I have list of parameters that take from a property file looks like that:
db_instanceid=i-0c2b12ae02d454018
db_secgrp=sg-8c2efcf3
backend_instanceid=i-0199621ba358d1814
backend_secgrp=sg-5e508221
frontend_instanceid=i-0199621ba358d1814
frontend_secgrp=sg-e152809e
in bash, I want to use an array to perform actions using those parameters,
Something like that:
declare -a arr=("frontend" "backend" "db")
for i in "${arr[#]}"
do
inter=$i
echo "Get PublicIp for $inter server"
echo "$inter security group - $inter_secgrp" ;
done
But $inter_secgrp itself will be read as a parameter,
so I get an empty string (which make sense). How can I read this parameter
the right way so I will get the value of all "secgrp"
needed output:
Get PublicIp for frontendserver
frontendserver security group - sg-e152809e
Get PublicIp for backend
backendsecurity group - sg-5e508221
Get PublicIp for db
dbsecurity group - sg-8c2efcf3
Could do it with indirect expansion
. parameterfile
declare -a arr=("frontend" "backend" "db")
for inter in "${arr[#]}"
do
echo "Get PublicIp for $inter server"
inter_secgrp=${inter}_secgrp
echo "$inter security group - ${!inter_secgrp}" ;
done
The temptation to use eval is particularly strong here.
. parameterfile
declare -a arr=("frontend" "backend" "db")
for inter in "${arr[#]}"
do
echo "Get PublicIp for $inter server"
eval inter_secgrp=\$${inter}_secgrp
echo "$inter security group - $inter_secgrp" ;
done
And then $inter_secgrp is what you want.
To be simple, eval tells Bash to evaluate the line for an extra time before executing it. Here after the first evaluation, the eval line turns to (for example
eval inter_secgrp=$db_secgrp
And then things after eval is evaluated again, so $db_secgrp gets expanded, giving what you want.
Related
I have this idea in mind:
I have this number: CN=20
and a list=( "xa1-" "xa2-" "xb1-" "xb2-")
and this is my script:
for a in "${list[#]}"; do
let "CN=$(($CN+1))"
echo $CN
Output:
21
22
23
24
I am trying to create a loop where it creates the following variables, which will be referenced later in my script:
fxp0_$CN="fxp-$a$CN"
fxp0_21="fxp-xa1-21"
fxp0_22="fxp-xa2-22"
fxp0_23="fxp-xb1-23"
fxp0_24="fxp-xb2-24"
However, I have not been able to find a way to change the variable name within my loop. Instead, I was trying myself and I got this error when trying to change the variable name:
scripts/srx_file_check.sh: line 317: fxp0_21=fxp0-xa2-21: command not found
After playing around I found the solution!
for a in "${list[#]}"; do
let "CN=$(($CN+1))"
fxp_int="fxp0-$a$CN"
eval "fxp0_$CN=${fxp_int}"
done
echo $fxp0_21
echo $fxp0_22
echo $fxp0_23
echo $fxp0_24
echo $fxp0_25
echo $fxp0_26
echo $fxp0_27
echo $fxp0_28
Output:
fxp0-xa1-21
fxp0-xa2-22
fxp0-xb1-23
fxp0-xb2-24
fxp0-xc1-25
fxp0-xc2-26
fxp0-xd1-27
fxp0-xd2-28
One common method for maintaining a dynamically generated set of variables is via arrays.
When the variable names vary in spelling an associative array comes in handy whereby the variable 'name' acts as the array index.
In this case since the only thing changing in the variable names is a number we can use a normal (numerically indexed) array, eg:
CN=20
list=("xa1-" "xa2-" "xb1-" "xb2-")
declare -a fxp0=()
for a in "${list[#]}"
do
(( CN++ ))
fxp0[${CN}]="fxp-${a}${CN}"
done
This generates:
$ declare -p fxp0
declare -a fxp0=([21]="fxp-xa1-21" [22]="fxp-xa2-22" [23]="fxp-xb1-23" [24]="fxp-xb2-24")
$ for i in "${!fxp0[#]}"; do echo "fxp0[$i] = ${fxp0[$i]}"; done
fxp0[21] = fxp-xa1-21
fxp0[22] = fxp-xa2-22
fxp0[23] = fxp-xb1-23
fxp0[24] = fxp-xb2-24
As a general rule can I tell you that it's not a good idea to modify names of variables within loops.
