需要在unix中提取特定单词

Question

我想从下面的文本文件中提取特定单词，即PassThroughHandlerJob: 2299 succeeded, 0 failed, 0 pending。

在日志文件2299中，成功计数每天都在变化。并且计数可能每天都有所不同，因为失败，未决和成功。

NewFile.txt

2014-07-10 22:28:13,193 INFO [mainScheduler_Worker-28]     [com.myvest.sps.scheduler.jobs.AbstractDistributedJob] Checked the status of 2299 work units for job PassThroughHandlerJob:  2299 succeeded, 0 failed, 0 pending

2014-07-10 22:28:21,859 INFO [mainScheduler_Worker-28]     [com.myvest.sps.scheduler.jobs.success.PassThroughHandlerJob] Job PassThroughHandlerJob.PassThroughHandlerJob finished

Answer 1

对于此特定任务，您应该能够使用grep;

$ grep PassThroughHandlerJob: log.txt
PassThroughHandlerJob: 2299 succeeded, 0 failed, 0 pending 2014-07-10 22:28:21,859 INFO

Answer 2

你可以：

egrep -o 'PassThroughHandlerJob:\s+[0-9]+ succeeded, [0-9]+ failed, [0-9]+ pending' file

或者只是正常grep：

grep -o 'PassThroughHandlerJob:[[:space:]]\+[0-9]\+ succeeded, [0-9]\+ failed, [0-9]\+ pending' file