How to grep with alternation

Question

I would like to grep lines which include a comma followed by four identical digits followed by a comma followed by an alphabetic character.

I tried

grep -E ,'1111|2222|3333|4444|5555|6666|7777|8888|9999',[[:alpha:]] file

This doesn't seem to do what I describe. The problem is that it doesn't handle the commas and [[:alpha:]] properly it seems.

How can you do this?

Answer 1

It's because your alternation is not applied in the way you are expecting. To make it behave as you want, you need to use groups:

grep -E ,'(1111)|(2222)|(3333)|(4444)|(5555)|(6666)|(7777)|(8888)|(9999)',[[:alpha:]] file

Alternatively, this could be expressed more succinctly using a backref:

grep -E ,'([[:digit:]])\1{3},[[:alpha:]]' file

which basically means the same digit 4 times. This also includes 0, however, so it may or may not help you.

EDIT:

Of course... to make it only 1-9, you could

grep -E ,'([1-9])\1{3},[[:alpha:]]' file

Answer 2

为您尝试此正则表达式

',(1111|2222|3333|4444|5555|6666|7777|8888|9999|0000),\w'