How do I use shell variables in an awk script?

BashShellAwk

Bash Problem Overview


I found some ways to pass external shell variables to an awk script, but I'm confused about ' and ".

First, I tried with a shell script:

$ v=123test
$ echo $v
123test
$ echo "$v"
123test

Then tried awk:

$ awk 'BEGIN{print "'$v'"}'
$ 123test
$ awk 'BEGIN{print '"$v"'}'
$ 123

Why is the difference?

Lastly I tried this:

$ awk 'BEGIN{print " '$v' "}'
$  123test
$ awk 'BEGIN{print ' "$v" '}'
awk: cmd. line:1: BEGIN{print
awk: cmd. line:1:             ^ unexpected newline or end of string 

I'm confused about this.

Bash Solutions


Solution 1 - Bash

> #Getting shell variables into awk may be done in several ways. Some are better than others. This should cover most of them. If you have a comment, please leave below.                                                                                    v1.5


Using -v (The best way, most portable)

Use the -v option: (P.S. use a space after -v or it will be less portable. E.g., awk -v var= not awk -vvar=)

variable="line one\nline two"
awk -v var="$variable" 'BEGIN {print var}'
line one
line two

This should be compatible with most awk, and the variable is available in the BEGIN block as well:

If you have multiple variables:

awk -v a="$var1" -v b="$var2" 'BEGIN {print a,b}'

Warning. As Ed Morton writes, escape sequences will be interpreted so \t becomes a real tab and not \t if that is what you search for. Can be solved by using ENVIRON[] or access it via ARGV[]

PS If you like three vertical bar as separator |||, it can't be escaped, so use -F"[|][|][|]"

> Example on getting data from a program/function inn to awk (here date is used)

awk -v time="$(date +"%F %H:%M" -d '-1 minute')" 'BEGIN {print time}'

> Example of testing the contents of a shell variable as a regexp:

awk -v var="$variable" '$0 ~ var{print "found it"}'

Variable after code block

Here we get the variable after the awk code. This will work fine as long as you do not need the variable in the BEGIN block:

variable="line one\nline two"
echo "input data" | awk '{print var}' var="${variable}"
or
awk '{print var}' var="${variable}" file
  • Adding multiple variables:

awk '{print a,b,$0}' a="$var1" b="$var2" file

  • In this way we can also set different Field Separator FS for each file.

awk 'some code' FS=',' file1.txt FS=';' file2.ext

  • Variable after the code block will not work for the BEGIN block:

echo "input data" | awk 'BEGIN {print var}' var="${variable}"


Here-string

Variable can also be added to awk using a here-string from shells that support them (including Bash):

awk '{print $0}' <<< "$variable"
test

This is the same as:

printf '%s' "$variable" | awk '{print $0}'

P.S. this treats the variable as a file input.


ENVIRON input

As TrueY writes, you can use the ENVIRON to print Environment Variables. Setting a variable before running AWK, you can print it out like this:

X=MyVar
awk 'BEGIN{print ENVIRON["X"],ENVIRON["SHELL"]}'
MyVar /bin/bash

ARGV input

As Steven Penny writes, you can use ARGV to get the data into awk:

v="my data"
awk 'BEGIN {print ARGV[1]}' "$v"
my data

To get the data into the code itself, not just the BEGIN:

v="my data"
echo "test" | awk 'BEGIN{var=ARGV[1];ARGV[1]=""} {print var, $0}' "$v"
my data test

Variable within the code: USE WITH CAUTION

You can use a variable within the awk code, but it's messy and hard to read, and as Charles Duffy points out, this version may also be a victim of code injection. If someone adds bad stuff to the variable, it will be executed as part of the awk code.

This works by extracting the variable within the code, so it becomes a part of it.

If you want to make an awk that changes dynamically with use of variables, you can do it this way, but DO NOT use it for normal variables.

variable="line one\nline two"
awk 'BEGIN {print "'"$variable"'"}'
line one
line two

Here is an example of code injection:

variable='line one\nline two" ; for (i=1;i<=1000;++i) print i"'
awk 'BEGIN {print "'"$variable"'"}'
line one
line two
1
2
3
.
.
1000

You can add lots of commands to awk this way. Even make it crash with non valid commands.

