用grep读取日志文件中的型式1和打印包含型式1只行。停止搜索时PATTERN2在文件中被发现 [英] Using grep to read a log for pattern1 in the file and print only lines containing pattern1. Stop searching when pattern2 is found in the file

问题描述

使用的grep / SED / AWK 我要搜索样式1 直到 PATTERN2 在文件中被发现，打印出含有结果仅样式1 。

Using grep/sed/awk I want to search for pattern1 until pattern2 is found in a file and print the results containing only pattern1.

我不想对此有在这个网站很多解决方案样式1的范围和PATTERN2之间的界限。请帮助。

I don't want the lines between the range of pattern1 and pattern2 for which there are many solutions in this site. Please help.

我试过没有结果如下：

亚行logcat | grep的-i样式1| grep的-m 1'PATTERN2

adb logcat | grep -i 'pattern1' | grep -m 1 'pattern2'

看起来|使它成为键，条件和我要找的字符串，不在同一行上。他们在不同的线路。

Looks like '|' makes it an AND condition and the strings that I am looking for, are not in the same line. They are in different lines.

更新：的解决方案，在终端，但不是在一个脚本工作。有人能找出为什么没有在脚本的工作帮助吗？

Updated: The solutions work in the terminal but not in a script. Can someone help with finding out why it's not working in a script?

推荐答案

使用sed将停止打印PATTERN2被发现后：

Use sed to stop printing after pattern2 is found:

adb logcat | sed  '/pattern2/q' | grep -i 'pattern1'

或者使用SED既打印样式1和PATTERN2后退出：

Or use sed to both print pattern1 and quit after pattern2:

adb logcat | sed  -n '/pattern1/p;/pattern2/q'

这篇关于用grep读取日志文件中的型式1和打印包含型式1只行。停止搜索时PATTERN2在文件中被发现的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！