如何grep只有一个数字字符

时间:2013-10-24 13:46:49

标签: bash

我有一个文件如下所示,我想grep只有数字“1”但是当我给命令grep并且grep每个字符都有“1”

 0 turks-g         DD-Production-Server Incremental-Bac pls-i085-g      10/23/2013 18:04:53
 1 pls-w099-g      Production-PCI-Serve Incremental-Bac pls-i097-g      10/23/2013 18:05:36
 0 longjawed-g     DD-Production-Server Incremental-Bac pls-i086-g      10/23/2013 18:09:09
 1 pls-i085-g      NetBackup-Servers    Incremental-Bac pls-i085-g      10/23/2013 18:09:28

orion #cat tts | grep“1”

0 turks-g DD-Production-Server Incremental-Bac pls-i085-g 10/23/2013 18:04:53  1 pls-w099-g Production-PCI-Serve Incremental-Bac pls-i097-g 10/23/2013 18:05:36  0 longjawed-g DD-Production-Server Incremental-Bac pls-i086-g 10/23/2013 18:09:09  1 pls-i085-g NetBackup-Servers Incremental-Bac pls-i085-g 10/23/2013 18:09:28

orion #cat tts | grep'1'

0 turks-g DD-Production-Server Incremental-Bac pls-i085-g 10/23/2013 18:04:53  1 pls-w099-g Production-PCI-Serve Incremental-Bac pls-i097-g 10/23/2013 18:05:36  0 longjawed-g DD-Production-Server Incremental-Bac pls-i086-g 10/23/2013 18:09:09  1 pls-i085-g NetBackup-Servers Incremental-Bac pls-i085-g 10/23/2013 18:09:28

请建议我如何才能只使用“1”

谢谢

SA

2 个答案:

答案 0 :(得分:2)

使用字边界搜索:

grep "\<1\>" file

或者使用grep -w

grep -w "1" file

答案 1 :(得分:0)

您可以使用感兴趣文本周围的图案,即线条的开头和空白区域。

cat tts | grep "^ 1 "