Using variable within awk

Shell Programming and Scripting
1

Hi All,

I have a tab-delimited input file in which the 7th column contains the date in the format given below

Aug 29 2013 12:00AM

I have assigned this date value to a variable like

d1="Aug 29 2013 12:00AM"

I want to compare the 7th column in the file with this variable d1 and if it matches, I want the whole line to be copied to a separate file.

I am doing the below but not getting the desired result

cat input_file | awk -F"\t" '{if ($7=="$d1") print $0}'

Can you help?

Thanks

2

Hi,
You can try:

cat input_file | awk -F"\t" '{if ($7=="'$d1'") print $0}'

Regards.

3

it's giving the below error:

 syntax error The source line is 1.
 The error context is
                {if >>>  ($7=="Aug <<<
 awk: The statement cannot be correctly parsed.
 The source line is 1.
        awk: There is a missing } character.
        awk: There is a missing ) character.
4

Use -v for using variable in awk. and avoid use of cat if you are using awk.

d1="Aug 29 2013 12:00AM"

 awk -F"\t" -v D1="$d1" '{if ($7==D1) print $0}' input_file
5

Yep,
I forget to protect variable:

cat input_file | awk -F"\t" '{if ($7=="'"$d1"'") print $0}'

First double quote for awk
simple quote to come back shell for interpreting variable d1
second double quote for protect space in variable value

Regards.

6

You should not add variable to awk this way, do like pamu shows, or like this.
awk -F"\t" '{if ($7==D1) print $0}' D1="$d1" input_file

If yo do not have space, only tab, you can remove the field separator too.
awk '{if ($7==D1) print $0}' D1="$d1" input_file

7

You get several Useless Use of Cat Awards

There are two reliable ways to get a variable into awk, that is neither of them.

# method 1
awk -v VAR=value '{ ... }'

# method 2
awk '{ ... }' VAR=value

Your method is prone to syntax errors, should your variable accidentally contain any quotes or awk code.

8

As a side note:

  1. Some old awk implementations don't support the -v var... syntax:
bash-3.00$ awk -v x=1 '{ print x }' <<< a
awk: syntax error near line 1
awk: bailing out near line 1
bash-3.00$ awk '{ print x }' x=1  <<< a
1
bash-3.00$ type awk; uname -sr
awk is hashed (/usr/bin/awk)
SunOS 5.10

You should use either nawk or the X/Open version of awk (/usr/xpg4/bin/awk) on Solaris anyway.

  1. When you're using the syntax below, the value(s) is/are not available in the BEGIN block:
bash-3.00$ awk 'BEGIN { print x }' x=1 <<< a

bash-3.00$ /usr/xpg4/bin/awk 'BEGIN { print x }' x=1

bash-3.00$ nawk 'BEGIN { print x }' x=1

bash-3.00$
9

@Corona688: I do not disagree, I only correct the error of the single quote on the syntax of the example.
But, it is true that I have also fixed the method.:o

Regards.