Propagate all arguments in a bash shell script

Question

I am writing a very simple script that calls another script  and I need to propagate the parameters from my current script to the script I am executing   For instance  my script name is foo sh and calls bar sh  foo sh   bar  1  2  3  4   How can I do this without explicitly specifying each parameter

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My SUN Unix has a lot of limitations  even      was not interpreted as desired  My workaround is        For example      bin ksh find    -type f   xargs grep          By the way  I had to have this particular script because my Unix also does not support grep -r

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quot   array     quot  is the right way for passing any array in bash  I want to provide a full cheat sheet  how to prepare arguments  bypass and process them  pre sh - gt  foo sh - gt  bar sh     bin bash  args   quot --a b c quot   quot --e f g quot   args    quot --q w e quot   quot --a s   quot  d   quot  quot      foo sh  quot   args     quot      bin bash    bar sh  quot    quot      bin bash  echo  1 echo  2 echo  3 echo  4  result  --a b c --e f g --q w e --a s  quot  d  quot

User · Answer

Use  quot    quot  instead of plain    if you actually wish your parameters to be passed the same  Observe    cat no quotes sh    bin bash echo args sh       cat quotes sh    bin bash echo args sh  quot    quot     cat echo args sh    bin bash echo Received   1 echo Received   2 echo Received   3 echo Received   4      no quotes sh first second Received  first Received  second Received  Received       no quotes sh  quot one quoted arg quot  Received  one Received  quoted Received  arg Received       quotes sh first second Received  first Received  second Received  Received       quotes sh  quot one quoted arg quot  Received  one quoted arg Received  Received  Received

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A lot answers here recommends    or    with and without quotes  however none seems to explain what these really do and why you should that way  So let me steal this excellent summary from this answer    -------- ---------------------------    Syntax        Effective result        -------- ---------------------------                  1  2  3       N         -------- ---------------------------                  1  2  3       N         -------- ---------------------------      quot    quot         quot  1c 2c 3c   c  N  quot        -------- ---------------------------      quot    quot      quot  1 quot   quot  2 quot   quot  3 quot       quot   N  quot     -------- ---------------------------   Notice that quotes makes all the difference and without them both have identical behavior  For my purpose  I needed to pass parameters from one script to another as-is and for that the best option is    file  parent sh   we have some params passed to parent sh    which we will like to pass on to child sh as-is    child sh     Notice no quotes and    should work as well in above situation

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For bash and other Bourne-like shells   java com myserver Program

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usr bin env bash while     1           do   echo  Received    1    amp  amp  shift  done    Just thought this may be a bit more useful when trying to test how args come into your script

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If you include    in a quoted string with other characters the behavior is very odd when there are multiple arguments  only the first argument is included inside the quotes   Example      bin bash set -x bash -c  true foo       Yields     bash test sh bar baz   bash -c  true foo bar  baz   But assigning to a different variable first      bin bash set -x args      bash -c  true foo  args    Yields     bash test sh bar baz   args  bar baz    bash -c  true foo bar baz

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bar      will be equivalent to bar   1    2    3    4   Notice that the quotation marks are important                  or    will each behave slightly different regarding escaping and concatenation as described in this stackoverflow answer    One closely related use case is passing all given arguments inside an argument like this    bash -c  bar    1      2      3      4      I use a variation of  kvantour s answer to achieve this   bash -c  bar   printf --    s

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Use       works for all POSIX compatibles               bash features the      variable  which expands to all command-line parameters separated by spaces    From Bash by example

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I realize this has been well answered but here s a comparison between              and     Contents of test script     cat   test sh    usr bin env bash echo                                      echo  Quoted DOLLAR-AT  for ARG in       do     echo  ARG done  echo                                      echo  NOT Quoted DOLLAR-AT  for ARG in     do     echo  ARG done  echo                                      echo  Quoted DOLLAR-STAR  for ARG in       do     echo  ARG done  echo                                      echo  NOT Quoted DOLLAR-STAR  for ARG in     do     echo  ARG done  echo                                       Now  run the test script with various arguments       test sh   arg with space one   arg2  arg3                                   Quoted DOLLAR-AT arg with space one arg2 arg3                                   NOT Quoted DOLLAR-AT arg with space one arg2 arg3                                   Quoted DOLLAR-STAR arg with space one arg2 arg3                                   NOT Quoted DOLLAR-STAR arg with space one arg2 arg3

User · Answer

Sometimes you want to pass all your arguments  but preceded by a flag  e g  --flag     bar --flag   1  --flag   2  --flag   3    You can do this in the following way     bar   printf --   --flag   s           note  to avoid extra field splitting  you must quote  s and     and to avoid having a single string  you cannot quote the subshell of printf

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Works fine  except if you have spaces or escaped characters  I don t find the way to capture arguments in this case and send to a ssh inside of script   This could be useful but is so ugly   command opts    echo        awk -F -  BEGIN   OFS   -      for  i 2 i lt  NF i      gsub    a-z      amp     i    gsub          i   gsub           i     print  0      tr

[bash] Propagate all arguments in a bash shell script

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