I have three models, Which has some common but not exact fields, from a single view ex. home I am retrieving like this
interviews = Interviews.objects.all().order_by('pub_date')[:3]
publications = Publications.objects.all().order_by('pub_date')[:3]
published = Published.objects.all().order_by('pub_date')[:3]
From home view I want them to show in template in such order that all the latest/New entries associated with these models will be in top.
like if interview entry 10 is the most recent entry in these all models, then it will be first, then if the published one is second recent it will be second ... etc .etc
Can any one tell me how to do it?
Depending on your other requirements it may be a good idea to make them into sublcasses of a common superclass - containing the common elements.
It's hard to say if it's valid, but if you do you can query the different types of objects separately or together by calling
SuperClass.objects.all().order_by('-pub_date')[:9]
which will render the 9 first objects regerdless of what sublcass they are. Of course assuming the superclass is named SuperClass. Of course this does not insure that there are 3 of each model.
Another simple way of solving it - although admittedly not using a query would simply be to sort the lists.
entries = sorted(list(interviews) + list(publications) + list(published), key=lambda x: x.pub_date, reverse=True)
should work - basically turning them into lists and sorting them.
One possible way is using lambda to sort data, but costs are hiher since you will doing this to python, not DBMS...
lst = []
lst.extend(list(Interviews.objects.order_by('pub_date')[:10]))
lst.extend(list(Publications.objects.order_by('pub_date')[:10]))
lst.extend(list(Published.objects.order_by('pub_date')[:10]))
# take 10 records for each, since you could not know how many records will be picked from which table
# now order them...
lst.sort(lambda x, y: cmp(x.pub_date, y.pub_date))
# and reverse the order, so newset is the first...
lst.reverse()
this will give you a list of objects ordered py pub_date, so you can slice the final list to get any number of records you want...
lst = lst[:10]
Related
Can anyone help, I want to return an ordered list based on forloop in Django using a field in the model that contains both integer and string in the format MM/1234. The loop should return the values with the least interger(1234) in ascending order in the html template.
Ideally you want to change the model to have two fields, one integer and one string, so you can code a queryset with ordering based on the integer one. You can then define a property of the model to return the self.MM+"/"+str( self.nn) composite value if you often need to use that. But if it's somebody else's database schema, this may not be an option.
In which case you'll have to convert your queryset into a list (which reads all the data rows at once) and then sort the list in Python rather than in the database. You can run out of memory or bring your server to its knees if the list contains millions of objects. count=qs.count() is a DB operation that won't.
qs = Foo.objects.filter( your_selection_criteria)
# you might want to count this before the next step, and chicken out if too many
# also this simple key function will crash if there's ever no "/" in that_field
all_obj = sorted( list( qs),
key = lambda obj: obj.that_field.split('/')[1] )
I'm implementing search functionality with an option of looking for a record by matching multiple tables and multiple fields in these tables.
Say I want to find a Customer by his/her first or last name, or by ID of placed Order which is stored in different model than Customer.
The easy scenario which I already implemented is that a user only types single word into search field, I then use Django Q to query Order model using direct field reference or related_query_name reference like:
result = Order.objects.filter(
Q(customer__first_name__icontains=user_input)
|Q(customer__last_name__icontains=user_input)
|Q(order_id__icontains=user_input)
).distinct()
Piece of a cake, no problems at all.
But what if user wants to narrow the search and types multiple words into search field.
Example: user has typed Bruce and got a whole lot of records back as a result of search.
Now he/she wants to be more specific and adds customer's last name to search.So the search becomes Bruce Wayne, after splitting this into separate parts I'm having Bruce and Wayne. Obviously I don't want to search Orders model because order_id is a single-word instance and it's sufficient to find customer at once so for this case I'm dropping it out of query at all.
Now I'm trying to match customer by both first AND last name, I also want to handle the scenario where the order of provided data is random, to properly handle Bruce Wayne and Wayne Bruce, meaning I still have customers full name but the position of first and last name aren't fixed.
And this is the question I'm looking answer for: how to build query that will search multiple fields of model not knowing which of search words belongs to which table.
I'm guessing the solution is trivial and there's for sure an elegant way to create such a dynamic query, but I can't think of a way how.
You can dynamically OR a variable number of Q objects together to achieve your desired search. The approach below makes it trivial to add or remove fields you want to include in the search.
from functools import reduce
from operator import or_
fields = (
'customer__first_name__icontains',
'customer__last_name__icontains',
'order_id__icontains'
)
parts = []
terms = ["Bruce", "Wayne"] # produce this from your search input field
for term in terms:
for field in fields:
parts.append(Q(**{field: term}))
query = reduce(or_, parts)
result = Order.objects.filter(query).distinct()
The use of reduce combines the Q objects by ORing them together. Credit to that part of the answer goes to this answer.
The solution I came up with is rather complex, but it works exactly the way I wanted to handle this problem:
search_keys = user_input.split()
if len(search_keys) > 1:
first_name_set = set()
last_name_set = set()
for key in search_keys:
first_name_set.add(Q(customer__first_name__icontains=key))
last_name_set.add(Q(customer__last_name__icontains=key))
query = reduce(and_, [reduce(or_, first_name_set), reduce(or_, last_name_set)])
else:
search_fields = [
Q(customer__first_name__icontains=user_input),
Q(customer__last_name__icontains=user_input),
Q(order_id__icontains=user_input),
]
query = reduce(or_, search_fields)
result = Order.objects.filter(query).distinct()
I looked at this thread, with an example of sorting dictionaries.
