Replace a string with pattern within that string
 

I want to replace a part of string with a pattern within that string.

for eg:
if my string is "abcd#25 - efgh ijkl" it should print output as "abcd#efgh"

if my string is "xxxx#100 - yyyy zzzz" it should print output as "xxxx#yyyy"

The pattern is # followed by integer should be replaced with first word after # integer -

Sorry for my bad english.

Try the following:
See if that's what you want.

Use the -r option to remove the need to backslash everything.

With grep and sed, their basic regular expressions level treats parens and many other regex characters as literal. But as a gnu extension you can backslash escape them to enable their special meanings.

But when you use the extended regular expressions (sed -r/grep -E/egrep), everything is reversed. All regex characters are enabled by default, and escaping disables them.

Check out the grep man page for details (and the info page for even more).

I keep forgetting about the existence or ERE:)