Extracting data from a text file using awk [duplicate]

This question already has answers here: Closed 11 years ago.

Possible Dupl开发者_StackOverflow中文版icate:

Extract data between two points in a text file

For example:

Reply: [200/OK] bytes=29086 time=583ms

I would want to extract the value between "time=" and "ms"

Expected Result:

"583"

---

I would use sed for that, but since you ask for awk:

echo "Reply: [200/OK] bytes=29086 time=583ms" | awk -F'time=|ms' '{print $2}'

The -F defines extended regexp for field separator. So we define that "time=" or "ms" separates the fields, and then print second field.

using sed, it would be:

echo "Reply: [200/OK] bytes=29086 time=583ms" | sed 's/.*time=\([0-9]*\)ms.*/\1/'

---

echo "Reply: [200/OK] bytes=29086 time=583ms" | sed "s/.*time=\(.*\)ms/\1/"

---

Ugly but works:

$ echo "Reply: [200/OK] bytes=29086 time=583ms" | 
    awk '{print $4'} | sed -e 's/[a-z=]//g'
583

Or without sed:

$ echo "Reply: [200/OK] bytes=29086 time=583ms" | 
    awk '{ split($4,a,"="); gsub(/ms/,"", a[2]); print a[2] }'
583

---

Try this

printf "Reply: [200/OK] bytes=29086 time=583ms\n" \
| awk '/^Reply:/{sub(/^.*time=/,"",$0) ;sub(/ms$/,"",$0); print $0}'

I hope this helps.

P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, and/or give it a + (or -) as a useful answer.