not able to evaluate variable

mrajdeep · 09-26-2013, 10:56 AM

serv1=`hostname`
a=1

when I use following

echo $'serv'"$a"

I get

serv1

and not the value of $serv1

please help.

suicidaleggroll · 09-26-2013, 10:56 AM

You'll need to use eval if you want to dynamically build the variable name

edit: or this

Code:

serv1=`hostname`
a=1
var=serv$a
echo ${!var}

mrajdeep · 09-26-2013, 11:11 AM

i tried this...but I get output as 1

Pap · 09-26-2013, 11:13 AM

I don't see why using quotes here. This should work:

Code:

$ serv1='hostname'
$ a=1
$ echo $serv1$a
hostname1

What shell are you using? (find out by typing

Code:

ps -p $$

You should see something like

Code:

  PID TTY          TIME CMD
23811 pts/0    00:00:00 bash

The important part is "bash", or whatever it says there in your system.

Pap · 09-26-2013, 11:19 AM

Oh wait, I get it now... You want to evaluate the hostname command. Then use this:

Code:

$ serv=`hostname`
$ a=1
$ echo $(hostname)$a
debian1

mrajdeep · 09-26-2013, 11:22 AM

its KSH

PID TTY TIME CMD
26986 pts/4 00:00:00 ksh

echo $serv1$a gives me output as hostname suffixed by 1

but i want to evaluate $serv1 dynamically..

mrajdeep · 09-26-2013, 11:26 AM

Quote:

Originally Posted by Pap

Oh wait, I get it now... You want to evaluate the hostname command. Then use this:

Code:

$ serv=`hostname`
$ a=1
$ echo $(hostname)$a
debian1

there are different hostname i have stored as serv1, serv2, serv3 and so on

now i want to evaluate hostname based on choice made through case statement

something like if some one chose 1 then serv1 will be considered..

colucix · 09-26-2013, 11:28 AM

Code:

serv1=`hostname`
a=1
var=serv$a
eval echo \$$var

mrajdeep · 09-26-2013, 11:31 AM

Quote:

Originally Posted by colucix

Code:

serv1=`hostname`
a=1
var=serv$a
eval echo \$$var

Thank you very much. This works!

I'd really appreciate if you explai what happens here.

Thanks again.

colucix · 09-26-2013, 11:37 AM

Quote:

Originally Posted by mrajdeep

I'd really appreciate if you explai what happens here.

eval evaluates a command line after all the substitutions are made. In this case $var is substituted by its current value, while the escaped $ is protected by the substitution. At the end the command results in

Code:

echo $serv1

and this one is evaluated. There are some caveats about the usage of eval, anyway. For the sake of curiosity, You can read about them here: http://mywiki.wooledge.org/BashFAQ/048

suicidaleggroll · 09-26-2013, 11:45 AM

Quote:

Originally Posted by mrajdeep

i tried this...but I get output as 1

Probably because you're using ksh, I believe ${!var} is bash-specific. When writing scripts, you should always add the "#!/bin/xxxx" at the beginning to control which shell is used to interpret the script, with that set to "#!/bin/bash" what I posted should work.

Code:

$ cat script
#!/bin/bash

serv1=`hostname`
a=1
var=serv$a
echo ${!var}

$ ./script
fry

mrajdeep · 09-26-2013, 01:01 PM

Thank you very much colucix.

Thank you very much suicidaleggroll for providing such useful information.

jpollard · 09-26-2013, 05:57 PM

Actually, I think he is trying to use an array. If so, look at the bash manpage for "arrays".