awk word boundaries not working

Shell Programming and Scripting
1

Hi,

I am trying below code but the word boundaries not seem to be working. What am I doing incorrectly?

 echo " ECHO " | awk '{ q="ECHO" ; if ( $0 ~ /\bq\b/) print "HELLO" ; }'

OR

 echo " ECHO " | awk '{ q="ECHO" ; if ( $0 ~ /\b'$q'\b/) print "HELLO" ; }'

Or 

 echo " ECHO " | awk '{ q="ECHO" ; if ( $0 ~ /\<'$q'\>/) print "HELLO" ; }'
2

echo " ECHO " | awk '{ q="ECHO" ; if ( $0 ~ q) print "HELLO" ; }'
HELLO

3

That's not what I was looking for. Your example wouldd match the string present anywhere. Hence wanted to check the word boundaries

4

Try:

awk '/(^|[[:blank:]])ECHO([[:blank:]]|$)/{print "HELLO"}'

If you want to try this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk , /usr/xpg6/bin/awk , or nawk .

5

Seems like variables cannot be used as part of a regex enclosed in slashes. Apart from that, \b is a escape sequence for backspace ...

Try this:

$ echo " ECHO" | awk '{ q="ECHO" ; if ( $0 ~ " " q " " )  print "HELLO" ; }'
$ echo "ECHO " | awk '{ q="ECHO" ; if ( $0 ~ " " q " " )  print "HELLO" ; }'
$ echo " ECHO " | awk '{ q="ECHO" ; if ( $0 ~ " " q " " )  print "HELLO" ; }'
HELLO
$ echo " ECHO  " | awk '{ q="ECHO" ; if ( $0 ~ " " q " " )  print "HELLO" ; }'
HELLO
$ echo " ECHO  " | awk '{ q="ECHO" ; if ( $0 == " " q " " )  print "HELLO" ; }'
$

Hope this helps.

6

Don Cragun, I was working with variables and wanted to check the whole words only. Do you know how to do it?

Junior - helper , That's exactly what I was struggling with.

Apart from concatenation and checking the begining and end of strings, isnt there any other way of doing this?

7 
awk -v word="$variable" '$0 ~ "(^|[[:blank:]])" word "([[:blank:]]|$)"{print "found"}'

will print found if the expansion of $variable (treated as an extended regular expression) appears at the start of a line or follows a space or tab AND a space of tab follows it or it appears at the end of a line.