Suppose I have a file like this (disclaimer: this is not fixed I can have more than 7 rows, and more than 4 columns)
R H A 23
S E A 45
T E A 34
U A 35
Y T A 35
O E A 353
J G B 23
I want the output to select second column if third column is A but keeping newline or whitespace character.
output should be:
HEE TE
I tried this:
awk '{if ($3=="A") print $2}' file | awk 'BEGIN{ORS = ""}{print $1}'
But this gives:
HEETE%
Which has a weird % and is missing the space.
You may use this gnu-awk solution using FIELDWIDTHS:
awk 'BEGIN{ FIELDWIDTHS = "1 1 1 1 1 1 *" } $5 == "A" {s = s $3}
END {print s}' file
HEE TE
awk splits each record using width values provided in this variable FIELDWIDTHS.
1 1 1 1 1 1 * means each of first 6 columns will have single character length and remaining text will be filled in 7th column. Since you have a space after each value so $2,$4,$6 will be filled with a single space and $1,$3,$5 will be filled with the provided values in input.
$5 == "A" {s = s $3}: Here we are checking if $5 is A and if that condition is true then we keep appending value of $3 in a variable s. In the END block we just print variable s.
Without using fixed width parsing, awk will treat A in 4th row as $2.
Or else if we let spaces part of column value then use:
awk '
BEGIN{ FIELDWIDTHS = "2 2 2 *" }
$3 == "A " {s = s substr($2,1,1)}
END {print s}
' file