bash: string concatenation inside a parameter of a command

Question

I would like to run a command in a bash script according to the following code sample. Despite the fact that j and i variables contain the filenames (these filenames were read out from the samples.list.txt), these strings can't be inserted into the command. What could be wrong? Many thanks!

cat work/sample.list.txt | while read line;
do
    arrIN=(${line// / })
    i=${arrIN[0]}
    j=${arrIN[1]}
    k=${arrIN[2]}

   java -jar some.jar ./input1directory/"$i" ./input2directory/"$j" 
done

What's the point of replacing space with space? Why not just arrIn=($line)? — Barmar
– Barmar, Commented May 11, 2015 at 9:47
Well, your default delimiter is ok. Please run your script as follows: bash -x ./yourscript.sh and post here the output. Thanks — Samuel
– Samuel, Commented May 11, 2015 at 10:20

hek2mgl · Accepted Answer · 2015-05-11 09:39:41Z

4

You can read multiple variables at once, like this:

while read i j k ; do
   java -jar some.jar ./input1directory/"$i" ./input2directory/"$j" 
done < "work/sample.list.txt"

Every line read reads will get subject of word splitting and the individual fields will get assigned to the variables you specify on the command line. If there are more fields than variables, the whole remaining line gets assigned to that last variable specified.

answered May 11, 2015 at 9:39

hek2mgl

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