CODEWITHSUNDEEP
awk
Tryer
Tryer

Reputation: 4050

Accessing two fields of a line before a matched line

Given the following in a file, file.txt:

Line with asdfgh output1 45 output2 80
Special Header
Line with output1 38 output2 99
Special Header

I would like to print to a file:

45 80
38 99

i.e., in the line immediately preceding lines whose first column is Special, store the number (which can be a float type with decimals) after output1 and output2

I tried:

awk '($1 == "Special" ) {printf f; printf ("\n");} {f=$0}' file.txt > output.txt

This captures the entirety of the previous line and I get output.txt which looks like this:

Line with output1 45 output2 80
Line with output1 38 output2 99

Now, within the captured variable f, how can I access the specific values after output1 and output2?

Upvotes: 1

Views: 64

Answers (3)

Ed Morton
Ed Morton

Reputation: 203229

$ awk '$1=="Special"{print x, y} {x=$(NF-2); y=$NF}' file
45 80
38 99

Upvotes: 1

Sundeep
Sundeep

Reputation: 23667

With GNU awk:

$ awk '$1=="Special"{print m[1], m[2]}
       {match($0, /output1\s+(\S+).*output2\s+(\S+)/, m)}' ip.txt
45 80
38 99

With perl:

$  perl -ne 'print "@d\n" if /^Special/; @d = /output[12]\s+(\S+)/g' ip.txt
45 80
38 99

Upvotes: 1

Gilles Quénot
Gilles Quénot

Reputation: 185015

Like this:

$ awk '{for (i=1; i<NF; i++)
  if ($i == "output1") {arr[NR]=$(i+1)}
  else if ($i == "output2") {arr[NR]=arr[NR]" "$(i+1)}}
  ($1 == "Special") {print arr[NR-1]}
' file

Output

45 80
38 99

Upvotes: 1

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