How to store a command in a variable in a shell script

Question

I would like to store a command to use at a later period in a variable  not the output of the command  but the command itself   I have a simple script as follows   command  ls   echo  Command   command    Output is  Command  ls  b   command   echo  b   Output is  public html REV test     command worked successfully    However  when I try something a bit more complicated  it fails  For example  if I make  command  ls   grep -c         The output is   Command  ls   grep -c     ls  cannot access    No such file or directory ls  cannot access grep  No such file or directory ls  cannot access      No such file or directory   Any idea how I could store such a command  with pipes multiple commands  in a variable for later use

User · Accepted Answer

Use eval   x  ls   wc  eval   x  y   eval   x   echo   y

User · Answer

For bash  store your command like this   command  ls   grep -c        Run your command like this   echo  command   bash

User · Answer

bin bash  Note  this script works only when u use Bash  So  don t remove the first line   TUNECOUNT   ifconfig  grep -c -o tune0   Some command with  Grep   echo  TUNECOUNT                          This will return 0                                       if you don t have tune0 interface                                       Or count of installed tune0 interfaces

User · Answer

First of all there are functions for this  But if you prefer vars then your task can be done like this    cmd ls     cmd   works file  file2  test    cmd  ls   grep file      cmd   not works ls  cannot access      No such file or directory ls  cannot access  grep   No such file or directory  file    bash -c  cmd   works file  file2  test    bash -c  quot  cmd quot    also works file file2    bash  lt  lt  lt   cmd file file2    bash  lt  lt  lt   quot  cmd quot  file file2  Or via tmp file   tmp   mktemp    echo  quot  cmd quot   gt   quot  tmp quot    chmod  x  quot  tmp quot     quot  tmp quot  file file2    rm  quot  tmp quot

User · Answer

var   echo  asdf   echo  var     gt  asdf   Using this method  the command is immediately evaluated and it s return value is stored   stored date   date  echo  stored date     gt  Thu Jan 15 10 57 16 EST 2015    wait a few seconds  echo  stored date     gt  Thu Jan 15 10 57 16 EST 2015   Same with backtick  stored date  date  echo  stored date     gt  Thu Jan 15 11 02 19 EST 2015    wait a few seconds  echo  stored date     gt  Thu Jan 15 11 02 19 EST 2015   Using eval in the        will not make it evaluated later  stored date   eval  date   echo  stored date     gt  Thu Jan 15 11 05 30 EST 2015    wait a few seconds  echo  stored date     gt  Thu Jan 15 11 05 30 EST 2015   Using eval  it is evaluated when eval is used  stored date  date     lt  storing the command itself echo   eval   stored date       gt  Thu Jan 15 11 07 05 EST 2015    wait a few seconds  echo   eval   stored date       gt  Thu Jan 15 11 07 16 EST 2015                          Time changed   In the above example  if you need to run a command with arguments  put them in the string you are storing  stored date  date -u          For bash scripts this is rarely relevant  but one last note  Be careful with eval  Eval only strings you control  never strings coming from an untrusted user or built from untrusted user input    Thanks to  CharlesDuffy for reminding me to quote the command

User · Answer

Be Careful registering an order with the  X   Command   This one is still executed Even before being called  To check and confirm this  you cand do   echo test  X   for   c 0  c lt  5  c      do sleep 2  done   echo note the 5 seconds elapsed

User · Answer

I tried various different methods   printexec       printf --   033 1 37m  033 0m    printf --    q         printf --   n    eval --        eval --                        Output     printexec echo  -e  foo n  bar   echo -e foo  n bar foon bar foon bar foo  bar bash  echo -e foo n bar  command not found   As you can see  only the third one       gave the correct result

User · Answer

Its is not necessary to store commands in variables even as you need to use it later  just execute it as per normal  If you store in variable  you would need some kind of eval statement or invoke some unnecessary shell process to  execute your variable

User · Answer

Do not use eval  It has a major risk of introducing arbitrary code execution  BashFAQ-50 - I m trying to put a command in a variable  but the complex cases always fail  Put it in an array and expand all the words with double-quotes  quot   arr     quot  to not let the IFS split the words due to Word Splitting  cmdArgs    cmdArgs   date     H  M  S    and see the contents of the array inside  The declare -p allows  you see the contents of the array inside with each command parameter in separate indices  If one such argument contains spaces  quoting inside while adding to the array will prevent it from getting split due to Word-Splitting  declare -p cmdArgs declare -a cmdArgs    0   quot date quot   1   quot   H  M  S quot     and execute the commands as  quot   cmdArgs     quot  23 15 18   or  altogether use a bash function to run the command  cmd        date    H  M  S     and call the function as just cmd  POSIX sh has no arrays  so the closest you can come is to build up a list of elements in the positional parameters  Here s a POSIX sh way to run a mail program   POSIX sh   Usage  sendto subject address  address      sendto         subject  1     shift     first 1     for addr  do         if    quot  first quot    1    then set --  first 0  fi         set --  quot    quot  --recipient  quot  addr quot      done     if    quot  first quot    1    then         echo  quot usage  sendto subject address  address      quot          return 1     fi     MailTool --subject  quot  subject quot   quot    quot     Note that this approach can only handle simple commands with no redirections  It can t handle redirections  pipelines  for while loops  if statements  etc Another common use case is when running curl with multiple header fields and payload  You can always define args like below and invoke curl on the expanded array content curlArgs   -H   quot keyheader  value quot   -H   quot 2ndkeyheader  2ndvalue quot   curl  quot   curlArgs     quot   Another example  payload      hostURL  http   google com  authToken  someToken  authHeader  Authorization Bearer  quot   quot  authToken quot   quot    now that variables are defined  use an array to store your command args curlCMD  -X POST  quot  hostURL quot  --data  quot  payload quot  -H  quot Content-Type application json quot  -H  quot  authHeader quot    and now do a proper quoted expansion curl  quot   curlCMD     quot

[linux] How to store a command in a variable in a shell script?

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