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Format of date

2002-12-19T11:07:48-04:00

Ok, test it with

Code:

proc pub:test {nick host hand chan arg} {set tomorrow [clock scan "tomorrow"]set dayaftertomorrow [expr $tomorrow + (24 * 60 * 60)]putserv "PRIVMSG $chan : Tomorrow: [clock format $tomorrow -format "%b %d"] "putserv "PRIVMSG $chan : The day after: [clock format $dayaftertomorrow -format "%b %d"] "}

...and it works

Statistics: Posted by Guest — Thu Dec 19, 2002 11:07 am

Re: Format of date

2002-12-18T23:23:00-04:00

Hi,

today is eg 12/19/2002 then the following line
Code:
clock format [clock seconds] -format "%b %d"
will post Dec 19

so far so good....

But how must the line looks like to get the date from tormorrow (Dec 20) and the day after (Dec 21)?

Thanks
jagg

Well if [clock seconds] is today, and knowing that [clock seconds] returns the time in unix format (that being the number of seconds since a certain date), then logically it would follow you would want [clock seconds] + (24 * 60 * 60) [where 24 is hours in a day, 60 is minutes in an hour, and 2nd 60 is the seconds in a minute -> which gives the number of seconds in a day]..

So to find the date x amount of days after today you would use the formula... [clock seconds] + ($x * 24 * 60 * 60) ...

It's just simple math.

I've given you the pseudo code, I'll let you figure out the actual code, which isn't much to change.

Statistics: Posted by strikelight — Wed Dec 18, 2002 11:23 pm

Format of date

2002-12-18T22:24:47-04:00

Hi,

today is eg 12/19/2002 then the following line

Code:

clock format [clock seconds] -format "%b %d"

will post Dec 19

so far so good....

But how must the line looks like to get the date from tormorrow (Dec 20) and the day after (Dec 21)?

Thanks
jagg

Statistics: Posted by Guest — Wed Dec 18, 2002 10:24 pm