CODEWITHSUNDEEP
bashheredoc
Julio Diaz
Julio Diaz

Reputation: 9437

How to avoid printing quotation marks in cat <<

I am trying to generate a file given a template with this code in test.sh:

#!/bin/sh

#define parameters which are passed in.
ID=$1
DIRECTORY=$2

#define the template.
cat  <<EOF
./pgap.py -D singularity -r -o $DIRECTORY/$ID $DIRECTORY/$ID'_input.yaml'

I used ' in the last line to avoid the text trailing the $ID variable to be mixed with it. But when running the code prints both '. In other words, when I run:

bash test.sh id dir

I get

./pgap.py -D singularity -r -o dir/id dir/id'_input.yaml'

, but I'd like to get:

./pgap.py -D singularity -r -o dir/id dir/id_input.yaml

Upvotes: 1

Views: 108

Answers (2)

Cyrus
Cyrus

Reputation: 88583

Replace

$DIRECTORY/$ID'_input.yaml'

with

$DIRECTORY/${ID}_input.yaml

Upvotes: 3

Paul Barrett
Paul Barrett

Reputation: 41

You can achieve what you're trying to do with the single quotes by enclosing the variable name in curly braces, i.e. $DIRECTORY/${ID}_input.yaml

Upvotes: 1

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