How to pass BASH shell variables into AWK statement

Question

I want to pass 2 shell (bash) variables from into an awk statement. In this example the variables are marker1 and marker2

ubuntu@ubuntutest:/tmp$ echo $marker1
###___showhost___###
ubuntu@ubuntutest:/tmp$ echo $marker2
###___showhostset___###
ubuntu@ubuntutest:/tmp$

The script will extract all the lines of text between to different strings/markers ($marker1 and $marker2) from text file - file.in

ubuntu@ubuntutest:/tmp$ cat file.in 
###___showhost___###
 Id Name        Persona -WWN/iSCSI_Name- Port 
115 server5295 VMware  5005638024D3E8E6 1:2:4
116 server5296 VMware  5005638024D3EA86 1:2:4
117 server5297 VMware  5005638024D3F5D2 1:2:4
142 server5302 VMware  5005638024D3E8B6 1:2:4
143 server5303 VMware  5005638024D3E9C6 1:2:4
144 server5304 VMware  5005638024D3F4F2 1:2:4
###___showhostset___###
Id Name              Members    
 0 Mgt_Stack         server5295
 1 Cluster1          server5302
                     server5304
ubuntu@ubuntutest:/tmp$

If I run the awk statement, where I use the actual string values for variables marker1/marker2 ###___showhost___### and ###___showhostset___### respectively instead of using the variables themselves (within awk) I get a positive result. However, I want to put the awk statement in a bash scripts and need to read the variables marker1/marker2.

ubuntu@ubuntutest:/tmp$ awk '/^###___showhost___###/{flag=1;next}/^###___showhostset___###/{flag=0}flag' file.in 
 Id Name        Persona -WWN/iSCSI_Name- Port 
115 server5295 VMware  5005638024D3E8E6 1:2:4
116 server5296 VMware  5005638024D3EA86 1:2:4
117 server5297 VMware  5005638024D3F5D2 1:2:4
142 server5302 VMware  5005638024D3E8B6 1:2:4
143 server5303 VMware  5005638024D3E9C6 1:2:4
144 server5304 VMware  5005638024D3F4F2 1:2:4
ubuntu@ubuntutest:/tmp$

However, when i use the 2 variables $marker1 and $marker2 in the awk statement I get no output.

ubuntu@ubuntutest:/tmp$ awk '/^"'$marker1'"/{flag=1;next}/^"'$marker2'"/{flag=0}flag' file.in 
ubuntu@ubuntutest:/tmp$

Any help greatly appreciated !!!

Reading the man page and/or manual would have yielded this information pretty quickly — grail
– grail, Commented Mar 16, 2017 at 5:25

Akshay Hegde · Accepted Answer · 2017-03-15 19:02:32Z

8

You have to use -v, Getting shell variables into awk may be done in several ways. Some are better than others.

This is the best way to do it (Please note use a space after -v or it will be less portable. E.g., awk -v var= not awk -vvar)

# your shell variables
marker1="###___showhost___###"
marker2="###___showhostset___###"

# and passing them to awk
awk -v market1="$market1" -v market2="$market2" '
       $0 ~ "^"market1{found=1;next}
       $0 ~ "^"market2{found=0}found' file.in

Example:

akshay@db-3325:~$ foo="this is my shell variable"
akshay@db-3325:~$ awk -v myvar="$foo" 'BEGIN{print myvar}'
this is my shell variable

edited Mar 15, 2017 at 19:02

answered Mar 15, 2017 at 15:56

Akshay Hegde

17.1k2 gold badges25 silver badges39 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

vtecdec Over a year ago

Thanks Akshay, thats working now for me. The awk statement differs slightly from the original one, can you explain how it works (to extract the text between variables market1 & market2) now. thanks

Akshay Hegde Over a year ago

@vtecdec: I just changed flag to found for better readability rest all same, $0 means current record/line/row , $0 ~"^"marker1{} means if record starts with value of marker1 then make variable found=1 and go to next line, same way for market2 which makes found=0, and }found whenever found is non zero, then prints current record/row/line

Kaan · Accepted Answer · 2022-10-14 17:08:33Z

Here's the usage, according to awk man page (for awk -version "20200816"):

awk [ -F fs ] [ -v var=value ] [ 'prog' | -f progfile ] [ file ...  ]

Also from the man page, there are two ways of doing assignment. One like this:

Any file of the form var=value is treated as an assignment, not a filename, and is executed at the time it would have been opened if it were a filename.

And the other like this:

The option -v followed by var=value is an assignment to be done before prog is executed; any number of -v options may be present.

The question posted here is addressed by the second assignment option, using -v, which is described well in the answer posted by Akshay. Here's another simple example using -v assignment:

% X=aardvark                    
% echo | awk -v x=$X '{print x}'
aardvark

Collectives™ on Stack Overflow

How to pass BASH shell variables into AWK statement

2 Answers 2

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related