How does this awk command work?

Question

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?

Answer 1

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.