问题
I want to search a pattern in paragraph that begins with word1 and end with word2 and print the first line of the paragraph if the pattern match, I am not sure if I can do it using grep for example if I have the following file and I am looking for aaa
Word1 this is paragraph number 1
aaa
bbb
ccc
word2
Word1 this is paragraph number 2
bbb
ccc
ddd
word2
the answer should be like that
Word1 this is paragraph number 1
回答1:
You can try this awk:
awk '/^Word1/{f=1;l="";hold=$0} /word2$/{f=0; if(l ~ /aaa/){print hold}} f{l = l RS $0}' file
回答2:
This might work for you (GNU sed):
sed -n '/^Word1/!b;:a;N;/^word2/M!ba;/^aaa/MP' file
Ignore any lines that do not begin Word1. Collect lines in the pattern space until a line beginning word2 or the end of the file. If a match is made then match also on the required string (in this case aaa). If a match is made print the first line and repeat.
EDIT: If paragraphs can end in other words i.e. word3, use this:
sed -n '/^Word1/!b;:a;N;/^$/Mb;/^word2/M!ba;/^aaa/MP' file
回答3:
This is the simple, idiomatic awk solution:
$ awk -v RS= -F'\n' '/^Word1.*aaa.*word2$/{print $1}' file
Word1 this is paragraph number 1
If that doesn't do what you want then edit your question to clarify your requirements.
回答4:
Try this one liner AWK:
awk '/Word1/{l=$0;flag=1;next}/word2/{flag=0}flag && $0 ~ /aaa/ && !c{print l; c++}' file
Input:
Word1 this is paragraph number 1
aaa
aaa
bbb
aaa
word2
Word1 this is paragraph number 2
bbb
ccc
ddd
word2
Output:
Word1 this is paragraph number 1
回答5:
A simple solution (that doesn't match exactly what you asked for):
awk -F'\n' -v RS= '/bbb/{print $1}' file
This skips finding Word1/Word2, and assumes that you have a blank line between records as in your example.
Of course, you could always force the blank line beforehand, with sed:
gsed 's/^Word1/\n&/' file | ...above...
来源：https://stackoverflow.com/questions/35335409/how-can-i-find-a-matching-pattern-between-two-words-using-sed-or-awk