There is, however, a way to do something like that, using the source command, as explained in this URL with some examples. It comes down to the fact that you treat a file as a piece of source code.
Good luck
i have a a couple of variables with a number in its names. e.g
SERVER_IP48_SUBNET
..
SERVER_IP60_SUBNET
And an additional variable
SERVER_IP
Im trying to expand/concatenate them in the following way:
ALLIPs=${SERVER_IP}
for i in {48..64}; do
ALLIPs=${ALLIPs},${SERVER_IP${i}_SUBNET}
done
as you can imagine this script fails saying:
Wrong substitution
Does anybody of you know a good solution for this problem?
Thanks so far
Use a nameref with bash version 4.3 +
ALLIPs=${SERVER_IP}
for i in {48..64}; do
declare -n tmp="SERVER_IP${i}_SUBNET"
ALLIPs+=",$tmp"
done
But you should really be using an array in the first place:
server_ip=0.0.0.0
subnet_ip=(
[48]=1.1.1.1
[49]=2.2.2.2
# ...
[64]=16.16.16.16
)
all_ips=( "$server_ip" )
for i in {48..64}; do
all_ips+=( "${subnet_ip[i]}" )
done
(
IFS=,
echo "ALLIPs = ${all_ips[*]}"
)
Get out of the habit of using ALLCAPS variable names, leave those as
reserved by the shell. One day you'll write PATH=something and then
wonder why
your script is broken.
I just noticed, if you just want a to join the IP addresses with commas, and you're using an array, you don't need a loop at all:
all_ips=$(
IFS=,
set -- "$server_ip" "${subnet_ip[#]}"
echo "$*"
)
You can use ${!varprefix#} or ${!varprefix*} to expand to all variables with that common prefix (the difference is the same as $# and $*):
SERVER_IP48_SUBNET=48sub
SERVER_IP49_SUBNET=49sub
SERVER_IP50_SUBNET=50sub
SERVER_IP=1.2.3.4
# set this as empty since !SERVER_IP# also matches SERVER_IP
ALLIPS=""
for var in "${!SERVER_IP#}"; do
ALLIPS=$ALLIPS,${!var}
done
This would probably be more practical if you could invert the names like this, since we can only match prefixes:
SERVER_IP_SUBNET_48=48sub
SERVER_IP_SUBNET_49=49sub
SERVER_IP_SUBNET_50=50sub
SERVER_IP=1.2.3.4
ALLIPS=$SERVER_IP
for var in "${!SERVER_IP_SUBNET_#}"; do
ALLIPS=$ALLIPS,${!var}
done
More info on this feature in the bash manual.
One idea:
SERVER_IP48_SUBNET=48sub
SERVER_IP49_SUBNET=49sub
SERVER_IP50_SUBNET=50sub
SERVER_IP=1.2.3.4
ALLIPs=${SERVER_IP}
for i in {48..50}
do
tmpvar="SERVER_IP${i}_SUBNET" # build the variable name
ALLIPs="${ALLIPs},${!tmpvar}" # indirect variable reference via tmpvar
done
echo "ALLIPs = $ALLIPs}"
This generates:
ALLIPs = 1.2.3.4,48sub,49sub,50sub
I have a script that is (supposed to be) assigning a dynamic variable name (s1, s2, s3, ...) to a directory path:
savedir() {
declare -i n=1
sn=s$n
while test "${!sn}" != ""; do
n=$n+1
sn=s$n
done
declare $sn=$PWD
echo "SAVED ($sn): ${!sn}"
}
The idea is that the user is in a directory they'd like to recall later on and can save it to a shell variable by typing 'savedir'. It -does- in fact write out the echo statement successfully: if I'm in the directory /home/mrjones and type 'savedir', the script returns:
SAVED (s1): /home/mrjones
...and I can further type:
echo $sn
and the script returns:
s1
...but typing either...
> echo $s1
...or
echo ${!sn}
...both return nothing (empty strings). What I want, in case it's not obvious, is this:
echo $s1
/home/mrjones
Any help is greatly appreciated! [apologies for the formatting...]
