Shell Programming and Scripting

I have been reading several posts regarding how to extract text from a file, but none of those have helped me for what I need.
This is my problem: I need to extract the text after my pattern
So my line is:
485.74 6589.5 Log likelihood: 1485.79

My pattern is 'Log likelihood:'

and I need only the rest of the line: 1485.79

I can use "grep -w 'Log likelihood:' file" but get the whole line

How can I do it if i want only the beginning of the line: "485.74 6589.5" ?

Thanks guys

To get only stuff after ":" try this:

Code:

echo "485.74 6589.5 Log likelihood: 1485.79" | cut -d":" -f2

or

Code:

echo "485.74 6589.5 Log likelihood: 1485.79" | awk -F: '{print $2}'

or

Code:

echo "485.74 6589.5 Log likelihood: 1485.79" | sed 's/.*://'

---------- Post updated at 00:54 ---------- Previous update was at 00:52 ----------

To get stuff before Log..., try this:

Code:

echo "485.74 6589.5 Log likelihood: 1485.79" | sed 's/Log.*//'

---------- Post updated at 00:55 ---------- Previous update was at 00:54 ----------

or

Code:

echo "485.74 6589.5 Log likelihood: 1485.79" | awk '{print $1" "$2}'

Awesome!!

Thanks a lot!