Unix：使用sed删除每行的一部分

Question

I am trying to remove the Who: ,What: ,When: ,Where: and the proceeding space from each line in a text file. Below is a example:

Who: Tester1+Password
What: Authentication Success
When: Tues March 20, 2015 08:15:02 UTD
Where: 198.192.1.2

Answer 1

If you want to remove anything upto first : , then you can do:

sed -re 's/(^[^:]+: )(.*)/\2/' file
Tester1+Password
Authentication Success
Tues March 20, 2015 08:15:02 UTD
198.192.1.2

As Glenn suggested we can avoid capture groups completely by removing the portion we don't need.

sed 's/[^:]\+: //' file

Answer 2

使用cut排除第一个以空格分隔的字段：

cut -d " " -f 2- file

Answer 3

Put the pattern which matches What, where, when , who inside a capturing group followed by a colon. Then replace the matched chars with an empty string. Add i flag at the last if you want do a case-insensitive match.

$ sed 's/^Wh\(ere\|en\|at\|o\):[[:blank:]]*//' file
Tester1+Password
Authentication Success
Tues March 20, 2015 08:15:02 UTD
198.192.1.2

For general case, you could use

sed 's/^[^[:blank:]]\+[[:blank:]]\+//' file

Answer 4

sed 's/^[^ ]* *//' YourFile

假设只有这四个字是可能的，并且作为您的样本