CODEWITHSUNDEEP
bash
Homap
Homap

Reputation: 2214

Passing an array as bash variable

I am submitting a series of batch jobs in a bash script.

The batch script (script.sh) is as follows:

samtools merge $1 $2

$1 is the output and it is only one single file. So, it is no problem.

$2 is the input but it is not only one file, it is several files and it is different in every directory. The bash script is as follows: 

for i in directory1 directory2 
do
cd $i
d=$(echo *specific_prefix.txt)
echo $d
sbatch script.sh ${i}.output $d
cd ..
done

Running as above, the batch script only takes the first element of $d as its variable. Is there a way for a bash variable ($2) to take varying number of variables, for example for it to be a list whose length can be determined in another variable?

Edit:

I have the sbatch job script (merge.sh) as:

#!/bin/bash -l

arg1="$1"
shift 1
arg2=("$@")

samtools merge arg1 arg2

And I run it using a bash script:

#!/bin/bash -l

for i in directory
do
cd $i
d=$(echo *RG.sort.dedup.bam)
sbatch merge.sh ${i}.output $d
cd ..
done

The error is: samtools merge: fail to open "arg2": No such file or directory

What could be the error in the script? Thanks!

Upvotes: 2

Views: 128

Answers (1)

anubhava
anubhava

Reputation: 784958

You can use shift command after storing fixed arguments and use $@ as below script. You can use a shell array to store variable number of arguments starting from 3rd position.

# store first 2 arguments in variables
arg1="$1"
arg2="$2"

# using shift 2 discard first 2 arguments
shift 2

# now store all remaining arguments in an array
arg3=("$@")

# examine our variables
declare -p arg1 arg2 arg3

Now run it as:

./script.sh merge abc foo bar baz
declare -- arg1="merge"
declare -- arg2="abc"
declare -a arg3='([0]="foo" [1]="bar" [2]="baz")'

Upvotes: 2

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