To set a variable using a name stored in another variable I use printf -v, in this example:
printf -v "$sn" '%s' "$PWD"
declare here is creating a variable local to the function, which doesn't seem to be what you want. Quoting from help declare:
When used in a function, declare makes NAMEs local, as with the local
command. The -g option suppresses this behavior.
so you can either try the -g or with the printf
Use an array instead.
savedir() {
s+=("$PWD")
echo "SAVED (s[$((${#s[#]}-1))]): ${s[${#s[#]}-1]}"
}
I have a ksh script that reads a profile script with a number of sessions defined. Each session defines its own parameters as such:
SESSION_ONE_USER=...
SESSION_ONE_PWD=...
SESSION_TWO_USER=...
...
The script gets the SESSION parameter from the command line, but I simply set it for the example.
I want to let the SESSION parameter value define part of another parameter name, that I need the value from, like:
SESSION="SESSION_ONE"
USER=${${SESSION}_USER}
PASS=${${SESSION}_PWD}
That gives me a compile error.
I also tried
GET_USER_PARAM(){
echo ${SESSION}_USER
}
echo $`GET_USER_PARAM`
But that returns $SESSION_ONE_USER
I want it to return the value of the parameter SESSION_ONE_USER instead.
Does anyone have any solutions?
This is what eval is for:
SESSION=SESSION_ONE
eval echo \$${SESSION}_USER
should display the value of $SESSION_ONE_USER.
Don't monkey with variable names, use associative arrays instead
typeset -A users
typeset -A pwd
session=SESSION_ONE
users[$session]=joe
pwd[$session]=secret
for key in "${!users[#]}"; do
echo "user for session $key is ${users[$key]}"
echo "pwd for session $key is ${pwd[$key]}"
done
Try this:
SESSION="SESSION_ONE"
SESSION_ONE_USER="foo"
SESSION_ONE_PWD="bar"
SESSION_USER=${SESSION}_USER
SESSION_PWD=${SESSION}_PWD
USER=${!SESSION_USER}
PASS=${!SESSION_PWD}
echo $USER
echo $PASS
The "!" does a level of indirection. See Shell Parameter Expansion.
If this is ksh, then this is a job for nameref
alias nameref='typeset -n'
Example Solution
function session_parameters { set -u
typeset session=${1:?session name}
nameref user=SESSION_${session}_USER
nameref pass=SESSION_${session}_PASS
print session=$session user=$user pass=$pass
}
SESSION_ONE_USER="User1"
SESSION_ONE_PASS="Pass1"
SESSION_TWO_USER="User2"
SESSION_TWO_PASS="Pass2"
for s in ONE TWO THREE; do
session_parameters $s
done
Sample output
session=ONE user=User1 pass=Pass1
session=TWO user=User2 pass=Pass2
test_session_parameters[12]: session_parameters: line 5:
SESSION_THREE_USER: parameter not set
Note the usage of set -u to force the error message on line 3.
nameref usage: (from the builtin help text)
NAME
typeset - declare or display variables with attributes
SYNOPSIS
typeset [ options ] [name[=value]...]
-n
Name reference.
The value is the name of a variable that name references. name cannot contain a ... Cannot be use with any other options.
I want to save the variable name and its contents easily from my script.
Currently :-
LOGFILE=/root/log.txt
TEST=/file/path
echo "TEST : ${TEST}" >> ${LOGFILE}
Desired :-
LOGFILE=/root/log.txt
function save()
{
echo "$1 : $1" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Obviously the above save function just saves TEST : TEST
Want I want it to save is TEST : /file/path
Can this be done? How? Many thanks in advance!
You want to use Variable Indirection. Also, don't use the function keyword, it is not POSIX and also not necessary as long as you have () at the end of your function name.
LOGFILE=/root/log.txt
save()
{
echo "$1 : ${!1}" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Proof of Concept
$ TEST=foo; save(){ echo "$1 : ${!1}"; }; save TEST
TEST : foo
Yes, using indirect expansion:
echo "$1 : ${!1}"
Quoting from Bash reference manual:
The basic form of parameter expansion is ${parameter} [...] 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
Consider using the printenv function. It does exactly what it says on the tin, prints your environment. It can also take parameters
$ printenv
SSH_AGENT_PID=2068
TERM=xterm
SHELL=/bin/bash
LANG=en_US.UTF-8
HISTCONTROL=ignoreboth
...etc
You could do printenv and then grep for any vars you know you have defined and be done in two lines, such as:
$printenv | grep "VARNAME1\|VARNAME2"
VARNAME1=foo
VARNAME2=bar