Dear all

I want to grep all mail ids from /var/maillog file to check who is using mail service

i don't know what will be the syntex for this query in sed command

pls tell me the sed syntex for this

GREP and SED are two different commands.

To simply find a particular expression in a file, use GREP. If you want to change something, then SED is one of the tools you can use.

For example, to display all lines in filename which contain keyword:
cat filename | grep keyword

For more and better help, post a sample of the file format and the exact output you would like

Good manuals are at http://tldp.org ---I would start with Bash Guide for Beginners

The maillog file has

cat maillog|more
Oct 1 12:25:37 ns5 popper[11080]: (v3.0.2-mysql-0.10.2) POP login by user "user1
@example.com" at (59.94.251.158) 59.94.251.158
Oct 1 12:25:37 ns5 popper[10997]: I/O Error from at 59.94.211.8 (59.94.211.8):
[-1] 104 (Connection reset by peer); 0 (Success)
Oct 1 12:25:37 ns5 popper[10997]: (null) at 59.94.211.8 (59.94.211.8): -ERR POP
EOF or I/O Error: 104 (Connection reset by peer); 0 (Success)
Oct 1 12:25:37 ns5 popper[11080]: Stats: user2@example.com 0 0 0 0 59.94.25
1.158 59.94.251.158
Oct 1 12:25:38 ns5 popper[11085]: (v3.0.2-mysql-0.10.2) POP login by user "user3@example.com" at (59.94.251.158) 59.94.251.158
Oct 1 12:25:39 ns5 popper[11088]: (v3.0.2-mysql-0.10.2) POP login by user user4@example.com at (90.0.0.70) 90.0.0.70
Oct 1 12:25:39 ns5 popper[11088]: Stats: user5@example.com 0 0 0 0 90.0.0.70

now i want output like this
user1@example.com
user2@example.com
user3@example.com
user4@example.com
user5@example.com

First, this will be easiest if the desired string is always in the same format--eg _"user@example.com"_ (where "_" means space. Let's assume that is the case---here is a crude beginning:

"names" is the filename is used for the test.

cat names | sed -n 's/.*\( ".*@.*\..*" \).*/\1/p'

This uses the "backreference" feature of SED. Everything inside the escaped parentheses-- \(...)\ -- is substituted using \1

So....... It says call sed with no print option (-n), substitute the first occurence of the first expression with the backreference, and print that result.

The first expression reads as follows: "any # of characters, followed by the back-referenced expression, then followed by any number of characters."

In turn, the back-referenced expression is: "a space, then a ", then any # of characters, a @, any # of characters, a ., any #of characters, a ", and a space"

This will make a lot more sense after reading a good tutorial---eg: http://www.grymoire.com/Unix/Sed.html#uh-8