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Solve : measuring elapsed time of a task evoked in a dos batch file?

Answer» <html><body><p>Antonio <a href="https://interviewquestions.tuteehub.com/tag/posted-2944562" style="font-weight:bold;" target="_blank" title="Click to know more about POSTED">POSTED</a> a 2nd link that talks about the 24 hour problem.<br/>Dave just added <a href="https://interviewquestions.tuteehub.com/tag/one-241053" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> line of code.<br/><a href="http://www.dostips.com/forum/viewtopic.php?f=3&amp;t=6271">http://www.dostips.com/forum/viewtopic.php?f=3&amp;t=6271</a><br/> Quote from: miskox</p><blockquote>What would be the best solution if I <a href="https://interviewquestions.tuteehub.com/tag/start-239994" style="font-weight:bold;" target="_blank" title="Click to know more about START">START</a> the process before midnight and when time changes from 23:xx:xx to 00:xx:00 to have a <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> time difference? Maybe adding 24 to hours? Or maybe adding dates to start time and end time?<br/><br/>Saso</blockquote> Quote from: dbenham<blockquote>Yes, assuming your process takes less than 24 hours to complete, then all you need to do is conditionally at 24 hours if elapsed is less than 0.<br/> Code: <a>[Select]</a>if %elapsed% lss 0 set /a elapsed+=24*60*60*100<br/><br/>Dave Benham</blockquote></body></html>


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