referencing a ${variable} value inside a sed substitution

Question

I am trying to replace a string within a couple of similar sql scripts. in my toolbox I have a for loop as well as grep and sed inside a bash script.

#!/usr/bin/env bash
for efa_instance in efa_bauen_bb efa_bauen_bw
do
        echo ${efa_instance}
        secret=$(grep ${efa_instance}_user efa_secret.csv | cut -d; -f3)
        echo ${secret}
        echo replacing password for ${efa_instance}
        sed -i 's/<via Mail>/${secret}/' ${efa_instance}.service.sql
done

this, however replaces the string <via Mail> with literally the string ${secret} instead of the value stored in the ${secret} variable.

the components of this work as expected, just trying to use sed to actually replace the placeholder fails.

Am i ot seing something obvious or is my approach flawed by design.

maybe worth mentioning (just if somebody was tempted to recommend that) ... I have never used awk and have no knowledge ot if whatsoever

Note that you need to be careful if the variable contains slashes or backslashes (at least) see How to ensure that string interpolated into sed substitution escapes all metachars — ilkkachu, Nov 14 '22 at 16:34

score 2 · Accepted Answer · answered Nov 14 '22 at 15:42

2

Use " instead of ':

sed -i "s/<via Mail>/${secret}/" ${efa_instance}.service.sql

answered Nov 14 '22 at 15:42

jonasmike

169

great thx. I wasn't aware I was actually that close :-) – vrms Nov 14 '22 at 15:45

referencing a ${variable} value inside a sed substitution

1 Answers1