Process all arguments except the first one in a bash script

Question

I have a simple script where the first argument is reserved for the filename  and all other optional arguments should be passed to other parts of the script   Using Google I found this wiki  but it provided a literal example   echo       -1     I can t get anything else to work  like   echo      2     or  echo      2 1     I get  Bad substitution  from the terminal   What is the problem  and how can I process all but the first argument passed to a bash script

User · Answer

http   wiki bash-hackers org scripting posparams  It explains the use of shift  if you want to discard the first N parameters  and then implementing Mass Usage

User · Answer

Use this   echo      2       The following syntax   echo      2     would work as well  but is not recommended  because as  Gordon already explained  that using    it runs all of the arguments together as a single argument with spaces  while   preserves the breaks between them  even if some of the arguments themselves contain spaces   It doesn t make the difference with echo  but it matters for many other commands

User · Answer

If you want a solution that also works in  bin sh try  first arg   1  shift echo First argument    first arg  echo Remaining arguments         shift  n  shifts the positional parameters n times  A shift sets the value of  1 to the value of  2  the value of  2 to the value of  3  and so on  decreasing the value of    by one

User · Answer

Working in bash 4 or higher version     bin bash echo  quot  0 quot             quot bash quot  bash --version       quot GNU bash  version 5 0 3 1 -release  x86 64-pc-linux-gnu  quot   In function  echo                    quot p1 quot   quot p2 quot   quot p3 quot   quot p4 quot   quot p5 quot  echo      0      quot bash quot   quot p1 quot   quot p2 quot   quot p3 quot   quot p4 quot   quot p5 quot  echo      1             quot p1 quot   quot p2 quot   quot p3 quot   quot p4 quot   quot p5 quot  echo      2                  quot p2 quot   quot p3 quot   quot p4 quot   quot p5 quot  echo      2 1                quot p2 quot  echo      2 2                quot p2 quot   quot p3 quot  echo      -2                           quot p4 quot   quot p5 quot  echo      -2 1                         quot p4 quot   Notice the space between     and  -   otherwise it means different     var -word  If var is null or unset   word is substituted for var  The value of var does not change    var  word  If var is set   word is substituted for var  The value of var does not change   Which is described in Unix   Linux - Shell Substitution

User · Answer

Came across this looking for something else  While the post looks fairly old  the easiest solution in bash is illustrated below  at least bash 4  using set --          where   is the starting number of the array element we want to preserve forward          bin bash      someVar    1       someOtherVar    2       set --      3       input                 input             someword       amp  amp  someNewVar  trigger       echo -e    someVar  n  someOtherVar  n  someNewVar  n n        Basically  the set --      3   just pops off the first two elements in the array like perl s shift and preserves all remaining elements including the third   I suspect there s a way to pop off the last elements as well

[bash] Process all arguments except the first one (in a bash script)

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