Setting Up a Meeting

Last modified: 05/20/2006 07:57 AM

Revision:	meeting.txt 31197 2006-01-02 18:56:03Z dkuhlman

Contents

1 Blocked periods
2 Free periods
3 Recurrence
4 All-day events
5 Multi-day events
6 Transparent events

In this document we'll try to set up a meeting between various extremely busy people.

First, we're going to set up a calendar storage and create some people:

>>> from calcore import cal
>>> m = cal.StorageManager()
>>> m.setStorage(cal.MemoryStorage('storage'))
>>> s = cal.SimpleAttendeeSource(m)
>>> martijn = s.createIndividual('martijn', 'Martijn')
>>> lennart = s.createIndividual('lennart', 'Lennart')
>>> florent = s.createIndividual('florent', 'Florent')
>>> eric = s.createIndividual('eric', 'Eric')

1 Blocked periods

Now let's make them busy in the period march 9 till march 11:

>>> from datetime import datetime, timedelta, time
>>> a = martijn.createEvent(
...     dtstart=datetime(2005, 3, 10, 0, 0),
...     duration=timedelta(days=1),
...     status='TENTATIVE',
...     title="Martijn's day off")
>>> b = florent.createEvent(
...     dtstart=datetime(2005, 3, 9, 15, 00),
...     duration=timedelta(minutes=120),
...     title='Some meeting')
>>> c = eric.createEvent(
...     dtstart=datetime(2005, 3, 9, 10, 00),
...     duration=timedelta(minutes=90),
...     title='Another meeting')
>>> d = lennart.createEvent(
...     dtstart=datetime(2005, 3, 11, 11, 00),
...     duration=timedelta(minutes=60),
...     title='Yet another meeting')
>>> e = martijn.createEvent(
...     dtstart=datetime(2005, 3, 9, 15, 30),
...     duration=timedelta(minutes=30),
...     title='Another overlapping meeting')
>>> c.invite([lennart, florent])
>>> d.invite([florent])

The time is blocked off during five periods:

>>> people = [martijn, lennart, florent, eric]
>>> period = (datetime(2005, 3, 9), datetime(2005, 3, 12))
>>> time_period = (time(9, 0), time(17, 0))
>>> periods = m.getBlockedPeriods(people, period, time_period)
>>> len(periods)
5

On day 9, 0:00-9:00, blocked off start of the day:

>>> periods[0]
(datetime.datetime(2005, 3, 9, 0, 0), datetime.datetime(2005, 3, 9, 9, 0))

On day 9, 10:00-11:30:

>>> periods[1]
(datetime.datetime(2005, 3, 9, 10, 0), datetime.datetime(2005, 3, 9, 11, 30))

From day 9 15:00 until the start of day 11:

>>> periods[2]
(datetime.datetime(2005, 3, 9, 15, 0), datetime.datetime(2005, 3, 11, 9, 0))

On day 11, from 11:00 until 12:00:

>>> periods[3]
(datetime.datetime(2005, 3, 11, 11, 0), datetime.datetime(2005, 3, 11, 12, 0))

Finally, on day 11, from 17:00 until the end of the day:

>>> periods[4]
(datetime.datetime(2005, 3, 11, 17, 0), datetime.datetime(2005, 3, 12, 0, 0))

2 Free periods

Now let's ask for all the time periods free on march 9 till 11, between 9:00 till 17:00:

>>> periods = m.getFreePeriods(people, period, time_period)
>>> len(periods)
4

On day 9, 9:00-10:00:

>>> periods[0]
(datetime.datetime(2005, 3, 9, 9, 0), datetime.datetime(2005, 3, 9, 10, 0))

On day 9, 11:30-15:00:

>>> periods[1]
(datetime.datetime(2005, 3, 9, 11, 30), datetime.datetime(2005, 3, 9, 15, 0))

On day 11, from 9:00 until 11:00:

>>> periods[2]
(datetime.datetime(2005, 3, 11, 9, 0), datetime.datetime(2005, 3, 11, 11, 0))

And again on day 11, from 12:00 until 17:00:

>>> periods[3]
(datetime.datetime(2005, 3, 11, 12, 0), datetime.datetime(2005, 3, 11, 17, 0))

Now let's throw away these events again:

>>> m.deleteEvent(a)
>>> m.deleteEvent(b)
>>> m.deleteEvent(c)
>>> m.deleteEvent(d)
>>> m.deleteEvent(e)
>>> m.getEvents((datetime(2005, 3, 9), datetime(2005, 3, 12)))
[]

Let's try another scenario. Let's schedule some events for day 9 in march:

>>> a = martijn.createEvent(
...    dtstart=datetime(2005, 3, 9, 9, 0), duration=timedelta(minutes=60),
...    title="A")
>>> b = martijn.createEvent(
...    dtstart=datetime(2005, 3, 9, 10, 15), duration=timedelta(minutes=60),
...    title="B")
>>> c = martijn.createEvent(
...    dtstart=datetime(2005, 3, 9, 12, 0), duration=timedelta(minutes=90),
...    title="C")
>>> d = martijn.createEvent(
...    dtstart=datetime(2005, 3, 9, 16, 0), duration=timedelta(minutes=120),
...    title="D")

We're going to see what free time this means for that day, from 9:00 till 17:00:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 3, 9), datetime(2005, 3, 10)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
3
>>> periods[0]
(datetime.datetime(2005, 3, 9, 10, 0), datetime.datetime(2005, 3, 9, 10, 15))
>>> periods[1]
(datetime.datetime(2005, 3, 9, 11, 15), datetime.datetime(2005, 3, 9, 12, 0))
>>> periods[2]
(datetime.datetime(2005, 3, 9, 13, 30), datetime.datetime(2005, 3, 9, 16, 0))

