Shell Programming and Scripting

I want to get value of last row and 6 column from awk. Below is the format of my file. And RED one is my desired value. Actaully this stats usally update after every 1 hour so i want that every time i run the script i get the latest value.

Ending time - 01:00:58
HOURLY CALL ATTEMPTS (NORMALIZED) = 58227
Ending time - 02:00:59
HOURLY CALL ATTEMPTS (NORMALIZED) = 39987
Ending time - 03:00:58
HOURLY CALL ATTEMPTS (NORMALIZED) = 20922
Ending time - 04:00:56
HOURLY CALL ATTEMPTS (NORMALIZED) = 11385
Ending time - 05:00:57
HOURLY CALL ATTEMPTS (NORMALIZED) = 6441
Ending time - 06:00:56
HOURLY CALL ATTEMPTS (NORMALIZED) = 6598
Ending time - 07:00:59
HOURLY CALL ATTEMPTS (NORMALIZED) = 14472
Ending time - 08:00:59
HOURLY CALL ATTEMPTS (NORMALIZED) = 30849
Ending time - 09:00:58
HOURLY CALL ATTEMPTS (NORMALIZED) = 59088
Ending time - 10:00:58
HOURLY CALL ATTEMPTS (NORMALIZED) = 99476

When i use this command

Code:

more filename | awk '{print $6 }'

so it gives me the 6th column of all file but what i exactly want is last line and 6th column.

I hope, you understand.

lease HELP....

Regards,
Waqas Ahmed

Code:

tail -1 filename | awk '{print $6}'

Code:

awk -F'= ' 'END{print $NF}' file

Use /usr/xpg4/bin/awk on Solaris.

With nawk:

Code:

nawk -F'= ' '{_=$NF}END{print _}' file

Or:

Code:

sed -n '$s/.*= //p' file

If the leading space is not a problem:

Code:

tail -1 file|cut -d= -f2

...I want to capture the value of the third row, sixth column.
How do I do that in awk?

HOURLY CALL ATTEMPTS (NORMALIZED) = 20922

Thanks

Try the below one:

If you need the complete row, replace print $NF with print