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# 8  
Old 01-30-2015
Code:
awk '{for(I=1;I<=NF;I++) print $I}'

----- could print the $2 part of the output file but the issue is how to resolve the $1 part of the output file.....could you help with how I can print the first column of the output file

Moderator's Comments:
Mod Comment edit by bakunin: fixed CODE-tags. Please keep in mind to use them.
Last edited by bakunin; 01-30-2015 at 05:52 AM..
# 9  
Old 01-30-2015
Code:
awk ' /^[^ ]/  { 
		frst_fld=$1 
	       } 
      ! /^ *$/ { 
		for(i=NF==4?2:1;i<=NF;i++) { print frst_fld , $i }
	       }
' file

# 10  
Old 01-30-2015
Thanks anbu23... You are an expert and I need to understand your script now.
# 11  
Old 01-30-2015
Quote:
Originally Posted by anbu23
Code:
awk ' /^[^ ]/  { 
          [....]
	       }

Quote:
Originally Posted by Indra2011
Thanks anbu23... You are an expert and I need to understand your script now.
This means: when a line starts with anything save for a space char (this is what the regex "^[^ ]" means), execute whatever is between "{" and "}".

The second clause works similar but the "!" reverses the meaning: for every line not fitting the pattern do ...

I hope this helps.

bakunin
# 12  
Old 01-30-2015
/^[^ ]/ If the input row doesn't start with space then assign first field to frst_fld

! /^ *$/ Execute the below block if input row is not empty or only spaces

Code:
{ 
		for(i=NF==4?2:1;i<=NF;i++) { print frst_fld , $i }
	       }

for(i=NF==4?2:1;i<=NF;i++) { print frst_fld , $i } If input row has four fields then assign two else one to i
# 13  
Old 01-30-2015
Playing around with the field separator:
Code:
awk '$1 != "" {F1=$1} {for (i=2; i<=NF; i++) print F1, $i}' FS=" +" file
1 21
1 31
1 12
1 21
1 53
1 65
2 78
2 67
2 53
2 34
2 45
2 67

Caveat: some awks might not like this.
# 14  
Old 01-30-2015
Thanks all of u
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