在bash的第一个管道之间提取东西？

Question

I want to extract the stuff in between the first two pipes:

| a650f360-a29d-4562-b46f-024fe15aa7dc | emi-ubuntu-12.04.2-server-amd64-08152013           | ACTIVE |

The output should be:

 a650f360-a29d-4562-b46f-024fe15aa7dc 

I have a regex that I was going to use sed with: ^\\|\\s*(.*?)\\s*\\|

but according to a regex calculator it gets me the pipes as well. How do I get rid of it?

Answer 1

Cut is probably easier to understand than regex. Why not

cut -d\| -f2

?

Answer 2

sed 's/^[^|]*|\([^|]*\)|.*/\1/'

它与开头的所有非管道匹配（示例数据中未包含任何内容），一个管道，捕获不是管道的东西，匹配另一个管道以及末尾的任何东西，并将其替换为捕获的东西。

Answer 3

Bash only. In a loop this can be done using an array:

while IFS='|'; read -a line ; do
  echo ${line[1]}            # with leading/trailing blanks
  echo ${line[1]// /}        # without leading/trailing blanks
done < "$infile"