Question

I am running my script in different environments and therefore I store the path and name of program I want to start in variable $CreateSequenceDictionary.

This program uses parameters R= and O= for this reason I put the whole command in quotations “”. Without these, program is started without any paramaters.

And now my problem: instead of $reference I would like to pass content of this variable. I tried this:

"java -jar $CreateSequenceDictionary R=$reference O=output" >mlogfile

and many variations (such as “ “” eval \) with no success.

Please, how should I adjust my command so I could run program stored in variable with parameter stored in variable?

EDIT: If I should use command in terminal, it would look like this:

 java -jar /path/CreateSequenceDictionary.jar R=input O=output

However, I want to get name of the program from variable and also name of input file from variable.

Answer 1

You can do that without the surrounding quotes.

java -jar $CreateSequenceDictionary R=$reference O=output > mlogfile

Bash will expand the vars before running the command.

If you want to conditionally include the R= and O= options, you could try something like this:

CreateSequenceDictionary="your_Program"
r_option ="$reference"
o_option="output"

arguments=()
[[ "$r_option" ]] && arguments+=("$r_option")
[[ "$o_option" ]] && arguments+=("$o_option")

java -jar $CreateSequenceDictionary "${arguments[@]}" > mlogfile

This builds the additional arguments on-the-fly and skips those where the associated *_option variable is empty.