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use a variable (like $i) as part of another variables name in bash

Posted on 2013-06-20
5
235 Views
Last Modified: 2013-06-23
I have a file with variables like this...
server_1=
port_1=
user_1=

server_2=
port_2=
user_2=

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I am trying to write a script that can read them in and manipulate them and write them back out.  And it should be able to handle any number of them.  Below is a simplified version of what I want to be able to do.  (i.e. not actually trying to edit anything here.  Just write it back out)

Is this at all possible?


source filewithvariables
for (( i=1; i<=$number_of_servers; i++ ))
do

echo "server_${i}=${server_${i}}" >> output 
echo "port_${i}=${port_${i}}" >> output 
echo "user_${i}=${user_${i}}" >> output 

done

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Question by:Xetroximyn
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5 Comments
 
LVL 84

Accepted Solution

by:
ozo earned 500 total points
ID: 39263887
eval echo "server_${i}=\${server_${i}}" >> output
or
name=server_${i}; echo "server_${i}=${!name}" >> output
0
 
LVL 84

Expert Comment

by:ozo
ID: 39263920
or
for name in ${!server_*}  ; do  echo "$name=${!name}"  ; done
0
 

Author Comment

by:Xetroximyn
ID: 39266692
Thanks!

Just want to understand this a bit better... I think I understand the val method.  Baiscally eval replaces the ${i} with the "1" and leaves the first $ in there because it's escaped and thus bash is left to execute
server_1=${server_1}

And I think I understand the indirect referencing with the ${!name} syntax.  It seems like it takes the contents of $name and uses that for the variable name to get the contents of instead of just getting the contents of ${!name}

However - if it worked like I think it does - I don't quite understand why these two things would not be the same

echo $name_1
i=1
echo $name_${!i}

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any chance you can help me understand that?
0
 
LVL 84

Expert Comment

by:ozo
ID: 39267231
i=1
echo $name_${!i}
would have the same effect as
echo $name_$1
0
 

Author Closing Comment

by:Xetroximyn
ID: 39269814
TY!
0

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