CMD measure command execution time

Question

I'm stuck trying to print a custom command execution time. If I try to do something like this:

echo %time% & START "TestAgente" /W systeminfo  & echo %time%

I got same begin and end time:

I have tried with some combinations playing with delayed-expansion, but I think it fails (I guess) because in running 3 commands in a line instead of batch file.

Does anyone done something like this (without PowerShell). It has to be a single command line that someone could copy, paste and execute.

I read that before posting, but I couldn't find a proper answer for my problem, sorry. Could you please tell me what's the proper way to fix it in a single line without using power shell? — Carol
– Carol, Commented Oct 5, 2018 at 17:08
cmd /v:on /c "echo !TIME! & timeout 5 & echo !TIME!" - by Nathan Herring. — Stephan
– Stephan, Commented Oct 5, 2018 at 17:17
Possible duplicate of How do I measure execution time of a command on the Windows command line? — Stephan
– Stephan, Commented Oct 6, 2018 at 15:46

Carol · Accepted Answer · 2018-10-05 17:18:43Z

1

Like @Stephan said, the right way was in that post, and it worked!!

cmd /v:on /c "echo !time! & START "TestAgente" /W systeminfo & echo !time!"

answered Oct 5, 2018 at 17:18

Carol

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1 Comment

Stephan Over a year ago

Note: Start opens another cmd instance, which needs time, which gives you (slightly) wrong results. More accurate: cmd /v:on /c "echo !time! & systeminfo & echo !time!"

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