Extract string between newline and variable data in Bash

Question

As part of a load test, I've a file with a generated list of requests. The format is below, in that, the payload is in between two newline characters; the GET indicating the start of a new request. I would like to extract only the payload.

Connection: false
Content-Length: 25

foo=bar1&foo2=bar2

GET http://localhost:1080/dasistgut

I have tried several combinations without success:

grep -zoP "(?<=(\n)).*(?<=(\GET))"  sample.txt
grep -zoP "(?<=(\n)).*(?<=(\n))"  sample.txt
grep -zoP "(?<=(\n)).*(?<=\n)"  sample.txt

Expected output: foo=bar1&foo2=bar2

Answer 1

You may use this gnu grep :

var=$(grep -zoP '.+(?=\R{2}GET )' file)

echo "$var"
foo=bar1&foo2=bar2

Lookahead assertion (?=\\R{2}GET ) makes sure that input has exact 2 newlines ( \\R ) followed by "GET " ahead of current position.

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Alternative awk based solution:

awk '!NF{++n; next} n && /^GET /{print p} {p=$0; n=0}' file

foo=bar1&foo2=bar2

Answer 2

使用awk匹配空白行，获取下一行并打印，然后跳过下一行（应始终为空白）。

awk '/^$/ { getline; print; getline; }' sample.text