

1234567890123456789

$


A few simpler choices if you don't need to use seq:



$ perl -le 'print 1..9,0,1..9'

1234567890123456789

$ printf "%s" {1..9} {0..9} $'n'

1234567890123456789


Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.






share|improve this answer























  • Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
    – GreenMatt
    37 mins ago






  • 1




    @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
    – terdon♦
    33 mins ago








  • 1




    @GreenMatt of course, if you need portability, you should also avoid using echo.
    – terdon♦
    29 mins ago










  • Perl? <shudder>
    – GreenMatt
    16 mins ago



















2














You can remove newlines from any stream with tr -d 'n'. In your case



(seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d 'n'


While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.






share|improve this answer

















  • 1




    Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
    – terdon♦
    20 mins ago










  • Ah yes, I never think of tr.
    – GreenMatt
    16 mins ago



















1














Consider also:



printf '%d' $(seq -w 1 1 99 | cut -c2)


We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.



To get your desired output, use 20 instead:



$ printf '%d' $(seq -w 1 1 20 | cut -c2)

12345678901234567890





share|improve this answer































    0














    I would like to reinforce the answer of @terdon, but propose a solution that works with the tcsh shell as well.
    Unfortunately, I didn't have enough reputation to write this as a comment:



    printf "%sn" "seq 1 9` `seq 0 9`"`



    The backtick - like all methods of command-line substitution -, should cut the trailing newline. Then, printf's own backslash-escape interpretation is used, instead of the ANSI string expansion used in terdon's answer.



    It also works like this, by allowing printf to interpret all escape sequences after substitution with the format specifiers (%b specifier):



    printf "%b" "seq 1 9` `seq 0 9`n"`






    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









      3














      Assuming you just want to print 1234567890123456789, you can do it with:



      $ printf "%s" $(seq 1 9) $(seq 0 9)
      
1234567890123456789$


      That won't have a trailing newline at all though, so maybe you prefer:



      $ printf "%s" $(seq 1 9) $(seq 0 9) $'n'
      
1234567890123456789
      
$


      A few simpler choices if you don't need to use seq:



      $ perl -le 'print 1..9,0,1..9'
      
1234567890123456789
      
$ printf "%s" {1..9} {0..9} $'n'
      
1234567890123456789


      Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.






      share|improve this answer























      • Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
        – GreenMatt
        37 mins ago






      • 1




        @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
        – terdon♦
        33 mins ago








      • 1




        @GreenMatt of course, if you need portability, you should also avoid using echo.
        – terdon♦
        29 mins ago










      • Perl? <shudder>
        – GreenMatt
        16 mins ago
















      3














      Assuming you just want to print 1234567890123456789, you can do it with:



      $ printf "%s" $(seq 1 9) $(seq 0 9)
      
1234567890123456789$


      That won't have a trailing newline at all though, so maybe you prefer:



      $ printf "%s" $(seq 1 9) $(seq 0 9) $'n'
      
1234567890123456789
      
$


      A few simpler choices if you don't need to use seq:



      $ perl -le 'print 1..9,0,1..9'
      
1234567890123456789
      
$ printf "%s" {1..9} {0..9} $'n'
      
1234567890123456789


      Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.






      share|improve this answer























      • Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
        – GreenMatt
        37 mins ago






      • 1




        @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
        – terdon♦
        33 mins ago








      • 1




        @GreenMatt of course, if you need portability, you should also avoid using echo.
        – terdon♦
        29 mins ago










      • Perl? <shudder>
        – GreenMatt
        16 mins ago














      3












      3








      3






      Assuming you just want to print 1234567890123456789, you can do it with:



      $ printf "%s" $(seq 1 9) $(seq 0 9)
      
1234567890123456789$


      That won't have a trailing newline at all though, so maybe you prefer:



      $ printf "%s" $(seq 1 9) $(seq 0 9) $'n'
      
1234567890123456789
      
$


      A few simpler choices if you don't need to use seq:



      $ perl -le 'print 1..9,0,1..9'
      
1234567890123456789
      
$ printf "%s" {1..9} {0..9} $'n'
      
1234567890123456789


      Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.






      share|improve this answer














      Assuming you just want to print 1234567890123456789, you can do it with:



      $ printf "%s" $(seq 1 9) $(seq 0 9)
      
1234567890123456789$


      That won't have a trailing newline at all though, so maybe you prefer:



      $ printf "%s" $(seq 1 9) $(seq 0 9) $'n'
      
