CODEWITHSUNDEEP
bashshell
olidev
olidev

Reputation: 20644

String concatenation in Bash script

I am writing this Bash script:

count=0   
result

for d in `ls -1 $IMAGE_DIR | egrep "jpg$"`
do

    if (( (count % 4) == 0 )); then
                result="abc $d"

                if (( count > 0 )); then
                    echo "$result;"
                fi

        else
            result="$result $d"
        fi

        (( count++ ))

done

if (( (count % 4) == 0 )); then
    echo $result
fi

The script is to concate part strings into a string when the value is divided by 4 and it should be larger than 0.

In the IMAGE_DIR, I have 8 images,

I got outputs like this: 

abc et004.jpg
abc et008.jpg

But I would expect to have:

abc et001.jpg et002.jpg et003.jpg et004.jpg;
abc et005.jpg et006.jpg et007.jpg et008.jpg;

How can I fix this?

Upvotes: 1

Views: 5987

Answers (3)

Dimitre Radoulov
Dimitre Radoulov

Reputation: 28000

Something like this (untested, of course):

count=0 result=

for d in "$IMAGE_DIR"/*jpg; do
   (( ++count % 4 == 0 )) &&
     result="abc $d"
   (( count > 0 )) &&
     printf '%s\n' "$result" ||
      result+=$d
done

Upvotes: 1

Kusalananda
Kusalananda

Reputation: 15613

Something like this?

count=0   

find $IMAGE_DIR -name "*.jpg" |
while read f; do
        if (( (count % 4) == 0 )); then
                result="abc $f"

                if (( count > 0 )); then
                        echo $result
                fi

        else
                result="$result $d"
        fi

        (( count++ ))
done

Upvotes: 2

Kilian Foth
Kilian Foth

Reputation: 14346

The = operator must always be written without spaces around it:

result="$result $d"

(Pretty much the most important difference in shell programming to normal programming is that whitespace matters in places where you wouldn't expect it. This is one of them.)

Upvotes: 7

Related Questions