Check if a variable exists in a list in Bash

Question

I am trying to write a script in bash that check the validity of a user input  I want to match the input  say variable x  to a list of valid values  what I have come up with at the moment is  for item in  list do     if    quot  x quot      quot  item quot     then         echo  quot In the list quot          exit     fi done  My question is if there is a simpler way to do this  something like a list contains x  for most programming languages  Say list is  list  quot 11 22 33 quot   my code will echo the message only for those values since list is treated as an array and not a string  all the string manipulations will validate 1 while I would want it to fail

User · Answer

The shell built-in compgen can help here. It can take a list with the -W flag and return any of the potential matches it finds.

# My list can contain spaces so I want to set the internal
# file separator to newline to preserve the original strings.
IFS=$'\n'

# Create a list of acceptable strings.
accept=( 'foo' 'bar' 'foo bar' )

# The string we will check
word='foo'

# compgen will return a list of possible matches of the 
# variable 'word' with the best match being first.
compgen -W "${accept[*]}" "$word"

# Returns:
# foo
# foo bar

We can write a function to test if a string equals the best match of acceptable strings. This allows you to return a 0 or 1 for a pass or fail respectively.

function validate {
  local IFS=$'\n'
  local accept=( 'foo' 'bar' 'foo bar' )
  if [ "$1" == "$(compgen -W "${accept[*]}" "$1" | head -1)" ] ; then
    return 0
  else
    return 1
  fi
}

Now you can write very clean tests to validate if a string is acceptable.

validate "blah" || echo unacceptable

if validate "foo" ; then
  echo acceptable
else 
  echo unacceptable
fi

User · Answer

Examples    in list super test me out NO    in list  super dude  test me out NO    in list  super dude  test me  super dude  YES    How to use in another script if     in list  1 OPTION1 OPTION2      NO    then   echo  UNKNOWN type for param 1  Should be OPTION1 or OPTION2    exit  fi   in list  function show help       IT   CAT  lt  lt EOF    usage  SEARCH FOR  ITEM1   ITEM2   ITEM3         e g      a b c d                    - gt  NO   a b a d                    - gt  YES    test me  how  test me     - gt  YES        echo   IT    exit    if     1      help    then   show help fi  if        -eq 0    then   show help fi  SEARCH FOR  1 shift   for ITEM in      do   if     SEARCH FOR       ITEM      then     echo  YES      exit    fi done  echo  NO

User · Answer

how about echo  list   grep -w -q  x  you could either check the output  or     of above line to make the decision  grep -w checks on whole word patterns  Adding -q prevents echoing the list

User · Answer

Thought I d add my solution to the list     Checks if element   1  is in array   2     NOTE      Be sure that array is passed in the form             ARR      elementIn            shopt -s nocasematch   Can be useful to disable case-matching     local e     for e in      2    do      e       1      amp  amp  return 0  done     return 1      Usage  list  11 22 33  item 22  if elementIn   item     list       then     echo TRUE  else     echo FALSE fi   TRUE  item 44 elementIn  item    list       amp  amp  echo TRUE    echo FALSE   FALSE

User · Answer

list          space     x      space         amp  amp  echo  yes     echo  no    or create a function   contains             1          space     2      space         amp  amp  exit 0     exit 1      to use it   contains aList anItem echo      0  match  1  failed

User · Answer

If it isn t too long  you can just string them between equality along a logical OR comparison like so  if    ITEM     quot item1 quot  -o  ITEM     quot item2 quot  -o  ITEM     quot item3 quot     then     echo In the list fi   I had this exact problem and while the above is ugly it is more obvious what is going on than the other generalized solutions

User · Answer

If your list of values is to be hard-coded in the script  it s fairly simple to test using case   Here s a short example  which you can adapt to your requirements   for item in  list do     case   x  in       item1 item2          echo  In the list                   not an item          echo  Error   gt  amp 2         exit 1                esac done   If the list is an array variable at runtime  one of the other answers is probably a better fit

User · Answer

Matvey is right  but you should quote  x and consider any kind of  spaces   e g  new line  with      list          space      x       space         amp  amp  echo  yes     echo  no     so  i e     list include item  10 11 12   2  function list include item     local list   1    local item   2    if     list          space      item       space          then       yes  list include item     result 0   else     result 1   fi   return  result     end then   list include item  10 11 12   12     amp  amp  echo  yes     echo  no    or  if  list include item  10 11 12   1     then   echo  yes  else    echo  no  fi   Note that you must use    in case of variables    list include item   my list    my item     amp  amp  echo  yes     echo  no