1234567890123456789
      
$


      A few simpler choices if you don't need to use seq:



      $ perl -le 'print 1..9,0,1..9'
      
1234567890123456789
      
$ printf "%s" {1..9} {0..9} $'n'
      
1234567890123456789


      Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 19 mins ago

























      answered 54 mins ago









      terdon♦

      128k31249423




      128k31249423












      • Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
        – GreenMatt
        37 mins ago






      • 1




        @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
        – terdon♦
        33 mins ago








      • 1




        @GreenMatt of course, if you need portability, you should also avoid using echo.
        – terdon♦
        29 mins ago










      • Perl? <shudder>
        – GreenMatt
        16 mins ago


















      • Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
        – GreenMatt
        37 mins ago






      • 1




        @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
        – terdon♦
        33 mins ago








      • 1




        @GreenMatt of course, if you need portability, you should also avoid using echo.
        – terdon♦
        29 mins ago










      • Perl? <shudder>
        – GreenMatt
        16 mins ago
















      Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
      – GreenMatt
      37 mins ago




      Thanks. How portable is that? It worked under Bash. Under tcsh I got an "Illegal variable name" error. Under sh, the 'n' was displayed.
      – GreenMatt
      37 mins ago




      1




      1




      @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
      – terdon♦
      33 mins ago






      @GreenMatt ah, that depends. For extreme portability, you may as well just use the perl one which is shell-agnostic. The printf "%s" $(seq 1 9) $(seq 0 9) will work in any POSIX shell, including sh and dash, but the printf "%s" $(seq 1 9) $(seq 0 9) $'n' will fail in dash. Note that it does work in sh (or at least in bash running as sh, so in POSIX mode), I assume you're running Ubuntu or a simmilar system whose sh is actually dash. tcsh is a strange one and not even trying to be POSIX so don't expect any of the shell-based approaches to work there.
      – terdon♦
      33 mins ago






      1




      1




      @GreenMatt of course, if you need portability, you should also avoid using echo.
      – terdon♦
      29 mins ago




      @GreenMatt of course, if you need portability, you should also avoid using echo.
      – terdon♦
      29 mins ago












      Perl? <shudder>
      – GreenMatt
      16 mins ago




      Perl? <shudder>
      – GreenMatt
      16 mins ago













      2














      You can remove newlines from any stream with tr -d 'n'. In your case



      (seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d 'n'


      While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.






      share|improve this answer

















      • 1




        Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
        – terdon♦
        20 mins ago










      • Ah yes, I never think of tr.
        – GreenMatt
        16 mins ago
















      2














      You can remove newlines from any stream with tr -d 'n'. In your case



      (seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d 'n'


      While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.






      share|improve this answer

















      • 1




        Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
        – terdon♦
        20 mins ago










      • Ah yes, I never think of tr.
        – GreenMatt
        16 mins ago














      2












      2








      2






      You can remove newlines from any stream with tr -d 'n'. In your case



      (seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d 'n'


      While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.






      share|improve this answer












      You can remove newlines from any stream with tr -d 'n'. In your case



      (seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d 'n'


      While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 31 mins ago









      Kamil Maciorowski

      1,2191625




      1,2191625








      • 1




        Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
        – terdon♦
        20 mins ago










      • Ah yes, I never think of tr.
        – GreenMatt
        16 mins ago














      • 1




        Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
        – terdon♦
        20 mins ago










      • Ah yes, I never think of tr.
        – GreenMatt
        16 mins ago








      1




      1




      Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
      – terdon♦
      20 mins ago




      Nice. This also has the benefit of being portable to most (all?) shells. I think the OP wants the final newline, so another option might be ( seq -s '' 1 9 ; echo -n 0 ; seq -s '' 1 9 ) | tr -d 'n' ; printf "n".
      – terdon♦
      20 mins ago












      Ah yes, I never think of tr.
      – GreenMatt
      16 mins ago




      Ah yes, I never think of tr.
      – GreenMatt
      16 mins ago











      1














      Consider also:



      printf '%d' $(seq -w 1 1 99 | cut -c2)


      We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.



      To get your desired output, use 20 instead:



      $ printf '%d' $(seq -w 1 1 20 | cut -c2)
      
12345678901234567890





      share|improve this answer




























        1














        Consider also:



        printf '%d' $(seq -w 1 1 99 | cut -c2)


        We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.



