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如何 grep 文本到下一个空格？

tag-icon • tag-icon command-line awk sed grep pcregrep
如何 grep 文本到下一个空格？ 
7/04/27 10:50:17 INFO Master: Driver submitted org.apache.spark.deploy.worker.DriverWrapper
17/04/27 10:50:17 INFO Master: Launching driver driver-20170427105017-0000 on worker worker-20170427103840-192.168.5.242-7078
17/04/27 10:50:22 INFO Master: 192.168.5.5:53156 got disassociated, removing it.
17/04/27 10:50:22 INFO Master: 192.168.5.5:37668 got disassociated, removing it.
17/04/27 10:50:22 INFO Master: 192.168.5.5:53154 got disassociated, removing it.
17/04/27 10:55:27 INFO Master: Registering app ETL DataPipeline App
17/04/27 10:55:27 INFO Master: Registered app ETL DataPipeline App with ID app-20170427105527-0000
17/04/27 10:55:27 INFO Master: Launching executor app-20170427105527-0000/0 on worker worker-20170427103842-192.168.5.175-7078
17/04/27 10:55:27 INFO Master: Launching executor app-20170427105527-0000/1 on worker worker-20170427103838-192.168.5.37-7078
17/04/27 11:08:25 INFO Master: Asked to kill driver driver-20170427105017-0000
17/04/27 11:08:25 INFO Master: Kill request for driver-20170427105017-0000 submitted
17/04/27 11:08:26 INFO Master: Received unregister request from application app-20170427105527-0000

我将如何获取 driver-20170427105017-0000 和相应的 192.168.5.242 以及类似地如何 grep app-20170427105527-0000/0 及其相应的 192.168.5.175 。

答案1

使用sed来获得全部 driver以及executor与“启动”相关的消息：

$ sed -n -E 's/^.*Launching (driver|executor) ([^ ]*).*worker-[0-9]*-([^-]*).*$/\2 \3/p' file.in
driver-20170427105017-0000 192.168.5.242
app-20170427105527-0000/0 192.168.5.175
app-20170427105527-0000/1 192.168.5.37
  • [^ ]*将匹配任意数量的任意字符（空格除外）。
  • \2和\3分别是对第二个和第三个括号匹配的内容的反向引用。第二个括号包含并将匹配或[^ ]*之后的文本，第三个括号包含并将匹配 IP 地址（直到终止地址）。Launching driverLaunching executor[^-]*-
  • ^in$将s/^...$/.../p正则表达式锚定在行的开头和结尾，而 whilep告诉sed“打印”替换的结果（如果进行了替换）。

或者，由于正则表达式的魔力较少，可能会更健壮，使用awk：

$ awk '/Launching/ { split($NF, a, "-"); print $7, a[3] }' file.in

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