variable not showing up in bash script when using an array

Question

I have a script to prepare some files to submit trough qsub to a cluster, I am creating an array based on a file and then using the elements in that array to create my qsub files. However, I can't append the variables to the $RawData/$i_1.fastq.gz part. This is my script:

> cat create.sh
#!/bin/bash

 RawData="/home/jfertaj/data/FASTQ" 

 # make an array of each sample id 
 mapfile -t myArray < array.txt

 for i in "${myArray[@]}"
 do
      cat > pbs.script.$i << EOF
 #!/bin/bash

 kallisto quant -t 16 -b 100 -o /home/jfertaj/data/results_kallisto/output_bootstrap_$i $RawData/$i_1.fastq.gz $RawData/$i_2.fastq.gz

 EOF
 done
 exit 0;

when I run the bash script and look the created files I see this:

...
kallisto quant -t 16 -b 100 -o /home/jfertaj/data/results_kallisto/output_bootstrap_INTP_993 /home/jfertaj/data/FASTQ/.fastq.gz /home/jfertaj/data/FASTQ/.fastq.gz

I have tried including "$i" but then this shows up on the resulting files: "INTP_993"_1.fastq.tz. Is there any way to fix it?

chepner · Accepted Answer · 2016-11-05 15:40:58Z

2

_ is a legal character for variable names, so you need to use the full syntax for parameter expansion to keep it from being treated as part of the parameter name. $i_1.fastq.gz is interpreted as ${i_1}.fastq.gz, not ${i}_1.fastq.gz.

for i in "${myArray[@]}"
do
    cat > "pbs.script.$i" << EOF
#!/bin/bash

kallisto quant -t 16 -b 100 \
  -o "/home/jfertaj/data/results_kallisto/output_bootstrap_$i" \
  "$RawData/${i}_1.fastq.gz" \
  "$RawData/${i}_2.fastq.gz"

EOF
done
exit 0

answered Nov 5, 2016 at 15:40

chepner

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Comments

janos · Accepted Answer · 2016-11-05 16:00:16Z

Replace $RawData/$i_1.fastq.gz $RawData/$i_2.fastq.gz with $RawData/${i}_1.fastq.gz $RawData/${i}_2.fastq.gz.

That is, change this:

kallisto quant -t 16 -b 100 -o /home/jfertaj/data/results_kallisto/output_bootstrap_$i $RawData/$i_1.fastq.gz $RawData/$i_2.fastq.gz

To this:

kallisto quant -t 16 -b 100 -o /home/jfertaj/data/results_kallisto/output_bootstrap_$i $RawData/${i}_1.fastq.gz $RawData/${i}_2.fastq.gz

The shell treats i_1 and i_2 as variable names, when you really want the variable i appended with "_1" and "_2". In such situations, when you need to use some variable somevar followed by some suffix starting with symbols that are valid in variable names, let's say _suffix, you need to use braces to identify the variable. So instead of $somevar_suffix you need to write ${somevar}_suffix.

This is mentioned in man bash, emphasis mine:

${parameter}

The value of parameter is substituted. The braces are required when parameter is a positional parameter with more than one digit, or when parameter is followed by a character which is not to be interpreted as part of its name. [...]

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