Get execution time from log using bash

Question

I have some application that produce log file and at the end of file I have total processing time like:

Processing done in 45.00031 sec

I can get this line using:

cat app.log | grep "Processing done in"

But how to parse string to get only number of seconds?

45.00031

techraf · Accepted Answer · 2016-03-09 13:20:58Z

1

If the pattern is set you can just use cut to select the 4th field from a space-delimited table:

cat app.log | grep "Processing done in" | cut -d " " -f 4

answered Mar 9, 2016 at 13:11

techraf

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Comments

Peter K · Accepted Answer · 2016-03-09 13:03:21Z

0

grep again for only numbers and dot

cat app.log | grep "Processing done in" | grep -oE '[0-9.]+'

Be careful, it might stop working in another locale, that uses , as decimal separator.

answered Mar 9, 2016 at 13:03

Peter K

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shellter · Accepted Answer · 2016-03-09 13:27:59Z

0

Or, in 1 process instead of 3

 sed -n 's/^Pr.*in //;s/ sec//p' app.log

output

45.00031

IHTH

answered Mar 9, 2016 at 13:27

shellter

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