Re: Still don't understand timeout
Re: Still don't understand timeout
- Subject: Re: Still don't understand timeout
- From: Emmanuel LEVY <email@hidden>
- Date: Fri, 19 Aug 2011 21:21:08 +0200
My 2 cents, to get rid of the past-midnight issue, if (nowTime -
startTime) ≥ 2 then exit repeat would work as well.
Emmanuel
On Aug 19, 2011, at 9:16 PM, Steve Mills wrote:
On Aug 19, 2011, at 14:01:13, Robert Poland wrote:
If I understand right "with timeout" doesn't end and stop the
block. Is there something that will do that?
A timeout is a per-command timeout. So if a *single* command doesn't
return control back to the script within the timeout, then it'll
force it to return control to the script. The same mechanism is
provided as a parameter to the "display dialog" command (giving up
after). Try it and you'll see.
If you want to force a loop to timeout once a given time limit is
reached, you'll have to measure the time limit yourself.
set startTime to current date
repeat while true
set nowTime to current date
if (time of nowTime) - (time of startTime) ≥ 2 then exit repeat
say "z"
end repeat
The above won't work if startTime is before midnight and nowTime is
after midnight. If you anticipate that happening, add the day of the
month multiplied by the number of seconds in a month. But then
that'll break if you start on the 31st and cross midnight into the
1st of the next month. Didn't there used to be a command called "the
ticks" that returned the number of ticks since system boot? That was
most useful.
--
Steve Mills
Drummer, Mac geek
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