        To get your desired output, use 20 instead:



        $ printf '%d' $(seq -w 1 1 20 | cut -c2)
        
12345678901234567890





        share|improve this answer


























          1












          1








          1






          Consider also:



          printf '%d' $(seq -w 1 1 99 | cut -c2)


          We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.



          To get your desired output, use 20 instead:



          $ printf '%d' $(seq -w 1 1 20 | cut -c2)
          
12345678901234567890





          share|improve this answer














          Consider also:



          printf '%d' $(seq -w 1 1 99 | cut -c2)


          We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.



          To get your desired output, use 20 instead:



          $ printf '%d' $(seq -w 1 1 20 | cut -c2)
          
12345678901234567890






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 38 mins ago









          terdon♦

          128k31249423




          128k31249423










          answered 45 mins ago









          Jeff Schaller

          38.8k1053125




          38.8k1053125























              0














              I would like to reinforce the answer of @terdon, but propose a solution that works with the tcsh shell as well.
              Unfortunately, I didn't have enough reputation to write this as a comment:



              printf "%sn" "seq 1 9` `seq 0 9`"`



              The backtick - like all methods of command-line substitution -, should cut the trailing newline. Then, printf's own backslash-escape interpretation is used, instead of the ANSI string expansion used in terdon's answer.



              It also works like this, by allowing printf to interpret all escape sequences after substitution with the format specifiers (%b specifier):



              printf "%b" "seq 1 9` `seq 0 9`n"`






              share|improve this answer


























                0














                I would like to reinforce the answer of @terdon, but propose a solution that works with the tcsh shell as well.
                Unfortunately, I didn't have enough reputation to write this as a comment:



                printf "%sn" "seq 1 9` `seq 0 9`"`



                The backtick - like all methods of command-line substitution -, should cut the trailing newline. Then, printf's own backslash-escape interpretation is used, instead of the ANSI string expansion used in terdon's answer.



                It also works like this, by allowing printf to interpret all escape sequences after substitution with the format specifiers (%b specifier):



                printf "%b" "seq 1 9` `seq 0 9`n"`






                share|improve this answer
























                  0












                  0








                  0






                  I would like to reinforce the answer of @terdon, but propose a solution that works with the tcsh shell as well.
                  Unfortunately, I didn't have enough reputation to write this as a comment:



                  printf "%sn" "seq 1 9` `seq 0 9`"`



                  The backtick - like all methods of command-line substitution -, should cut the trailing newline. Then, printf's own backslash-escape interpretation is used, instead of the ANSI string expansion used in terdon's answer.



                  It also works like this, by allowing printf to interpret all escape sequences after substitution with the format specifiers (%b specifier):



                  printf "%b" "seq 1 9` `seq 0 9`n"`






                  share|improve this answer












                  I would like to reinforce the answer of @terdon, but propose a solution that works with the tcsh shell as well.
                  Unfortunately, I didn't have enough reputation to write this as a comment:



                  printf "%sn" "seq 1 9` `seq 0 9`"`



                  The backtick - like all methods of command-line substitution -, should cut the trailing newline. Then, printf's own backslash-escape interpretation is used, instead of the ANSI string expansion used in terdon's answer.



                  It also works like this, by allowing printf to interpret all escape sequences after substitution with the format specifiers (%b specifier):



                  printf "%b" "seq 1 9` `seq 0 9`n"`







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 16 mins ago









                  Larry

                  463




                  463






























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                      Understanding the information contained in the Deep Space Network XML data?

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                      up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret
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                      Ross-on-Wye

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                      Ross-on-Wye Welsh: Rhosan ar Wy Town centre, looking north from Market House Ross-on-Wye Ross-on-Wye shown within Herefordshire Population 10,700 (2011) [1] OS grid reference SO597241 Unitary authority Herefordshire Ceremonial county Herefordshire Region West Midlands Country England Sovereign state United Kingdom Post town ROSS-ON-WYE Postcode district HR9 Dialling code 01989 Police West Mercia Fire Hereford and Worcester Ambulance West Midlands EU Parliament West Midlands UK Parliament Hereford and South Herefordshire List of places UK England Herefordshire 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Coordinates: 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Ross-on-Wye (Welsh: Rhosan ar Wy [2] ) is a small market town with a population of 10,700 (according to the 2011 census), in south eastern Herefordshire, England, located on the River Wye, and on
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                      Eastern Orthodox Church

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                      Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate
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