regex - bash + 如何从长输出中捕获单词

Question

我有以下命令的以下输出

zookeeper-shell.sh 19.2.6.4  get /brokers/ids/1010

输出是

Connecting to 19.2.6.4

WATCHER::

WatchedEvent state:SyncConnected type:None path:null
{"listener_security_protocol_map":{"PLAINTEXT":"PLAINTEXT"},"endpoints":["PLAINTEXT://kafka1.lulu.com:6667"],"rack":"/default-rack","jmx_port":9997,"port":6667,"host":"kafka1.lulu.com","version":4,"timestamp":"1630507307906"}

主要目标是捕获机器名称 kafka1 – 从上面的输出
所以我成功地执行了以下长命令语法

zookeeper-shell.sh 119.2.6.4  get /brokers/ids/1010 | sed s'/\/\// /g' | sed s'/:/ /g' | sed s'/,/ /g' | sed s'/"/ /g' | sed s'/\./ /g'| awk '{for (i=1;i<=NF;i++) print $i}' | grep -i kafka | sort | uniq

结果是:(如预期的结果)

kafka1

情况是我对我的方法感到不舒服 ，太长了，不够优雅
我们可以得到建议(使用 awk/sed/perl 一个衬垫)，从我的语法中得到更好的建议吗？

Answer 1

使用您显示的样本，请尝试关注 awk代码。由于我没有带 zookeeper 命令，因此我编写了此代码并仅根据您显示的输出对其进行了测试。

zookeeper-shell.sh 19.2.6.4  get /brokers/ids/1010 | 
awk '
/WatchedEvent state/{
  found=1
  next
}
found && match($0,/"PLAINTEXT:\/\/[^:]*/){
  print substr($0,RSTART+13,RLENGTH-13)
}
' 

说明:为以上添加详细说明 awk代码。

awk '                                         ##Starting awk program from here.
/WatchedEvent state/{                         ##Checking condition if line contains WatchedEvent state
  found=1                                     ##Then set found to 1 here.
  next                                        ##next will skip all further statements from here.
}
found && match($0,/"PLAINTEXT:\/\/[^:]*/){    ##Checking condition if found is SET then match regex "PLAINTEXT:\/\/[^:]* in match function of awk.
  print substr($0,RSTART+13,RLENGTH-13)       ##Printing sub string of matched regex used in match function above.
}
'