running bash command within script

Question

i have to run the following command within a script in bash and store its output in a variable. the db_name is another variable that i want to substitute too. Notice the name-db and true, i don't want my shell to start running them.

aws rds describe-db-snapshots --db-instance-identifier ${db_name} --query 'DBSnapshots[?contains(DBSnapshotIdentifier, `name-db`) == `true`]'.DBSnapshotIdentifier --output text | sort -k8 | tail -n1 | gawk '{print $4}'

I started by storing the whole command in a string and then run the string using eval or the string directly but it fails everytime. I guess it keeps expanding the true and the name-db bit. Any help ?

can you try resp=$( aws ..... ); echo $resp ? This will actually open a sub-shell and resp will contain the output. Edit: this is kinda the same as the answer from @ruakh. — tgo
– tgo, Commented Oct 12, 2015 at 15:24
Storing commands in strings does not work. See Bash FAQ 050. A function, as in ruakh's answer, is the correct solution. — Etan Reisner
– Etan Reisner, Commented Oct 12, 2015 at 16:49

ruakh · Accepted Answer · 2015-10-12 15:24:09Z

2

Rather than storing the command in a string, you're better off storing it in a shell function:

function get_latest_foo () {
    local db_name="$1"
    aws rds describe-db-snapshots \
      --db-instance-identifier "$db_name" \
      --query 'DBSnapshots[?contains(DBSnapshotIdentifier, `name-db`) == `true`].DBSnapshotIdentifier' \
      --output text \
    | sort -k8 \
    | tail -n1 \
    | gawk '{print $4}'
}

latest_foo="$(get_latest_foo "$db_name")"

(Note: I used foo in the function-name because I couldn't tell which field $4 is. You'll want to change the name to something more meaningful.)

answered Oct 12, 2015 at 15:24

ruakh

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