I have a dictionary of programme objects where the key is a programme object and the value is a lookup of the number of related Project objects.
def DepartmentDetail(request, pk):
department = Department.objects.get(pk=pk)
programmes = Programme.objects.all().filter(department=department).exclude(active=False).order_by('long_name')
combi = {}
for p in programmes:
prj = Project.objects.all().filter(programme=p)
combi[p] = str(len(prj))
return render(request, 'sysadmin/department.html',{'department': department, 'programmes': programmes, 'combi': sorted(combi.items())})
In the model, Programme returns a string 'long_name', so I believe that I am trying to sort a string key and a string value.
In the template I get to the keys and values as so,
{% for programme, n in combi %}
This gives me the error..
unorderable types: Programme() < Programme()
I don't really understand the error, in the python 3 documentation it states that the sorted() method accepts any iterable - So why does this happen?
I'm looking at collections.OrderedDict to solve the problem, but I want to know why this doesn't work.
Thanx.
Databases with index on columns are really good at sorting. There's almost never a need to sort on the client side. You can almost always do it in the server. The funny part it you apparently know how to do it too.
....exclude(active=False).order_by('long_name') # <--- this
Guess what, your data is already sorted there isn't a need to sort it again inside python!!
But there is a much bigger issue in your code. You are fetching a set of Project items and then looping through that set to retrieve them all over again one by one. So if you have 200 Project items you are doing 200 queries when one query does the job just as well. Just add select_related or prefetch_related depending on which direction you have the relationship.
Your code ideally should be something like this
department = Department.objects.get(pk=pk)
programmes = Programme.objects.all().filter(department=department).exclude(active=False).order_by('long_name')
return render(request, 'sysadmin/department.html',{'department': department, 'programmes': programmes,})
As far as I can see combi just contains duplicated data. The same thing can be accessed from programmes eg. programme.project_set.all()
(again this depends on which direction you have the relationship, your models are not shown)
Recommended reading: https://docs.djangoproject.com/en/1.10/ref/models/fields/#django.db.models.ForeignKey
The issue is that sorted expects a way to be able to order the items and by default there isn't any way to know how to order your objects. You can provide a key
sorted(combi.items(), key=lambda i: i.long_name)
I have a Django model called Collection that represents a collection of items (CollectionItem). Each Collection only contains items of a specific type. (CollectionItem has a foreign key to Collection).
I want to get all of the CollectionItems that are in public-flagged lists of a specific type and return them sorted by a particular field. Here's the query code that I use:
lists = Collection.objects.filter(is_public=True, type=7)
items = CollectionItem.objects.none()
for list in lists:
items |= CollectionItem.objects.filter(collection=list)
items = items.order_by('name')
I have to imagine that this will not scale well at all when I have a large database with tons of lists and items. Is there a better way to do this in Django? Or is the inefficiency involved in the query loop negligible enough compared to other options that I shouldn't worry too much?
Sounds like you just need:
items = CollectionItem.objects.filter(
collection__is_public=True, collection__type=7
).order_by('name')
I have a two models:
class Category(models.Model):
pass
class Item(models.Model):
cat = models.ForeignKey(Category)
I am trying to return all Categories for which all of that category's items belong to a given subset of item ids (fixed thanks). For example, all categories for which all of the items associated with that category have ids in the set [1,3,5].
How could this be done using Django's query syntax (as of 1.1 beta)? Ideally, all the work should be done in the database.
Category.objects.filter(item__id__in=[1, 3, 5])
Django creates the reverse relation ship on the model without the foreign key. You can filter on it by using its related name (usually just the model name lowercase but it can be manually overwritten), two underscores, and the field name you want to query on.
lets say you require all items to be in the following set:
allowable_items = set([1,3,4])
one bruteforce solution would be to check the item_set for every category as so:
categories_with_allowable_items = [
category for category in
Category.objects.all() if
set([item.id for item in category.item_set.all()]) <= allowable_items
]
but we don't really have to check all categories, as categories_with_allowable_items is always going to be a subset of the categories related to all items with ids in allowable_items... so that's all we have to check (and this should be faster):
categories_with_allowable_items = set([
item.category for item in
Item.objects.select_related('category').filter(pk__in=allowable_items) if
set([siblingitem.id for siblingitem in item.category.item_set.all()]) <= allowable_items
])
if performance isn't really an issue, then the latter of these two (if not the former) should be fine. if these are very large tables, you might have to come up with a more sophisticated solution. also if you're using a particularly old version of python remember that you'll have to import the sets module
I've played around with this a bit. If QuerySet.extra() accepted a "having" parameter I think it would be possible to do it in the ORM with a bit of raw SQL in the HAVING clause. But it doesn't, so I think you'd have to write the whole query in raw SQL if you want the database doing the work.
EDIT:
This is the query that gets you part way there:
from django.db.models import Count
Category.objects.annotate(num_items=Count('item')).filter(num_items=...)
The problem is that for the query to work, "..." needs to be a correlated subquery that looks up, for each category, the number of its items in allowed_items. If .extra had a "having" argument, you'd do it like this:
Category.objects.annotate(num_items=Count('item')).extra(having="num_items=(SELECT COUNT(*) FROM app_item WHERE app_item.id in % AND app_item.cat_id = app_category.id)", having_params=[allowed_item_ids])