Shell script variables to returning values

Question

I have a config file something like below.

_ispip=$_octet.129
_octet=10.89.2
_rxpip=$_octet.132

And when i try to echo the value its not printing the full values for the 1st variable. Is there a easy way to fix this?

# source test.cfg
# echo $_ispip
.129

# echo $_octet
10.89.2

# echo $_rxpip
10.89.2.132

Barmar · Accepted Answer · 2019-11-19 21:55:02Z

1

bash doesn't have lazy evaluation, it will try to replace $_octet when you refer to it. If you do this before the assignment, you get an empty string.

You need to put the _octet assignment before _ispip.

_octet=10.89.2
_ispip=$_octet.129
_rxpip=$_octet.132

answered Nov 19, 2019 at 21:55

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Sandosh Kumar P Over a year ago

Thanks for your response. My configuration file as 1000+ lines. Using the source command its loading the variables to memory and using it when needed. So rearranging the values is kind a tough job here. Any other ways you recommend?

Benjamin W. Over a year ago

@SandoshKumarP Automate rearranging the values ;)

Barmar Over a year ago

@SandoshKumarP How many of the variables are referenced in other variables? You only need to rearrange those variables, put them all at the top.

Collectives™ on Stack Overflow

Shell script variables to returning values

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related