CODEWITHSUNDEEP
arraysbash
MLMH
MLMH

Reputation: 141

Array won't return ref value (bash)

I've been trying to storage my data in an array as follows:

table="$1"
reference="$2"
directory="$3"

declare -A count

$reference | while read line; do
    ref=`echo $line | cut -d " " -f1`
    key=`echo $line | cut -d " " -f4`
    count=([$key]=$ref)
    echo $ref
    echo $key
    echo "${count[$key]}"
 done

This works, I do the prints and for each key I got the value I want. Then, I try to use with some keys:

cat $table | while read line; do
    sample=`echo $line | cut -d "_" -f1`
    id=${count[$line]}
    echo $sample
    echo $line
    echo $id
    echo "works"
done

Here is the problem: Sample is echoed perfectly, just as are $line and "works". But $id is not working, and I have no idea what I am missing here

Upvotes: 0

Views: 38

Answers (1)

P.P
P.P

Reputation: 121397

That's because whatever you store in count are gone as soon as the subshell that runs the while loop exits. The while loop that you pipe into runs in a subshell. So any variables you "set" won't be available outside if it. That means when you later use count, it doesn't have any elements.

Change the loop to:

while IFS= read -r line; do
    ref=`echo $line | cut -d " " -f1`
    key=`echo $line | cut -d " " -f4`
    count["$key"]="$ref"
    echo $ref
    echo $key
    echo "${count[$key]}"
 done < "${reference}"

Similarly change in the other loop.

I am assuming reference is a file. If reference is actually the text you want to read the "lines" from then you can use here strings:

while IFS= read -r line; do
    ...
    ...  
done <<< "${reference}"

Upvotes: 1

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