Print text at end of semicolon Sed/awk

Question

I've this strings in a file that looks like aaaaa;bbbbbbb;ccccc\\n and aaaaa;bbbbbb\\n the idea is only to show the text between the last semicolon and newline. Output will be

ccccc
bbbbb

etc.

Thought that I could do it with sed -e [:]$ | awk -F '\\n' {print $1} sed -e [:]$ | awk -F '\\n' {print $1}

Answer 1

There are many ways to solve the task:

sed 's/.*;//'
awk -F';' '{print $NF}'
rev | cut -d';' -f 1 | rev

Answer 2

只需删除所有内容，直到最后一个分号即可：

sed 's/.*;//'