I have bash variables defined in some file:
VAR1=....
VAR2="some value"
VAR3=....
....
How can I change the value of some variable and add one more variable in the specific line? I need to do it in the single shell script.
EDIT:
Expected output is:
VAR1=....
VAR2="another value"
VAR3=....
NEW_VAR=....
....
I believe that the following code would achieve the results you want, if I understood your question correctly:
#!/bin/bash
#change the value of a certain variable
sed -i -e 's/^VAR2=.*$/VAR2="another-value"/gi' /folder/file
#add a new variable to the variable declaration area, using one existing variable as a reference point
sed -i -e '/^VAR2=/i \NEW_VAR="another-value"' /folder/file
exit 0
That would substitute the value of a variable and add a new variable to the list using one existing variable as the reference point as to where to insert the new variable. This code would work in many Bash versions.
Related
in bash what does the -n parameter in local -n var... mean? - how does it differ from local var...
I can't find a good example / explanation for it. There are no man pages for keywords (it seems?). The closest I have found is a comment here: local: -n: invalid option - which suggests something about not using ! param expansion
Parameters to local are unfortunately not documented in help local, but in help declare:
`-n` make NAME a reference to the variable named by its value
How does it work?
#! /bin/bash
f () {
local -n x=y
y=12
x=42
echo $x $y # 42 42
}
f
You can achieve similar behaviour with the ! indirection (that's what the comment in the linked question means):
#! /bin/bash
f () {
x=y
y=12
echo ${!x} # 12
}
f
-n declares the variable to be a nameref:
A variable can be assigned the nameref attribute using the -n option to the declare or local builtin commands (see Bash Builtins) to create a nameref, or a reference to another variable. This allows variables to be manipulated indirectly. Whenever the nameref variable is referenced, assigned to, unset, or has its attributes modified (other than using or changing the nameref attribute itself), the operation is actually performed on the variable specified by the nameref variable’s value. A nameref is commonly used within shell functions to refer to a variable whose name is passed as an argument to the function. For instance, if a variable name is passed to a shell function as its first argument, running
declare -n ref=$1
inside the function creates a nameref variable ref whose value is the variable name passed as the first argument. References and assignments to ref, and changes to its attributes, are treated as references, assignments, and attribute modifications to the variable whose name was passed as $1.
It's worth noting that namerefs and -n were added in Bash 5 (January 2019).
I have a text file, called texto.txt in Documentos folder, with some values like the ones below:
cat ~/Documentos/texto.txt
65f8: Testado
a4a1: Testado 2
So I want to change a whole line by using a customized function which gets as parameters the new value.
The new value will always keep the first 6 characters, changing only what comes after them. Although I am testing only the first four.
Then I edited my .bashrc including my function like shown below.
muda()
{
export BUSCA="$(echo $* | cut -c 1-4)";
sed -i "/^$BUSCA/s/.*/$*/" ~/Documentos/texto.txt ;}
When I run the command below it works like a charm, but I feel it could be improved.
muda a4a1: Testado 3
Result:
cat ~/Documentos/texto.txt
65f8: Testado
a4a1: Testado 3
Is there a smarter way to do this? Maybe by getting rid of BUSCA variable?
I'd write:
muda() {
local new_line="$*"
local key=${newline:0:4}
sed -i "s/^${key//\//\\/}.*/${new_line//\//\\/}/" ~/Documentos/texto.txt
}
Notes:
using local variables, not exported environment variables
does not call out to cut, bash can extract a substring
escaping any slashes in the variable values so the sed code is not broken.
$var='system1'
data=C:/data/$var/current_extract/*
Output should be
data=C:/data/system1/current_extract/*"
but i still see the result C:/data/$var/current_extract/* $var value is
**system1** not showing the path
Remove the dollar in the assignment
var='system1'
For setting a variable in Linux variable_name=value. For displaying it you have to prefix the variable name with $ symbol. for example echo $variable_name
var='system1'
data=C:/data/$var/current_extract/*
I am trying to set up some variables using indirect expansion. According to the documentation I've read, the set up should be simple:
var1=qa
qa_num=12345
varname="${var1}_ci"
echo ${!varname}
I should be getting "12345". Instead, the output is "varname". If I remove the exclamation point, I end up with "qa_ci", not "12345"
This should be a relatively simple solution, so I'm not sure what I'm missing, if anything.
Your code defines qa_num, but the varname assignment references qa_ci. As a result, your echo was expanding nonexistent qa_ci, giving empty results. Changing the varname assignment fixes the problem on my system.
Example: foo.sh:
#!/bin/bash
var1=qa
qa_num=12345
varname="${var1}_num" # <=== not _ci
echo "${!varname}" # I also added "" here as a general good practice
Output:
$ bash foo.sh
12345
I am new to shell scripting just started off.
I have written this script
#!/bin/sh
profile_type= cat /www/data/profile.conf
echo $profile_type
the o/p of this script is
. /tmp/S_panicA1.txt
. /tmp/S_panicA0.txt
away_Def="panicA1 panicA0"
away_Byp=0
away_Sts=$((panicA1+panicA0-away_Byp))
In this i want to get panicA1 panicA0 and 0 and store it in other variable how to do this?
When you want to assign the output of a command to a variable, you use the dollar parenthesis syntax.
foo=$(cat /my/file)
You can also use the backticks syntax.
foo=`cat /my/file`
In your script, you simply run the command cat and assign its result, nothing, to your variable. Hence the output consisting of the content of your file, result of cat, followed by an empty line, result of echo with an empty variable.