Finding an elements in an array in bash

Question

I am using the following code to first split an array into 2 and then search if 2 elements "Alchemist" and "Axe" are present in each of the two split arrays.

tempifs=$IFS
    IFS=,
    match=($i)
    IFS=$tempifs
    team1=( "${match[@]:0:5}" )
    team2=( "${match[@]:5:5}" )
        if [ array_contains2 $team1 "Alchemist" "Axe" ]
    then
    echo "team1 contains"
        fi
    if [ array_contains2 $team2 "Alchemist" "Axe" ]
    then
    echo "team2 contains"
        fi  

array_contains2 () { 
    local array="$1[@]"
    local seeking=$2
    local seeking1=$3
    local in=0
    for element in "${array[@]}"; do
        if [[ $element == $seeking && $element == $seeking1]]
    then
            in=1
            break
        fi
    done
    return $in
}

But I am getting the following error -

/home/ashwin/bin/re: line 18: [: Alchemist: binary operator expected
/home/ashwin/bin/re: line 14: [: too many arguments

Lines 14 and 18 are if [ array_contains2 $team1 "Alchemist" "Axe" ] and if [ array_contains2 $team2 "Alchemist" "Axe" ] respectively.

Is the error because of IFS. If not what is the cause for the error?

Youre quoting the wrong stuff. "Alchemist" and "Axe" dont expand, but $element $seeking and $team do. — mikeserv
– mikeserv, Commented Apr 19, 2014 at 9:16

Community · Accepted Answer · 2017-05-23 11:33:33Z

3

I believe the problem has to do with your if statements. It looks like if you're using a function you don't need the square brackets. Please see this:

https://stackoverflow.com/questions/8117822/in-bash-can-you-use-a-function-call-as-a-condition-in-an-if-statement

I believe that you'll want to instead do:

if array_contains2 $team1 "Alchemist" "Axe"; then
    echo "This is true"
fi

edited May 23, 2017 at 11:33

CommunityBot

1

answered Apr 19, 2014 at 7:55

just some user

4832 silver badges9 bronze badges

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mikeserv · Accepted Answer · 2014-04-19 09:28:56Z

1

You're already using a function, why would you limit yourself to bash arrays rather than use the shell $@ array?

bash_array=(one two three)
set -- $bash_array
printf %s\\n "$@"
    #output
one
two
three

IFS=/ ; echo "$*" ; echo "$@"
    #output 
/one/two/three
one two three

unset IFS ; in=$* ; 

[ -n "${in#"${in%$2*}"}" ] && echo "$2 is in $@" || echo nope
    #output
two is in one two three

answered Apr 19, 2014 at 9:28

mikeserv

59.4k10 gold badges122 silver badges242 bronze badges

Would you explain your answer? What is the type of $in toward the end (a string, not an array, right?)? Are you finding an element by looking through a STRING representation of the array?

lmat - Reinstate Monica
– lmat - Reinstate Monica

2022-03-13 20:20:33 +00:00
Commented Mar 13, 2022 at 20:20

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