CODEWITHSUNDEEP
bashshellyaml
User19792255
User19792255

Reputation: 111

is there any way to get yaml values dynamically in shell script

I have a yaml file like follows,

dev:
  host_place: "some place"
qa:
  host_place: "another place"

In a shell script I am trying to get that value. If I hard corded dev or qa those values are showing. But if I get it as a input parameter it does not showing anything. My shell script like below

#! /bin/bash
. yaml_par.sh
eval $(yaml_par config.yml "con_")
host=$con_dev_host_place
echo $host

If I run this it will work without any issue

but if I do below modification it won't work.

#! /bin/bash
. yaml_par.sh
envo=$1
eval $(yaml_par config.yml "con_")
host=$con_$envo_host_place
echo $host

What could be the reason?

Upvotes: 0

Views: 1298

Answers (2)

jijinp
jijinp

Reputation: 2662

For bash: If a value in variable is used in an another variable name use eval to get the value:

eval host=\$con_${envo}_host_place

Upvotes: 0

Etan Reisner
Etan Reisner

Reputation: 80921

Your second example doesn't expand the way you think it does.

Think about it. You are expecting the shell to look at $con_$envo_host_place and realize that you meant for it to expand $envo first to get $con_dev_host_place and then expand that to get the value.

How is the shell supposed to know that?

You might think you can do ${con_${envo}_host_place} to force the issue but that doesn't work either. You just get a "bad substitution" error from the shell.

You can have indirect shell variable expansion however. This is covered in Bash FAQ 006.

# Bash
realvariable=contents
ref=realvariable
echo "${!ref}"   # prints the contents of the real variable

# ksh93 / mksh / Bash 4.3
realvariable=contents
typeset -n ref=realvariable
echo "${!ref} = $ref"      # prints the name and contents of the real variable

# zsh
realvariable=contents
ref=realvariable
echo ${(P)ref}   # prints the contents of the real variable

Upvotes: 1

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