use the max count as the index
test if all booked hours exceeds max if not then list pa is complete
else
add together object hours to not exceed max and make each list pa, pb etc
ending up with
pa = [(object, object, object)]
pb = [(object, object, object, object)] etc
then unpack the tuples in the template with if statements to position as to time and duration
Would that work better?
What code would achieve it?
Appreciate any help here
Thanks for response I’ll try to answer as below in red
From: django...@googlegroups.com [mailto:django...@googlegroups.com] On Behalf Of Andrew Beales
Sent: 12 January 2017 19:52
To: Django users
Subject: Re: list? queryet? joining together
Hi, just a couple of follow-up questions as having trouble following precisely.
Can you re-state the exact queries you’re looking for?
Your comment about ‘folding’ model objects makes sense so,
I thought some more about it and am thinking about creating a dict of all bookings for a tool for the day Booking.objects.all().filter(resource=1).filter(date=now).order_by('start').values() and now considering how to manipulate that.
Example:
[{'name': u'Test', 'resource_id': 1L, 'hours': 6L, 'start': u'10', 'date': datetime.date(2017, 1, 13), 'id': 31L, 'resource_quantity_booked': 1L}, {'name': u'Me', 'resource_id': 1L, 'hours': 1L, 'start': u'8', 'date': datetime.date(2017, 1, 13), 'id': 33L, 'resource_quantity_booked': 1L}, {'name': u'Testing', 'resource_id': 1L, 'hours': 4L, 'start': u'9', 'date': datetime.date(2017, 1, 13), 'id': 32L, 'resource_quantity_booked': 1L}]
If it is possible to manipulate that dict (or object) to determine that tool1 can be expressed as a booking of 1 hour at 9 and then booked from 10 for 6 hours and therefore the next booking has to be applied to tool2
The wrinkle is that more than 1 tool can be booked to a job but the logic to determine tool2 is required to accommodate booking object 3 then similar logic should be able to determine that if all hours are filled on 2 objects therefore 2 tools are required.
Kinda hoping to use this manipulated dict in the view to send html to the template like:
<tr><td>8</td><td>9</td>…….etc</tr>
<tr><td>BOOKING 1 determined tool1 has a job here for 1 hour</td><td>BOOKING 2 determined that tool1 has a job here</td><td>tool1 still in use</td>…….etc until hours used then unbooked hours<td><a href=”book?time=x&resource=1&date={{now}}”>Book</a></td>… etc
And if there are no bookings for a tool on a day to just output rows of <td><a href=”book?time=x&resource=1&date={{now}}”>Book</a></td> for the currently unused tools just in case another tool is required.
Hope that makes better sense
- Are you looking to check if there are tools of a given type available at a given time? Yes, by outputting the bookings on the tools in a tabular format by time
- And whether a tool of a given type can be booked in a given time window? Yes
- When you said you wanted to “fold” 2 bookings into the same time window - intuitively at first glance a booking app would want to do the opposite, ie. find empty time windows - could you elaborate on these booking use cases? Is this for a single customer who wants to book multiple tools at a time? I agree, kinda given up on that idea
‘Customer’ the shop can either use a single tool on a job for any number of hours in a day and could also require more than 1 tool for any number of hours
- Sharing the Tool and Booking models would be useful.
class Resource(models.Model): // various tools
resource = models.CharField(max_length=30)
resource_capacity = models.IntegerField()
seating = models.IntegerField()
def __unicode__(self):
return self.resource
class Booking(models.Model):
resource = models.ForeignKey(Resource)
resource_quantity_booked = models.IntegerField()
start = models.CharField(max_length=5, help_text="whole number like 8, 9, 10, 11, 12, 13, 14, 15, 16, 17")
hours = models.IntegerField(max_length=2, help_text="Number of hours as a whole number like 3")
date = models.DateField()
class Meta:
ordering = ('date',)
This may be too basic, but for time queries if you’re not already aware there are the keyword lookups gt, gte (greater than, greater than or equal to) and lt, lte, for example:
now = timezone.now()
active_bookings_for_tool_type = Booking.objects.filter(tool__resource=2, start_time__lte=now, finish_time__gte=now)
and timedelta from the datetime module is useful for handling time intervals.
Aware of these keywords thanks
Also there are some open source django booking apps out there which look pretty good, for example django-booking: https://github.com/bitlabstudio/django-booking - could be worth looking into?
Unfortunately this is an old site written in 1.3.1 with a suite of custom tools developed for 1.3.1 and would have to be completely re-written entirely.. good thought though.
Another thought occurred that may or may not be practical as I it is just a thought, currently I have set up the models so a tool is assigned to a booking and am wondering about what the possibilities would be of reversing that and making a booking assigned to a tool?
