bash: How to place variable into specific string

Question

I have the following command in a bash script:

$prefixBioAwk -c fastx '{ if(length($seq) > 600) { print ">"$name; print $seq }}' my.fasta > short.fasta

Now I want to make the number 600 flexible by inserting the var $myVar (which contains an iteger) there.

How do I do it?

If $prefixBioAwk is anything like Awk, awk -v variable="$value" 'length($seq) > variable { print ">" $name; print $seq }' my.fasta — tripleee
– tripleee, Commented Aug 31, 2016 at 12:06
hmmm... `-v' seems to have another function in bioawk. is there another way? — Michael
– Michael, Commented Aug 31, 2016 at 12:42
Looking at the github page, it seems -v serves same purpose as in awk — Fazlin
– Fazlin, Commented Aug 31, 2016 at 13:48

Eric · Accepted Answer · 2016-09-01 00:45:56Z

1

Since this is posted as a bash question, I'm answering with a bash response rather than something bioawk specific:

myVar=600 prefixBioAwk -c fastx '{ if(length($seq) > '$myVar') { print ">"$name; print $seq }}' my.fasta > short.fasta

The key is to get out of the single quotes so that the shell can interpret the variable and substitute its value.

answered Sep 1, 2016 at 0:45

Eric

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1 Comment

ups, surprisingly easy. I have to have a look at the logics of bash replacement, quotes, etc. Many thanks