Need to find a way to locate desired variables based on string combination
#!/bin/bash
DSPDSP="1234"
$A="DSP"
$B="DSP"
PORTLIST=$A$B
echo $PORTLIST
DSPDSP
I hope there is a smart function in bash that concerts a string into variable name
smart_echo $PORTLIST
1234
Please try something like:
smart_echo() {
local varname="$1"
echo "${!varname}"
}
DSPDSP="1234"
A="DSP"
B="DSP"
PORTLIST="$A$B"
smart_echo "$PORTLIST"
=> 1234
If your bash version is 4.3 or newer, you can also say as an alternative:
smart_echo() {
declare -n p="$1"
echo "$p"
}
Thanks, based on your input, I found my working version like this:
A="DSP"
B="DSP"
PORTLIST=$A$B
echo ${!PORTLIST}
=> 1234
Related
Hi I'm trying to make a function which should get a text file and then do some things on it and then echo. But when I try to execute it, it says syntax error near unexpected token `"$cat"'
#!/bin/usr/bash
cat=$(< cat_dialogue.txt)
function test_cat (){
echo $1
}
test_cat($cat)
desired output:
>meow meow
Your program may look like the following. Note all differences. Check your scripts with shellcheck.
#!/usr/bin/env bash
cat=$(< cat_dialogue.txt)
test_cat() {
echo "$1"
}
test_cat "$cat"
Here is an example BASH function that strips a branchname:
#create function
function strip () {
#create local variable that takes input and fills $TEXT
local TEXT=$1
#stips the branch number from the branchname
echo $TEXT | sed 's/-[0-9]*//2'
}
strip "testbranch-12345-28796"
hope it helps :) also check the BASH documentation as mentioned by #joshmeranda
I have this idea in mind:
I have this number: CN=20
and a list=( "xa1-" "xa2-" "xb1-" "xb2-")
and this is my script:
for a in "${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
Is it possible to first pass a parameter to the function and than change its value?
#!/bin/bash
name=old_name
echo $name #echoes "old_name"
alter () {
$1=new_name #throws error that says 'command not found'
}
alter name
echo $name #I would like to see "new_name" here
Yes, using a nameref:
alter () {
declare -n foo=$1
foo=new_name
}
See Bash FAQ 006 for more advice and warnings, as well as workarounds for versions of bash that predate nameref support (i.e., 4.2 or earlier).
You can't do it that way, unfortunately. Also, bash functions can't return values. The usual options are (1) set a global var in the function (yuck), or (2) echo the "return" value and use command substitution to call it. Something like this:
alter () {
echo "$1 but different"
}
name="Fred"
name=$(alter $name)
echo Name is now $name
returns: Name is now Fred but different
I have a script that is (supposed to be) assigning a dynamic variable name (s1, s2, s3, ...) to a directory path:
savedir() {
declare -i n=1
sn=s$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 the following code in my ~/.bashrc:
date=$(which date)
date() {
if [[ $1 == -R || $1 == --rfc-822 ]]; then
# Output RFC-822 compliant date string.
# e.g. Wed, 16 Dec 2009 15:18:11 +0100
$date | sed "s/[^ ][^ ]*$/$($date +%z)/"
else
$date "$#"
fi
}
This works fine, as far as I can tell. Is there a reason to avoid having a variable and a function with the same name?
It's alright apart from being confusing. Besides, they are not the same:
$ date=/bin/ls
$ type date
date is hashed (/bin/date)
$ type $date
/bin/ls is /bin/ls
$ moo=foo
$ type $moo
-bash: type: foo: not found
$ function date() { true; }
$ type date
date is a function
date ()
{
true*emphasized text*
}
$ which true
/bin/true
$ type true
true is a shell builtin
Whenever you type a command, bash looks in three different places to find that command. The priority is as follows:
shell builtins (help)
shell aliases (help alias)
shell functions (help function)
hashed binaries files from $PATH ('leftmost' folders scanned first)
Variables are prefixed with a dollar sign, which makes them different from all of the above. To compare to your example: $date and date are not the same thing. So It's not really possible to have the same name for a variable and a function because they have different "namespaces".
You may find this somewhat confusing, but many scripts define "method variables" at the top of the file. e.g.
SED=/bin/sed
AWK=/usr/bin/awk
GREP/usr/local/gnu/bin/grep
The common thing to do is type the variable names in capitals. This is useful for two purposes (apart from being less confusing):
There is no $PATH
Checking that all "dependencies" are runnable
You can't really check like this:
if [ "`which binary`" ]; then echo it\'s ok to continue.. ;fi
Because which will give you an error if binary has not yet been hashed (found in a path folder).
Since you always have to use $ to dereference a variable in Bash, you're free to use any name you like.
Beware of overriding a global, though.
See also:
http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_02.html
An alternative to using a variable: use bash's command keyword (see the manual or run help command from a prompt):
date() {
case $1 in
-R|--rfc-2822) command date ... ;;
*) command date "$#" ;;
esac
}