bash: expanding variable

Question

I'm trying to add a function to my .bashrc to ease prepending $PWD to environment variables. I'd like the function to take one argument -- the name of the variable on which to prepend the working directory. I'm thinking something like this...

function prependTo{ export $1=$PWD:\$$1 }

Is what I'm looking to do possible in bash?

SiegeX · Accepted Answer · 2011-03-04 19:41:14Z

3

Don't use the function keyword, it is deprecated and non-POSIX. Instead do this:

 prependTo(){ export $1=$PWD:${!1}; }

Explanation

From man bash

If the first character of parameter is an exclamation point, a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion.

edited Mar 4, 2011 at 19:41

answered Mar 4, 2011 at 19:33

SiegeX

142k25 gold badges150 silver badges159 bronze badges

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

Collectives™ on Stack Overflow

bash: expanding variable

1 Answer 1

Explanation

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Explanation

Comments

Your Answer

Sign up or log in

Post as a guest

Related