UNIX for Beginners Questions & Answers

One could still try:

Code:

awk '$1 ~ /^PS/ {for(i=3; i<=NF; i++) if($i == "<Ob>]"){print $1,substr($(i-1), 2); next}}' file

without needing to use split() (unless I misunderstood and you changed your input file format to remove the <space> before the <Ob>]).

I'm so sorry Scrutinizer, but as my input is many thousand lines long I did not notice a potential complicating issue that I was wondering if I could get your help addressing. There are time where the desired string between an initial "[" and "<Ob>] contains a space.

So for example, given:

Code:

 PS028,006 [KJ <Cj>] [CM< <Pr>] [QWL TXNWNJ- <Ob>]
 Lexeme     KJ      # CM<      # QWL TXNWN J      #
 PhraseType  6(6) 1(1:2) 2(2.1,2.1,7)
 PhraseLab  509[0]    501[0]     503[0]
 ClauseType xQt0

Which I would pare down with INPUT | awk '$1 ~/^ PS/' to get:

Code:

PS028,006 [KJ <Cj>] [CM< <Pr>] [QWL TXNWNJ- <Ob>]

In this case, the desired output would be:

Code:

PS028,006 QWL TXNWNJ-

or

Code:

PS028,006 [QWL TXNWNJ- <Ob>]

The code you helped me with only gives:

Code:

PS028,006 QWL

Again, I apologize that I did not see the possibility of the space within the desired string until I double-checked the output against INPUT | sed -e 's/.* \[\(.*\) <Ob>\].*/\1/' which gives me the desired string but not the $1 when $1 ~/^ PS/.

Would you be able to help me iron this out?

--- Post updated at 10:02 PM ---

This works well Don except that I represented the desired output strings as "ABC" and "XYZ" which it seems that you took at being a three character string. I should have been more specific and said that "ABC" and "XYZ" represents a string of any length. Thus something like ["some amount of text" <Ob>].

OK... One final attempt...

Based on your single sample latest input file, the following seems to do what you want and will at least show you lines where it wasn't able to match:

Code:

awk '
$1 ~ /^PS/ {
	if(match($0, /[[][^[]* <Ob>[]]/))
		print $1, substr($0, RSTART + 1, RLENGTH - 7)
	else
		print "No Match Found on line " NR, $0
}' file

Try also

Code:

awk -F"[][]" '/^ *PS.*<Ob>/ {sub(/ *<Ob>.*$/, ""); print $1, $NF}' file
 PS028,005  ABC 
 PS028,005  XYZ 
 PS028,006  QWL TXNWNJ-

That did it RubiC! Such a simple and elegant way to accomplish it! Thanks so much also to Scrutinizer and Don Cragun for your help!

If I may, could I please ask a question about the field separator value? The man AWK page seems to only imply rather than being explicit that the use of the square brackets when setting the field separator from the command line tells AWK to interpret what is between them as a regex rather than simply a fixed string which would otherwise be indicated by "..."? Is this correct? Thanks again!

--- Post updated at 04:30 PM ---

That did it RubiC! Such a simple and elegant way to accomplish it! Thanks so much also to Scrutinizer and Don Cragun for your help!

If I may, could I please ask two questions about how this code is working? The first is about the field separator value. The man AWK page seems to only imply rather than being explicit that the use of the square brackets when setting the field separator from the command line tells AWK to interpret what is between them as a regex rather than simply a fixed string which would otherwise be indicated by "..."? Is this correct?

Secondly, since the value for FS has been set to "][" how come when the print statement calls for {print $1} is does not print from the beginning of the line to the first instance of "][" but rather prints what would be $1 when FS is set to whitespace? In other words, given:

Code:

 PS028,006 [KJ <Cj>] [CM< <Pr>] [QWL TXNWNJ- <Ob>]
 Lexeme     KJ      # CM<      # QWL TXNWN J      #
 PhraseType  6(6) 1(1:2) 2(2.1,2.1,7)
 PhraseLab  509[0]    501[0]     503[0]
 ClauseType xQt0

Why does RudiC's code not give:PS028,006 [KJ <Cj> for {print $1} if FS is set to "]["?

Rather it gives the (desired) first field if FS was at default PS028,006?

Thanks again!

Hi jvoot,
The standards clearly state that the value of the awk FS variable is an extended regular expression and it doesn't matter whether it is set using the -F option, using the -v option, using an assignment statement between pathname operands, or using an assignment statement in the awk script itself. When the ERE is set to [][] that is a bracket expression that specifies that the <open-square-bracket> character ([) and the <close-square-bracket> character (]) are each to be treated as separate field separators.

With the FS value RudiC used, field 1 is everything that appears in the record before the 1st open or close square bracket character (including the leading and trailing <space>). I chose to use the default FS value because I didn't think you wanted the leading and trailing <space> characters at the start of lines in your input data to be included in your output.

Hope this helps,
Don

You are partly right, the field separator string will be interpreted as a regex, and always. In Scrutinizers proposal (from which I stole shamelessly), he uses the bracket expression [][].
man regex:
So awk splits the input line at any occurrence of either [ or ] .

BTW, awk's default FS is a bracket expression regular expression (/[ \t\n]+/) by itself.


It does. Please apply what has been said to the repective line:

Code:

 PS028,006 [KJ <Cj>] [CM< <Pr>] [QWL TXNWNJ- <Ob>]
^          ^       ^ ^        ^ ^                ^--- last separator; $NF is empty
|          +-------+-+--------+-+-------------------- all FS
+---------------------------------------------------- field 1

Is that clearer now? If you want to remove the leading space from field 1, additional measures must be taken.