I have a variable with multiple values seperated by ':' , how can I get those fetched seperately to receive values of abc, cde story, bjd in a bash script?
abc: 10
cde story: 123abc
bjd: I have some values
I am a little confused about your question. Are these variables stored in a separate file, like 'vars.txt'? It would be very helpful for you to provide some more context.
That said, check out https://linuxhint.com/bash_split_examples/ for some examples of how to split a string in bash. The TL;DR is this:
If you have a bash variable:
text="Hello:World"
and you want to split it by a delimiter, ":", you must use the IFS variable.
IFS=":"
This seems to be a special bash variable that affects how the read command works.
So for you, it might look something like this:
some_var="10:123abc:i have some values"
IFS=":"
read -a var_array <<< "$some_var"
abc=${var_array[0]}
cde_story=${var_array[1]}
bjd=${var_array[2]}
echo "$abc, $cde_story, $bjd"
The output of which is:
10, 123abc, i have some values
Related
This is my script code
#!/bin/bash
timestamp=$(date +%F-%T)
clinet_id="123"
STRING=s3://<bucketname>/folder/$client_id/$client_id_gdpr_access_report_$timestamp.csv
echo "$STRING"
$SHELL
If i run this code am getting timestamp value.csv file
how can i concatenate variable with string.
am expecting out put like below
s3://<bucketname>/folder/123/123_report_2022-01-25-14:55:47.csv
i can able to concatenateaccess_report_$timestamp.csv
if i add $client_id_ in the beginning, it will print
2022-01-25-14:55:47.csv
Expecting a better advice
You need to look better at the names of your variables; it's 'client_id' not 'clinet_id' ...
And you should take care of double quoting your string, and put braces around variables when in doubt:
STRING="s3://<bucketname>/folder/${client_id}/${client_id}_gdpr_access_report_${timestamp}.csv"
Working on a bash script. I'm reading a line from a properties file using grep and cut and have fetched some value in a variable role_portions something like this role_portions=role_1:10,role_2:25,role_3:75,role_4:50,role_5:75,role_6:25,role_7:50
Now, I get a few roles as csv input parameter in my bash script and I would want to change those roles values to 0.
For example, when I run modify_script.sh role_2,role_4,role_7, after reading the above value from the file, the script should provide as output role_1:10,role_2:0,role_3:75,role_4:0,role_5:75,role_6:25,role_7:0. Can someone help with this?
When the role names are without special characters (like & and /) you can use sed.
for role in role_2 role_4 role_7; do
role_portions=$(sed -r "s/(^|,)(${role}):[^,]*/\1\2:0/" <<< "${role_portions}")
done
When you are already using grep and cut you might be able to combine commands (maybe use awk).
In Shell, I need to parse a string that will consistently have three components divided by periods. The length will not be consistent.
Example given values:
1.0.5
11.5.13
0.12.0
I need to parse them into three variables. For a given value 1.0.5, I need:
major=1
minor=0
fix=5
I have seen a good example that apparently only works in bash, which I don't have bash as an option.
You've got a few options:
Use a heredoc
IFS=. read -r major minor fix <<EOF
$version
EOF
Change input field separator, assign to positional parameters using set
old_IFS=$IFS
IFS=.
set -- $version
major=$1
minor=$2
fix=$3
IFS=$old_IFS
I prefer the first option, since it's shorter and doesn't affect the positional parameters.
Hi I would like to set a variable by concatenating two other variables.
Example
A=1
B=2
12=C
echo $A$B
desired result being C
however the answer I get is always 12
Is it possible?
UPDATED
Example
A=X
B=Y
D=$A$B
xy=test
echo $D
desired result being "test"
It looks like you want indirect variable references.
BASH allows you to expand a parameter indirectly -- that is, one variable may contain the name of another variable:
# Bash
realvariable=contents
ref=realvariable
echo "${!ref}" # prints the contents of the real variable
But as Pieter21 indicates in his comment 12 is not a valid variable name.
Since 12 is not a valid variable name, here's an example with string variables:
> a='hello'
> b='world'
> declare my_$a_$b='my string'
> echo $my_hello_world
my string
What you are trying to do is (almost) called indirection: http://wiki.bash-hackers.org/syntax/pe#indirection
...I did some quick tests, but it does not seem logical to do this without a third variable - you cannot do concatenated indirection directly as the variables/parts being concatenated do not evaluate to the result on their own - you would have to do another evaluation. I think concatenating them first might be the easiest. That said, there is a chance you could rethink what you're doing.
Oh, and you cannot use numbers (alone or as the starting character) for variable names.
Here we go:
cake="cheese"
var1="ca"
var2="ke"
# this does not work as the indirection sees "ca" and "ke", not "cake". No output.
echo ${!var1}${!var2}
# there might be some other ways of tricking it to do this, but they don't look to sensible as indirection probably needs to work on a real variable.
# ...this works, though:
var3=${var1}${var2}
echo ${!var3}
I want to get an input from the user using read
read line
and the proper input would be a string and then a number:
a 5
b 6
+ 87
How do you separate the "a" and 5 into two variables, with the 5 into a integer variable?
read supports the command line option -a, to store the input in an array like that:
$ read -a line
a 4
$ echo ${line[0]}
a
$ echo ${line[1]}
4
That would be nicer than using two variables, in my opinion.
I suggest reading the documenentation of read to get you started:
help read
Here are the first two paragraphs:
Read a line from the standard input and split it into fields.
Reads a single line from the standard input, or from file descriptor FD
if the -u option is supplied. The line is split into fields as with word
splitting, and the first word is assigned to the first NAME, the second
word to the second NAME, and so on, with any leftover words assigned to
the last NAME. Only the characters found in $IFS are recognized as word
delimiters.
Note that bash doesn't have a notion of "integer variables" comparable to other programming languages. Bash variables are untyped. Declaring a variable as integer using declare -i only influences assignments to this variable -- everything that is not a valid integer is silently set to 0.
I presume you're operating in the shell since this post is tagged "bash" but you might want to make that explicit.
Anyway, the "read" command to the shell takes multiple variable names, not just one. You can give it two and it will hand each word on the line to you in the respective variables. (They're split on the field separator given by the IFS variable.)
The shell doesn't really have any distinction between "integer" variables and any other kind in the general case.
I suggest reading the man page for the shell fully if you really want to understand how to write shell scripts properly.
The read command will split your input along whatever is in $IFS. This, by default is whitespace, so simply doing this:
read my_string my_number
will split your input into two sections separated by the space. Sometimes, you'll see this:
read my_string my_number garbage
Because read will read in the entire rest of the line into the last variable no matter how many parameters you had. For example, if I had:
read my_string my_number
And the user put in:
this 1 foo foo foo!
$my_string will be this, but $my_number will be 1 foo foo foo!
By putting another variable (called garbage in this case), I eliminate this issue. If I put:
read my_string my_number garbage
And the user put in:
this 1 foo foo foo!
$my_string will be this, $my_number will be 1, and $garbage would be foo foo foo!.
A simple test program:
while read my_string my_number garbage
do
echo "The string is '$my_string'. The number is '$my_number'."
echo "Note there's no error checking of input."
echo "That's your job, Bunky."
done