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Print out just a word from the file

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# 1  
Old 02-20-2005
Print out just a word from the file
Hi,
Would like to find a more suitable solution for the following. I have a file eg test.log. In this file, i have to find the line that has "Final rating" which is the starting of the line. I need to print out only 5.75 instead of the whole line using "grep". May I know what suitable command shall i use?

Final rating = 5.75 / 10.00

Regards and Thank you in advance.
# 2  
Old 02-20-2005
Code:
sed -ne '/^Final rating/s#^\([^=][^=]*\)=\([^/][^/]*\).*#\2#p' file

# 3  
Old 02-20-2005
Hi,
Thank you. Can I check with you whether this command is applicable in Linux?
# 4  
Old 02-20-2005
it should be as long as you have 'sed' available.
# 5  
Old 02-20-2005
Sorry i try but nothing comes out.

sed -ne '/^Final rating/s#^\([^=][^=]*\)=\([^/][^/]*\).*#\2#p' file

For this command,
May I know wat does /s stands for. Why is tat two [^=] and [^/]?

I know #\2#p is Print. Copy the pattern space to the standard output. Am I right?
# 6  
Old 02-20-2005
Quote:
Originally Posted by Kinki
Sorry i try but nothing comes out.

sed -ne '/^Final rating/s#^\([^=][^=]*\)=\([^/][^/]*\).*#\2#p' file

For this command,
May I know wat does /s stands for. Why is tat two [^=] and [^/]?

I know #\2#p is Print. Copy the pattern space to the standard output. Am I right?
it works just fine for this file based on your description:
Code:
Final rating = 5.75 / 10.00
foo
barr Final rating = 6.75 / 10.00
Final rating = 7.75 / 10.00

check your input file and make sure it really starts with 'Final rating' without any leading blanks.
# 7  
Old 02-20-2005
oic. Yes u are right. if there is leading Blank, how do u go about doing it?
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