Compare array elements with an element - arrays

Question

I am trying to compare an array with a value and print an error statement if the match is not found.
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
For example, this stores values such as small, large and medium which are the files present in new-dir.
The value to be compared with will be entered by the user viz. var
I have tried something like following:
(for i in "${arraylist[#]}"; do [[ "$i" == "$var"]] && exit 0; done) && echo "found" || echo "not found"
Also tried, however, doesn't work:
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
count=0
for((i=0; i<${#arraylist[#]}; i++)); do
if [ "$arraylist[$i]" == "$var" ] ; then
count=1
fi
done
if [ $count -eq 0 ]; then
echo "Not found"
fi
Is there any other way to do this comparison?

Answer 1

This code works. Just replace arraylist with your array.
arraylist=('hello' 'world')
var='hello'
count=0
for i in "${arraylist[#]}"
do
if [ "$i" == "$var" ]; then
count=1
fi
done
if [ $count -eq 0 ]; then
echo "Not found"
fi

Answer 2

To extract the first "word" from each filename, I'd do
declare -A prefixes
cd new-dir
for file in *; do
read -r prefix _ <<< "$file"
prefixes["$prefix"]=1
done
cd -
then to look for a match:
# $var is somehow provided
result=""
for p in "${!prefixes[#]}"; do
if [[ $p == "$var" ]]; then
result=found
break
fi
done
echo "${result:-not found}"
This requires bash v4.0+ for the use of an associative array.
If you have bash v4.3+ you can do
[[ -v prefixes[$var] ]] && echo found || echo not found
Alternately:
shopt -s nullglob
files=( new-dir/"$var "* )
(( ${#files[#]} > 0 )) && echo found || echo not found

Answer 3

This is what shell programming is about. Keep it simple :
Array1=( "item1" "item2" "item3" "item-4" )
var="item3"
count=$(echo ${Array1[#]} | tr ' ' '\n' | awk '$1 == "'"$var"'"{print $0}' | wc -l)
[ $count -eq 0 ] && echo "Not found" || echo "found"

Answer 4

The arraylist might be an array with just one index.
The command:
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
might just assign whole "ls new-dir" output to ${arraylist[0]}
You can check the index value in an array by echo ${arraylist[1]} (to confirm).
Try changing the command to generate an array for a list of files/dirs to:
arraylist=($(ls new-dir/))
The complete script:
arraylist=($(ls new-dir/))
(for i in "${arraylist[#]}"; do [[ "$i" == "$var" ]] && exit 0; done) && echo "found" || echo "not found"
You can also ignore the whole array assignment and just use subshell-command in for loop, eg:
(for i in $(ls -1 new-dir/); do [[ "$i" == "$var" ]] && exit 0; done) && echo "found" || echo "not found"
Remember, no quotes " around $(ls -1 new-dir/)
Another one-liner without any array or loops:
ls -1 new-dir/ | grep -q "$var" && echo "found" || echo "not found"
Edit:
As #M. Nejat Aydin suggested, use [[ -e new-dir/$var ]] && echo "found" || echo "Not found" and do not use ls in the script.