Grep 提取匹配给定模式的单词 [英] Grep to extract the word matching the given pattern

问题描述

我有一个包含类似条目的日志文件

I have a log file with entries like

INFO 2013-08-16 13:46:48,660 Index=abc:12 insertTotal=11  
INFO 2013-08-16 13:46:48,660 Index=abcd:12 insertTotal=11  
INFO 2013-08-16 13:46:48,660 Index=def:134 insertTotal=11  
INFO 2013-08-16 13:46:48,660 Index=abkfe insertTotal=11
INFO 2013-08-16 13:46:48,660 Index=lmkfe insertTotal=11
INFO 2013-08-16 13:46:48,660 Index=lmkfe insertTotal=11

我想 grep 并提取与我的模式匹配的单词，即 abc:<some_number> 和 def:<some_number>.

I would like to grep and extract the words that match my pattern which is abc:<some_number> and def:<some_number>.

$ cat "log.txt" | grep -w "abc" -w "def" >> "failed_values.txt";

所以在这种情况下，我的failed_values.txt应该只有

So in this case, my failed_values.txt should only have

abc:12
def:134

要注意的关键是我的模式以 : 结尾，后跟一个数字，然后是一个空格 e.G.abc:122.

The key to note is that my pattern ends with a : followed by a number and then a space e. g. abc:122.

推荐答案

尝试以下操作:

$ grep  -Eio '[a-z]+:[0-9]+' log.txt 
abc:12
abcd:12
def:134

• -i 忽略大小写.
• -o 只打印匹配的部分.
• -i to ignore case.
• -o to print only matched part.

更新

只匹配abc/def:

$ grep  -Eio '\b(abc|def):[0-9]+\b' log.txt 
abc:12
def:134

• (abc|def)::匹配abc或(|) def后跟:.
• [0-9]+:匹配的数字.
• \b:匹配词边界
• (abc|def):: match abc or(|) def followed by :.
• [0-9]+: matched numbers.
• \b: match word boundary

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