求助：如何用awk模式匹配输出特定字段

jason680

回复 7# three_body

# awk 'function get(k,n, s,e,t){t=length(k)+1;s=index($0,k"=");e=index($0,n"=");if(s<e){return(substr($0,s, e-s-1))}return""}{print get("policy_id", "action"),get("src","src-xlated ip")}' FILE
policy_id=103 service=VDO Live (tcp) proto=6 src zone=Trust dst zone=Untrust src=10.100.1.43 dst=119.188.11.3 src_port=4044 dst_port=7001
   

kooleon

我觉得5楼最直接了

q1208c

回复 10# three_body

再加一个 -e 's/ /_/g'
   

yestreenstars

which one do you like?

1. $ awk '{print gensub(/.*(service=.*) proto.*/,"\\1",1)}' i
   
2. service=http
   
3. service=NETBIOS (NS)
   
4. service=VDO Live (tcp)
   
5. $ grep -oP 'service=.*(?= proto)' i
   
6. service=http
   
7. service=NETBIOS (NS)
   
8. service=VDO Live (tcp)
   
9. $ sed -r 's/.*(service=.*) proto.*/\1/' i
   
10. service=http
    
11. service=NETBIOS (NS)
    
12. service=VDO Live (tcp)

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three_body

q1208c 发表于 2014-03-20 12:31
回复 10# three_body

再加一个 -e 's/ /_/g'

非常感谢各位，最后用了这个：

1. sed -r 's/.*(policy_id=.*zone=[^ ]+).*(src=.*) src-xlated.*/\1 \2/;:a;s/(service=[^ ]+) (.* proto=)/\1_\2/;ta' file

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summeradva

还是这个看着整洁～～ 回复 5# ly5066113