需要在unix中提取特定单词

我想从下面的文本文件中提取特定单词，即

PassThroughHandlerJob:2299成功，0失败，0挂起

在日志文件中，2299成功计数每天都在变化。失败、等待和成功的计数每天都可能不同

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

对于这个特殊的任务，你应该能够使用grep

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

你可以有：

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

或者只是普通的
grep
：

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

grep-o'PassThroughHandlerJob:[[：space:]\+[0-9]\+成功，[0-9]\+失败，[0-9]\+pending'文件命令工作正常非常感谢