On Tuesday, January 10, 2017 at 3:05:33 PM UTC, MikeKJ wrote:
May be another way of doing this?
5 tools @ 9 hours = 45 hours maximum possible utilisation
1 tool = 9 hours maximum possible utilisation
each tool is a list
pa = [] (tool1)
pb = [] (tool2) etc
max = 9
need to know the count of objects
loop all the booking objects us
e
the max count as the index
test if all booked hours exceeds max
if not then list pa is complete
else
add together object hours to not exceed max and make each list pa, pb etc
ending up with
pa = [(object, object, object)]
pb = [(object, object, object, object)] etc
then unpack the tuples in the template with if statements to position as to time and duration
Would that work better?
What code would ach
ie
ve it?
Appreciate any help here
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Thanks for response I’ll try to answer as below in red
Subject: Re: list? queryet? joining together
Hi, just a couple of follow-up questions as having trouble following precisely.
Can you re-state the exact queries you’re looking for?
Your comment about ‘folding’ model objects makes sense so,
I thought some more about it and am thinking about creating a dict of all bookings for a tool for the day Booking.objects.all().filter(resource=1).filter(date=now).order_by('start').values() and now considering how to manipulate that.
Example:
[{'name': u'Test', 'resource_id': 1L, 'hours': 6L, 'start': u'10', 'date': datetime.date(2017, 1, 13), 'id': 31L, 'resource_quantity_booked': 1L}, {'name': u'Me', 'resource_id': 1L, 'hours': 1L, 'start': u'8', 'date': datetime.date(2017, 1, 13), 'id': 33L, 'resource_quantity_booked': 1L}, {'name': u'Testing', 'resource_id': 1L, 'hours': 4L, 'start': u'9', 'date': datetime.date(2017, 1, 13), 'id': 32L, 'resource_quantity_booked': 1L}]
If it is possible to manipulate that dict (or object) to determine that tool1 can be expressed as a booking of 1 hour at 9 and then booked from 10 for 6 hours and therefore the next booking has to be applied to tool2
The wrinkle is that more than 1 tool can be booked to a job but the logic to determine tool2 is required to accommodate booking object 3 then similar logic should be able to determine that if all hours are filled on 2 objects therefore 2 tools are required.
Kinda hoping to use this manipulated dict in the view to send html to the template like:
<tr><td>8</td><td>9</td>…….etc</tr>
<tr><td>BOOKING 1 determined tool1 has a job here for 1 hour</td><td>BOOKING 2 determined that tool1 has a job here</td><td>tool1 still in use</td>…….etc until hours used then unbooked hours<td><a href=”book?time=x&resource=1&date={{now}}”>Book</a></td>… etc
And if there are no bookings for a tool on a day to just output rows of <td><a href=”book?time=x&resource=1&date={{now}}”>Book</a></td> for the currently unused tools just in case another tool is required.
Hope that makes better sense
- Are you looking to check if there are tools of a given type available at a
given time? Yes, by outputting the bookings on the
tools in a tabular format by time
- And whether a tool of a given type can be booked in a given time window? Yes
- When you said you wanted to “fold” 2 bookings into the same time
window - intuitively at first glance a booking app would want to do the
opposite, ie. find empty time windows - could you elaborate on these booking
use cases? Is this for a single customer who wants to book multiple tools at a
time? I agree,
kinda given up on that idea
‘Customer’ the shop can either use a single tool on a job for any number of hours in a day and could also require more than 1 tool for any number of hours
- Sharing the Tool and Booking models would be useful.
class Resource(models.Model): // various tools
resource = models.CharField(max_length=30)
resource_capacity = models.IntegerField()
seating = models.IntegerField()
def __unicode__(self):
return self.resource
class Booking(models.Model):
resource = models.ForeignKey(Resource)
resource_quantity_booked = models.IntegerField()
start = models.CharField(max_length=5, help_text="whole number like 8, 9, 10, 11, 12, 13, 14, 15, 16, 17")
hours = models.IntegerField(max_length=2, help_text="Number of hours as a whole number like 3")
date = models.DateField()
class Meta:
ordering = ('date',)
This may be too basic, but for time queries if you’re not already aware there
are the keyword lookups gt, gte (greater than, greater than or equal to) and
lt, lte, for example:
now = timezone.now()
active_bookings_for_tool_type = Booking.objects.filter(tool__resource=2,
start_time__lte=now, finish_time__gte=now)
and timedelta from the datetime module is useful for handling time intervals.
Aware of these keywords thanks
Also there are some open source django booking apps out there which look pretty
good, for example django-booking: https://github.com/bitlabstudio/django-booking
- could be worth looking into?
Unfortunately this is an old site written in 1.3.1 with a suite of custom tools developed for 1.3.1 and would have to be completely re-written entirely.. good thought though.