topic Re: Another awk question - pattern matching an internal var in Operating System - HP-UX

Another awk question - pattern matching an internal var

Tony Walker — Thu, 19 Dec 2002 12:22:15 GMT

Hello again!

Yesterday I was helped out with some syntax and got the following awk statement which works FINE - adding a string to the end of a pattern matched line.

awk '{if (/^STRING:/) { sub($0,$0"uname:")} print $0 }' filename | more

Now, as I intend to perform this on a number of STRINGS and unames I intend to do something like this:

awk -v OPTION=STRING -v NAME=uname '{if (/^OPTION:/) { sub($0,$0"NAME:")} print $0 }' filename | more

Basically I need a hand with the syntax as it is simply trying to pattern match the literal string OPTION and not my variable. I've tried escape characters and $ signs but to no avail.. I'm sure there is a nice simple solution to this so I await your replies. Thanks in advance.

Tony

Re: Another awk question - pattern matching an internal var

James R. Ferguson — Thu, 19 Dec 2002 12:34:14 GMT

Hi Tony:

# awk -v OPTION=^STRING -v NAME=TAKE '{if ($0~OPTION) {sub($0,$0"NAME")}

Regards!

...JRF...

Re: Another awk question - pattern matching an internal var

James R. Ferguson — Thu, 19 Dec 2002 12:40:21 GMT

Hi (again):

Oops, I dropped the last part and forgot too unquote your "NAME":

# awk -v OPTION=^xxx -v NAME=TAKE '{if ($0~OPTION) {sub($0,$0 NAME)} print $0}' filename|more

Notice, too, that the anchor (^) is part of the variable value as passed. The 'NAME' is *not* quoted, too.

Regards!

...JRF...

Re: Another awk question - pattern matching an internal var

James R. Ferguson — Thu, 19 Dec 2002 13:00:20 GMT

Hi Tony:

Now that I've had coffee, let's shorten what you appear to want. It appears that upon a match to a STRING anchored to the beginning of the line you merely desire to append 'uname' to the line's end. If you *really* mean the evaluated 'uname' then enclose it in backticks like this:

awk -v OPTION=^STRING -v NAME=`uname` '{if ($0~OPTION) {print $0 NAME} print $0}' filename|more

Again, note the -v OPTION=^STRING and the NAME=`uname`. You don't need to 'sub' -- just 'print $0 NAME'.

Since 'uname' returns "HP-UX" you might want 'hostname' instead.

Regards!

...JRF...

Re: Another awk question - pattern matching an internal var

Christian Gebhardt — Thu, 19 Dec 2002 13:20:26 GMT

Hi

OPTION=STRING
awk -v NAME=uname '/^'$OPTION'/ {printf("%s:"NAME"\n",$0)}' file

Chris

Points ?? ;-)

Re: Another awk question - pattern matching an internal var

Tony Walker — Thu, 19 Dec 2002 13:55:03 GMT

James,

Thanks for the replies and confirming why you always get so many points;)

Good point with the lack of need for the sub command (I'm always looking to shorten my commands. Alas I wrote uname as a short-term for username - but the thought is appreciated.

Thanks again.

Tony

Re: Another awk question - pattern matching an internal var

Tony Walker — Thu, 19 Dec 2002 14:20:57 GMT

James, just tried this and found that it prints the line twice. Once with my amendment and once the unchanged $0...

Re: Another awk question - pattern matching an internal var

Jean-Louis Phelix — Thu, 19 Dec 2002 15:08:31 GMT

Hi,

Or perhaps do it with sed ?

#!/usr/bin/sh
OPTION=STRING
NAME=UNAME
FILE=i.data
sed 's;^'$OPTION':.*$;&'$NAME';' $FILE

Regards

Re: Another awk question - pattern matching an internal var

James R. Ferguson — Thu, 19 Dec 2002 15:12:51 GMT

Hi Tony:

Oops, this is really not my day!

# awk -v OPTION=^STRING -v NAME="<<<" '{if ($0~OPTION) {print $0 NAME} else {print $0}}' filename

...or...

# awk -v OPTION=^STRING -v NAME="<<<" '$0~OPTION {$0=$0 NAME};{print $0}' filename

Regards!

...JRF...