create variables inside unix script

Question

I am trying to create a variable in a script, based on another variable.

I just don't know what needs to be adjusted in my code, if it is possible. I am simplifying my code for your understanding, so this is not the original code.

The code goes like that:

#!/bin/csh -f

set list_names=(Albert Bela Corine David)
set Albert_house_number=1
set Bela_house_number=2
set Corine_house_number=3
set David_house_number=4

foreach name ($list_names)
  #following line does not work....
  set house_number=$($name\_house_number)
  echo $house_number
end

the desired output should be:

Thanks for your help.

May be helpful to read this: Is it possible to build variable names from other variables in bash? — fedorqui
– fedorqui, Commented May 4, 2015 at 10:35
@fedorqui Those all look bash-, or at least POSIX-specific. csh is a completely different Beast (that deserves the capital "B"). — Wintermute
– Wintermute, Commented May 4, 2015 at 11:59

Wintermute · Accepted Answer · 2015-05-04 11:57:27Z

1

Unfortunately, the bashism ${!varname} is not available to us in csh, so we'll have to go the old-fashioned route using backticks and eval. csh's quoting rules are different from those of POSIX-conforming shells, so all of this is csh specific. Here we go:

set house_number = `eval echo \" \$${name}_house_number\"`
echo "$house_number"

${name} is expanded into the backticked command, so this becomes equivalent to, say,

set house_number = `eval echo \" \$Albert_house_number\"`

which then evaluates

echo " $Albert_house_number"

and because of the backticks, the output of that is then assigned to house_number.

The space before \$$ is necessary in case the value of the expanded variable has special meaning to echo (such as -n). We could not simply use echo "-n" (it wouldn't print anything), but echo " -n" is fine.¹

The extra space is stripped by csh when the backtick expression is expanded. This leads us to the remaining caveat: Spaces in variable values are going to be stripped; csh's backticks do that. This means that if Albert_house_number were defined as

set Albert_house_number = "3   4"

house_number would end up with the value 3 4 (with only one space). I don't know a way to prevent this.

¹ Note that in this case, the echo "$house_number" line would have to be amended as well, or it would run echo "-n" and not print anything even though house_number has the correct value.

answered May 4, 2015 at 11:57

Wintermute

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1 Comment

Tom Sasson Over a year ago

Thank you Wintermute, that is a perfect explanation and I learnt a lot!

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