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Old 11-05-2015, 05:01 PM   #1
dr.x
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Registered: Jan 2013
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NEED small help with $ in the bash file


Hi Guys ,

i have code as below :
IP1=69.46.87.106
IP2=69.46.87.107
IP3=69.46.87.108
IP4=69.46.87.109
IP5=69.46.87.110
################################
D=$(cat /dev/urandom | tr -dc '1-5' | fold -w 1 | head -n 1)
=====================
D is from 1-5

as u see above i need to echo the output randomly

say i want to put
echo $IP$D

i want to be able to have IP1 , IP2 , IP3

as u see above
$IP$D is not working with

is there a method to identify variable with many $ like fourmula

echo $IP$D

??

what options do i have to fix my issue

cheers
 
Old 11-05-2015, 05:41 PM   #2
berndbausch
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Registered: Nov 2013
Location: Tokyo
Distribution: Mostly Ubuntu and Centos
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There is no variable named IP, thus $IP yields the empty string.

Your can achieve what you want (looks like a load balancer written in Bash ) with an array. For example:
Code:
IP=(69.46.87.106 69.46.87.107 69.46.87.108 69.46.87.109 69.46.87.110)
echo ${IP[$D]}
See also the array section in the bash reference manual.

EDIT: Array indexes start at 0, not 1. You would have to tweak your code so that D ranges from 0 to 4. Or insert a dummy address at the beginning of the array (which would be an ugly hack).

Last edited by berndbausch; 11-05-2015 at 07:30 PM.
 
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Old 11-05-2015, 07:16 PM   #3
rknichols
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For the simple case you described, an array is certainly the easiest way to go.

For the more general case, you can build up a variable name as a string by any means you like, and then use that string indirectly as a variable name:
Code:
VarName=IP$D
echo ${!VarName}
That "!" ahead of VarName says to treat that string as the name of the variable to be referenced.
 
2 members found this post helpful.
  


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