匹配模式 > 0 in linux 后获取行

Question

我有这个日志，我想在不使用脚本的情况下为“QTY REGS > 0”的作业打印状态为“完成”的行。

INFO  2021-01-19 02:00:00,001 JOB: 0970b711] starting
INFO  2021-01-19 02:00:01,001 JOB: 0970b711] adding records
INFO  2021-01-19 02:00:02,001 JOB: 0970b711] registered
INFO  2021-01-19 02:00:03,001 JOB: 0970b711] QTY REGS: 3 processing
INFO  2021-01-19 02:00:03,001 JOB: 32970g233] debug started
INFO  2021-01-19 02:00:04,001 JOB: 0970b711] finished
INFO  2021-01-19 02:00:05,001 JOB: 10d0g655] starting
INFO  2021-01-19 02:00:06,001 JOB: 10d0g655] QTY REGS: 0 processing
INFO  2021-01-19 02:00:07,001 JOB: 10d0g655] finished
INFO  2021-01-19 02:00:08,001 JOB: 0870a711] starting
INFO  2021-01-19 02:00:09,001 JOB: 0870a711] QTY REGS: 1 processing
INFO  2021-01-19 02:00:10,001 JOB: 0870a711] retry
INFO  2021-01-19 02:00:11,001 JOB: 0870a711] connect error
INFO  2021-01-19 02:00:12,001 JOB: 0870a711] finished

我想return

INFO 2021-01-19 02:00:03,001 JOB: 0970b711] QTY REGS: 3 processing
INFO 2021-01-19 02:00:09,001 JOB: 0870a711] QTY REGS: 1 processing

这些记录中 QTY REGS > 0 具有相同作业编号但已“完成”描述的行

INFO 2021-01-19 02:00:04,001 JOB: 0970b711] finished
INFO 2021-01-19 02:00:12,001 JOB: 0870a711] finished

Answer 1

我建议使用数组：

awk '=="QTY" && >0{a[]=1}; =="finished" && a[]==1{print}' file

输出：

INFO 2021-01-19 02:00:04,001 JOB: 0970b711] finished
INFO 2021-01-19 02:00:12,001 JOB: 0870a711] finished

{print}也可以省略

匹配模式 > 0 in linux 后获取行

Get line after matching pattern > 0 in linux

search

logging

awk

grep

command-line-interface