Re: Still don't understand timeout
Re: Still don't understand timeout
- Subject: Re: Still don't understand timeout
- From: Steve Mills <email@hidden>
- Date: Fri, 19 Aug 2011 14:16:58 -0500
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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