Now we'll select only the times that are longer than 30 minutes:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 3, 9), datetime(2005, 3, 10)),
...   (time(9, 0), time(17, 0)),
...   timedelta(minutes=30))
>>> len(periods)
2
>>> periods[0]
(datetime.datetime(2005, 3, 9, 11, 15), datetime.datetime(2005, 3, 9, 12, 0))
>>> periods[1]
(datetime.datetime(2005, 3, 9, 13, 30), datetime.datetime(2005, 3, 9, 16, 0))

Now let's look for the free periods longer than 60 minutes:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 3, 9), datetime(2005, 3, 10)),
...   (time(9, 0), time(17, 0)),
...   timedelta(minutes=60))
>>> len(periods)
1
>>> periods[0]
(datetime.datetime(2005, 3, 9, 13, 30), datetime.datetime(2005, 3, 9, 16, 0))

Let's look for the free periods longer than 3 hours; we expect to get nothing:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 3, 9), datetime(2005, 3, 10)),
...   (time(9, 0), time(17, 0)),
...   timedelta(minutes=180))
>>> len(periods)
0

3 Recurrence

Let's check whether this works with recurrent events. Here's recurrent event for one participant:

>>> from calcore import recurrent
>>> a = martijn.createEvent(
...    dtstart=datetime(2005, 6, 9, 11, 0), duration=timedelta(minutes=60),
...    title="A",
...    recurrence=recurrent.DailyRecurrenceRule())

Now let's get free time over day 9 and 10. We should get free time for everything but 11:00 - 12:00 on both days:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 6, 9), datetime(2005, 6, 11)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
4
>>> periods[0]
(datetime.datetime(2005, 6, 9, 9, 0), datetime.datetime(2005, 6, 9, 11, 0))
>>> periods[1]
(datetime.datetime(2005, 6, 9, 12, 0), datetime.datetime(2005, 6, 9, 17, 0))
>>> periods[2]
(datetime.datetime(2005, 6, 10, 9, 0), datetime.datetime(2005, 6, 10, 11, 0))
>>> periods[3]
(datetime.datetime(2005, 6, 10, 12, 0), datetime.datetime(2005, 6, 10, 17, 0))

Let's throw away the recurrent event again:

>>> m.deleteEvent(a)

4 All-day events

We're going to create a two events, one of which is all day:

>>> from calcore import recurrent
>>> a = martijn.createEvent(
...    dtstart=datetime(2005, 7, 9, 11, 0), duration=timedelta(minutes=60),
...    title="A")
>>> b = martijn.createEvent(
...    dtstart=datetime(2005, 7, 11), duration=timedelta(days=1),
...    title="B", allday=True)

Now let's get free time over day 9 - 12:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 7, 9), datetime(2005, 7, 13)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
4
>>> periods[0]
(datetime.datetime(2005, 7, 9, 9, 0), datetime.datetime(2005, 7, 9, 11, 0))
>>> periods[1]
(datetime.datetime(2005, 7, 9, 12, 0), datetime.datetime(2005, 7, 9, 17, 0))
>>> periods[2]
(datetime.datetime(2005, 7, 10, 9, 0), datetime.datetime(2005, 7, 10, 17, 0))
>>> periods[3]
(datetime.datetime(2005, 7, 12, 9, 0), datetime.datetime(2005, 7, 12, 17, 0))

5 Multi-day events

We're going to create a multi-day spanning event:

>>> from calcore import recurrent
>>> a = martijn.createEvent(
...    dtstart=datetime(2005, 8, 9, 14, 0), duration=timedelta(hours=23),
...    title="A")

Now let's get free time over day 9 - 11:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 8, 9), datetime(2005, 8, 12)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
3
>>> periods[0]
(datetime.datetime(2005, 8, 9, 9, 0), datetime.datetime(2005, 8, 9, 14, 0))
>>> periods[1]
(datetime.datetime(2005, 8, 10, 13, 0), datetime.datetime(2005, 8, 10, 17, 0))
>>> periods[2]
(datetime.datetime(2005, 8, 11, 9, 0), datetime.datetime(2005, 8, 11, 17, 0))

Let's make an event that spans 3 days:

>>> from calcore import recurrent
>>> a = martijn.createEvent(
...    dtstart=datetime(2005, 9, 9, 14, 0), duration=timedelta(hours=47),
...    title="A")

Now let's get free time over day 9 - 11:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2005, 9, 9), datetime(2005, 9, 12)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
2
>>> periods[0]
(datetime.datetime(2005, 9, 9, 9, 0), datetime.datetime(2005, 9, 9, 14, 0))
>>> periods[1]
(datetime.datetime(2005, 9, 11, 13, 0), datetime.datetime(2005, 9, 11, 17, 0))

6 Transparent events

Let's create a transparent event:

>>> a = martijn.createEvent(
...    dtstart=datetime(2006, 1, 1, 14, 0), duration=timedelta(hours=4),
...    title="A",
...    transparent=True)

It won't block a period for Martijn:

>>> periods = m.getFreePeriods(
...   [martijn],
...   (datetime(2006, 1, 1), datetime(2006, 1, 2)),
...   (time(9, 0), time(17, 0)))
>>> len(periods)
1
>>> periods[0]
(datetime.datetime(2006, 1, 1, 9, 0), datetime.datetime(2006, 1, 1, 17, 0))