How can I pass a file argument to my bash script using a Terminal command in Linux

Question

So my question is how can I pass a file argument to my bash script using a Terminal command in Linux  At the moment I m trying to make a program in bash that can take a file argument from the Terminal and use it as a variable in my program  For example I run myprogram --file  path to file in Terminal  My Program    bin bash     File  the path from the argument    externalprogram  File  other parameters   How can I achieve this with my program

User · Answer

It ll be easier  and more  proper   see below  if you just run your script as  myprogram  path to file   Then you can access the path within the script as  1  for argument  1  similarly  2 is argument  2  etc    file   1  externalprogram   file   other parameters    Or just  externalprogram   1   otherparameters    If you want to extract the path from something like --file  path to file  that s usually done with the getopts shell function  But that s more complicated than just referencing  1  and besides  switches like --file  are intended to be optional  I m guessing your script requires a file name to be provided  so it doesn t make sense to pass it in an option

User · Answer

Assuming you do as David Zaslavsky suggests  so that the first argument simply is the program to run  no option-parsing required   you re dealing with the question of how to pass arguments 2 and on to your external program  Here s a convenient way      bin bash ext program   1  shift   ext program         The shift will remove the first argument  renaming the rest   2 becomes  1  and so on      refers to the arguments  as an array of words  it must be quoted     If you must have your --file syntax  for example  if there s a default program to run  so the user doesn t necessarily have to supply one   just replace ext program   1  with whatever parsing of  1 you need to do  perhaps using getopt or getopts   If you want to roll your own  for just the one specific case  you could do something like this   if        -gt 0 -a    1 0 6       --file     then     ext program    1 7   else     ext program  default program  fi

User · Answer

Bash supports a concept called  Positional Parameters   These positional parameters represent arguments that are specified on the command line when a Bash script is invoked   Positional parameters are referred to by the names  0   1   2     and so on   0 is the name of the script itself   1 is the first argument to the script   2 the second  etc     represents all of the positional parameters  except for  0  i e  starting with  1    An example      bin bash FILE   1  externalprogram   FILE   lt other-parameters gt

User · Answer

you can use getopt to handle parameters in your bash script  there are not many explanations for getopt out there  here is an example      bin sh  OPTIONS   getopt -o hf gb -l help file  foo bar --        if      -ne 0    then   echo  getopt error    exit 1 fi  eval set --  OPTIONS  while true  do   case   1  in     -h --help  HELP 1        -f --file  FILE   2    shift        -g --foo   FOO 1        -b --bar   BAR 1        --         shift   break                   echo  unknown option   1    exit 1      esac   shift done  if      -ne 0    then   echo  unknown option s         exit 1 fi  echo  help   HELP  echo  file   FILE  echo  foo   FOO  echo  bar   BAR    see also    the  canonical  example  http   software frodo looijaard name getopt docs getopt-parse bash a blog post  http   www missiondata com blog system-administration 17 17  man getopt

[linux] How can I pass a file argument to my bash script using a Terminal command in Linux?

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