1

Consider the file test.txt:

pfg025G
pfg025T
pfg034T
pfg039G

Now consider the following awk command and its output:

awk '(NR>1) {print "s/"p"/"$1"/g"}{p=$1}' test.txt

s/pfg025G/pfg025T/g
s/pfg025T/pfg034T/g
s/pfg034T/pfg039G/g

NR>1 is true for lines greater than one, this is valid for the "$1" term but not for "p" which takes the value of the first line. Why is that? Does NR>1 evaluate just the first block {} and not the second {p=$1}? Why does the first "$1" have double quotes "" while second $1 does not?

1
  • someone is apparently using an awk script to generate a sed script - that's not necessary since awk can do the substitutions directly itself and generally a bad idea as it's hard to do robustly. – Ed Morton Feb 4 at 16:40
3

you (splited) awk command looks like

awk '(NR>1) {print "s/"p"/"$1"/g"}
            {p=$1}'

which means

  • do {print "s/"p"/"$1"/g"} when NR>1
  • do {p=$1} always
  • "s/"p"/"$1"/g" quote splits (1): "s/" + p + "/" + $1 + "/g" , neither p nor $1 are quoted
  • (1) + for concatenation, note that awk use space (no space) as implicit concatenation operator

on first line, only {p=$1} is executed.

on second line {print "s/"p"/"$1"/g"} is executed first, and value of p is initialized from first line.

on last line, p end up with pfg039G that is discarded.


Use a semicolon in the same code block in order to create a series of commands:

awk '(NR>1) {print "s/"p"/"$1"/g" ; p=$1}' test.txt
s//pfg025T/g
s/pfg025T/pfg034T/g
s/pfg034T/pfg039G/g

Now the result is as expected using $1 from the previous match - in the first line, $1 is empty.

3
  • what about the use of double quotes and no quotes on "$1" $1 – alex Feb 4 at 11:09
  • @alex The double quotes are not around $1 but around / and /g. I would have written the statement as printf "s/%s/%s/g\n", p, $1. – Kusalananda Feb 4 at 11:12
  • ok nice I understand now, thanks. – alex Feb 4 at 11:23

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