User · Answer

You can use     wildcards  outside a case statement  too  if you use double brackets   string  My string    if     string     My     then echo  It s there    fi

User · Answer

IMHO easiest solution is to prepend and append the original string with a space and check against a regex with        haystack  foo bar  needle  bar   if       haystack           needle         then         fi   this will not be false positive on values with values containing the needle as a substring  e g  with a haystack foo barbaz    The concept is shamelessly stolen form JQuery s hasClass  -Method

User · Answer

This is almost your original proposal but almost a 1-liner  Not that complicated as other valid answers  and not so depending on bash versions  can work with old bashes    OK 0   MP FLAVOURS  vanilla lemon hazelnut straciatella  for FLAV in  MP FLAVOURS   do    FLAV     FLAVOR    amp  amp    OK 1   break      done    OK -eq 0    amp  amp    echo   FLAVOR not a valid value   MP FLAVOURS     exit 1       I guess my proposal can still be improved  both in length and style

User · Answer

An alternative solution inspired by the accepted response  but that uses an inverted logic  MODE  quot   1  quot   echo  quot  lt   MODE  gt  quot      quot   MODE  quot       preview live both       amp  amp  echo  quot OK quot     echo  quot Uh  quot   Here  the input   MODE  must be one of the options in the regular expression   preview    live   or  both    contrary to matching the whole options list to the user input  Of course  you do not expect the regular expression to change

User · Answer

If the list is fixed in the script  I like the following the best   validate         grep -F -q -x   1   lt  lt EOF item 1 item 2 item 3 EOF     Then use validate   x  to test if  x is allowed   If you want a one-liner  and don t care about whitespace in item names  you can use this  notice -w instead of -x    validate     echo  11 22 33    grep -F -q -w   1       Notes    This is POSIX sh compliant  validate does not accept substrings  remove the -x option to grep if you want that   validate interprets its argument as a fixed string  not a regular  expression  remove the -F option to grep if you want that     Sample code to exercise the function   for x in  item 1   item2   item 3   3       do     echo -n    x  is       validate   x   amp  amp  echo  valid     echo  invalid  done

User · Answer

Assuming TARGET variable can be only  binomial  or  regression   then following would do     Check for modeling types known to this script if      echo    TARGET     egrep -c    binomial regression      -eq 0    then     echo  This scoring program can only handle  binomial  and  regression  methods now    gt  amp 2     usage fi   You could add more strings into the list by separating them with a    pipe  character   Advantage of using egrep  is that you could easily add case insensitivity  -i   or check more complex scenarios with a regular expression

User · Answer

Consider exploiting the keys of associative arrays  I would presume this outperforms both regex pattern matching and looping  although I haven t profiled it   declare -A list    one  1  two  two  three   any non-empty value    for value in one two three four do     echo -n   value is         a missing key expands to the null string         and we ve set each interesting key to a non-empty value        -z    list  value        amp  amp  echo -n   not        echo  a member of      list        done   Output    one is a member of   one two three   two is a member of   one two three   three is a member of   one two three   four is  not  a member of   one two three

User · Answer

I find it s easier to use the form echo  LIST   xargs -n1 echo   grep  VALUE as illustrated below   LIST  ITEM1 ITEM2  VALUE  ITEM1  if   -n   echo  LIST   xargs -n1 echo   grep -e     VALUE        then         fi   This works for a space-separated list  but you could adapt it to any other delimiter  like    by doing the following   LIST  ITEM1 ITEM2  VALUE  ITEM1  if   -n   echo  LIST   sed  s     n g    grep -e     VALUE          then        fi   Note that the   are required for the test to work

User · Answer

Prior answers don t use tr which I found to be useful with grep  Assuming that the items in the list are space delimited  to check for an exact match  echo  mylist   tr       n    grep -F -x -q  quot  myitem quot   This will return exit code 0 if the item is in the list  or exit code 1 if it isn t  It s best to use it as a function   contains         Check if space-separated list  1 contains line  2   echo  quot  1 quot    tr       n    grep -F -x -q  quot  2 quot     mylist  quot aa bb cc quot     Positive check if  contains  quot   mylist  quot   quot   myitem  quot   then   echo  quot in list quot  fi    Negative check if    contains  quot   mylist  quot   quot   myitem  quot   then   echo  quot not in list quot  fi

[linux] Check if a variable exists in a list in Bash

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