why does this only run once instead of twice after making it as variable? bash

Question

I have something like this...

I have three files. two files which contains same variables but different values and a third file which executes commands such as cd to a certain place

my third file is called init.sh and looks something like

###############
# works perfectly if I do it this way
source db/abc.sh  # changes variable
source db/commands.sh

source db/def.sh  # changes variable
source db/commands.sh


###############
# does not work
RUN_COMMANDS=$(source db/commands.sh)

source db/1107556.sh  # changes variable
${RUN_COMMANDS}  # this does not run

source db/kunyuan.sh # changes variable
${RUN_COMMANDS}  # this runs

Reason I am doing this is so I can change the variables then run the same commands since the commands are the same but just different values for the variable.

Hopefully I am not doing anything stupid to think this way of using bash.

Thanks in advance for any help or suggestions.

ruakh · Accepted Answer · 2019-03-22 00:15:31Z

3

This:

RUN_COMMANDS=$(source db/commands.sh)

means "run the command source db/commands.sh, and store its output (= everything it printed to standard output) in the variable RUN_COMMANDS".

Instead, you seem to want:

RUN_COMMANDS=(source db/commands.sh)     # note -- no '$'

source db/1107556.sh
"${RUN_COMMANDS[@]}"

source db/kunyuan.sh
"${RUN_COMMANDS[@]}"

which sets RUN_COMMANDS to an array containing source and db/commands.sh, and then runs that array as a command (which is equivalent to running the command source db/commands.sh).

That said, you might want to consider writing a shell function:

function run-db-commands() {
    source "$1" && source db/commands.sh
}

run-db-commands db/1107556.sh
run-db-commands db/kunyuan.sh

answered Mar 22, 2019 at 0:15

ruakh

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1 Comment

Tsuna Over a year ago

ah! got it, and really thanks for the suggestions, the shell function you suggested would be way better and cleaner to look at too. Thanks a lot~!

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