One valid use of this approach, though, is when you want to pass a symbol to awk to be applied to some input, e.g. a simple calculator:

$ calc() { awk -v x="$1" -v z="$3" 'BEGIN{ print x '"$2"' z }'; }

$ calc 2.7 '+' 3.4
6.1

$ calc 2.7 '*' 3.4
9.18

There is no way to do that using an awk variable populated with the value of a shell variable, you NEED the shell variable to expand to become part of the text of the awk script before awk interprets it.


Extra info:

Use of double quote

It's always good to double quote variable "$variable"
If not, multiple lines will be added as a long single line.

Example:

var="Line one
This is line two"

echo $var
Line one This is line two

echo "$var"
Line one
This is line two

Other errors you can get without double quote:

variable="line one\nline two"
awk -v var=$variable 'BEGIN {print var}'
awk: cmd. line:1: one\nline
awk: cmd. line:1:    ^ backslash not last character on line
awk: cmd. line:1: one\nline
awk: cmd. line:1:    ^ syntax error

And with single quote, it does not expand the value of the variable:

awk -v var='$variable' 'BEGIN {print var}'
$variable
More info about AWK and variables

Read this faq.

Solution 2 - Bash

It seems that the good-old ENVIRON [tag:awk] built-in hash is not mentioned at all. An example of its usage:

$ X=Solaris awk 'BEGIN{print ENVIRON["X"], ENVIRON["TERM"]}'
Solaris rxvt

Solution 3 - Bash

Use either of these depending how you want backslashes in the shell variables handled (avar is an awk variable, svar is a shell variable):

awk -v avar="$svar" '... avar ...' file
awk 'BEGIN{avar=ARGV[1];ARGV[1]=""}... avar ...' "$svar" file

See http://cfajohnson.com/shell/cus-faq-2.html#Q24 for details and other options. The first method above is almost always your best option and has the most obvious semantics.

Solution 4 - Bash

You could pass in the command-line option -v with a variable name (v) and a value (=) of the environment variable ("${v}"):

% awk -vv="${v}" 'BEGIN { print v }'
123test

Or to make it clearer (with far fewer vs):

% environment_variable=123test
% awk -vawk_variable="${environment_variable}" 'BEGIN { print awk_variable }'
123test

Solution 5 - Bash

You can utilize ARGV:

v=123test
awk 'BEGIN {print ARGV[1]}' "$v"

Note that if you are going to continue into the body, you will need to adjust ARGC:

awk 'BEGIN {ARGC--} {print ARGV[2], $0}' file "$v"

Solution 6 - Bash

I just changed @Jotne's answer for "for loop".

for i in `seq 11 20`; do host myserver-$i | awk -v i="$i" '{print "myserver-"i" " $4}'; done

Solution 7 - Bash

I had to insert date at the beginning of the lines of a log file and it's done like below:

DATE=$(date +"%Y-%m-%d")
awk '{ print "'"$DATE"'", $0; }' /path_to_log_file/log_file.log

It can be redirect to another file to save

Solution 8 - Bash

Pro Tip

It could come handy to create a function that handles this so you dont have to type everything every time. Using the selected solution we get...

awk_switch_columns() {
     cat < /dev/stdin | awk -v a="$1" -v b="$2" " { t = \$a; \$a = \$b; \$b = t; print; } "
}

And use it as...

echo 'a b c d' | awk_switch_columns 2 4

Output:
a d c b

Solution 9 - Bash

example:

in.txt:

foo
bar

variable:

var=$(awk '{print $1}' in.txt) 

command:

echo -e "$var" > out.txt

out.txt

foo
bar

another:

in.txt

foo,aaa
bar,bbb

variable:

var=$(awk -F "," '{print $1}' in.txt) 

out.txt

foo
bar

or:

var=$(awk -F "," '{print $2}' in.txt) 

out.txt

aaa
bbb

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionhqjmaView Question on Stackoverflow
Solution 1 - BashJotneView Answer on Stackoverflow
Solution 2 - BashTrueYView Answer on Stackoverflow
Solution 3 - BashEd MortonView Answer on Stackoverflow
Solution 4 - BashjohnsywebView Answer on Stackoverflow
Solution 5 - BashZomboView Answer on Stackoverflow
Solution 6 - BashedibView Answer on Stackoverflow
Solution 7 - BashSinaView Answer on Stackoverflow
Solution 8 - BashLuis LView Answer on Stackoverflow
Solution 9 - BashacgboxView Answer on Stackoverflow