User user = User.find("byName", name).).).first();
what is the use of .first() method.
Almost certainly because, when you request a name, there may be duplicates so it gives you a list of records (try to think of searching for a name like "John Smith").
The .first() would then give you the first record from that list.
Related
I have a Feed model with the following field:
class Feed(models.Model)
authority=models.ForeignKey(Authority,blank=True,null=True)
I have a queryset of authority called followed_authority in which I want to get the corresponding feeds from each of the authority in followed_authority
The obvious thing for me is to use a for loop through followed_authority which I think its inefficient as the instanaces in following_authority and their corresponding feeds are very large.Kindly help me out
The correct thing to do is to always start from the model you want to get.
Feed.objects.filter(authority__in=followed_authority)
I have this query:
checkins = CheckinAct.objects.filter(time__range=[start, end], location=checkin.location)
Which works great for telling me how many checkins have happened in my date range for a specific location. But I want know how many checkins were done by unique users. So I tried this:
checkins = CheckinAct.objects.filter(time__range=[start, end], location=checkin.location).values('user').distinct()
But that doesn't work, I get back an empty Array. Any ideas why?
Here is my CheckinAct model:
class CheckinAct(models.Model):
user = models.ForeignKey(User)
location = models.ForeignKey(Location)
time = models.DateTimeField()
----Update------
So now I have updated my query to look like this:
checkins = CheckinAct.objects.values('user').\
filter(time__range=[start, end], location=checkin.location).\
annotate(dcount=Count('user'))
But I'm still getting multiple objects back that have the same user, like so:
[{'user': 15521L}, {'user': 15521L}, {'user': 15521L}, {'user': 15521L}, {'user': 15521L}]
---- Update 2------
Here is something else I tried, but I'm still getting lots of identical user objects back when I log the checkins object.
checkins = CheckinAct.objects.filter(
time__range=[start, end],
location=checkin.location,
).annotate(dcount=Count('user')).values('user', 'dcount')
logger.info("checkins!!! : " + str(checkins))
Logs the following:
checkins!!! : [{'user': 15521L}, {'user': 15521L}, {'user': 15521L}]
Notice how there are 3 instances of the same user object. Is this working correctly or not? Is there a difference way to read out what comes back in the dict object? I just need to know how many unique users check into that specific location during the time range.
The answer is actually right in the Django docs. Unfortunately, very little attention is drawn to the importance of the particular part you need; so it's understandably missed. (Read down a little to the part dealing with Items.)
For your use-case, the following should give you exactly what you want:
checkins = CheckinAct.objects.filter(time__range=[start,end], location=checkin.location).\
values('user').annotate(checkin_count=Count('pk')).order_by()
UPDATE
Based on your comment, I think the issue of what you wanted to achieve has been confused all along. What the query above gives you is a list of the number of times each user checked in at a location, without duplicate users in said list. It now seems what you really wanted was the number of unique users that checked in at one particular location. To get that, use the following (which is much simpler anyways):
User.objects.filter(checkinat__location=location).distinct().count()
UPDATE for non-rel support
checkin_users = [(c.user.pk, c.user) for c in CheckinAct.objects.filter(location=location)]
unique_checkins = len(dict(checkin_users))
This works off the principle that dicts have unique keys. So when you convert the list of tuples to a dict, you end up with a list of unique users. But, this will generate 1*N queries, where N is the total amount of checkins (one query each time the user attribute is used. Normally, I'd do something like .select_related('user'), but that too requires a JOIN, which is apparently out. JOINs not being supported seems like a huge downside to non-rel, if true, but if that's the case this is going to be your only option.
You don't want DISTINCT. You actually want Django to do something that will end up giving you a GROUP BY clause. You are also correct that your final solution is to combine annotate() and values(), as discussed in the Django documentation.
What you want to do to get your results is to use annotate first, and then values, such as:
CheckinAct.objects.filter(
time__range=[start, end],
location=checkin.location,
).annotate(dcount=Count('user').values('user', 'dcount')
The Django docs at the link I gave you above show a similarly constructed query (minus the filter aspect, which I added for your case in the proper location), and note that this will "now yield one unique result for each [checkin act]; however, only the [user] and the [dcount] annotation will be returned in the output data". (I edited the sentence to fit your case, but the principle is the same).
Hope that helps!
checkins = CheckinAct.objects.values('user').\
filter(time__range=[start, end], location=checkin.location).\
annotate(dcount=Count('user'))
If I am not mistaken, wouldn't the value you want be in the input as "dcount"? As a result, isn't that just being discarded when you decide to output the user value alone?
Can you tell me what happens when you try this?
checkins = CheckinAct.objects.values('user').\
filter(time__range=[start, end], location=checkin.location).\
annotate(Count('user')).order_by()
(The last order_by is to clear any built-in ordering that you may already have at the model level - not sure if you have anything like that, but doesn't hurt to ask...)
How can I retrieve the last record in a certain queryset?
Django Doc:
latest(field_name=None) returns the latest object in the table, by date, using the field_name provided as the date field.
This example returns the latest Entry in the table, according to the
pub_date field:
Entry.objects.latest('pub_date')
EDIT : You now have to use Entry.objects.latest('pub_date')
You could simply do something like this, using reverse():
queryset.reverse()[0]
Also, beware this warning from the Django documentation:
... note that reverse() should
generally only be called on a QuerySet
which has a defined ordering (e.g.,
when querying against a model which
defines a default ordering, or when
using order_by()). If no such ordering
is defined for a given QuerySet,
calling reverse() on it has no real
effect (the ordering was undefined
prior to calling reverse(), and will
remain undefined afterward).
The simplest way to do it is:
books.objects.all().last()
You also use this to get the first entry like so:
books.objects.all().first()
To get First object:
ModelName.objects.first()
To get last objects:
ModelName.objects.last()
You can use filter
ModelName.objects.filter(name='simple').first()
This works for me.
Django >= 1.6
Added QuerySet methods first() and last() which are convenience methods returning the first or last object matching the filters. Returns None if there are no objects matching.
When the queryset is already exhausted, you may do this to avoid another db hint -
last = queryset[len(queryset) - 1] if queryset else None
Don't use try...except....
Django doesn't throw IndexError in this case.
It throws AssertionError or ProgrammingError(when you run python with -O option)
You can use Model.objects.last() or Model.objects.first().
If no ordering is defined then the queryset is ordered based on the primary key. If you want ordering behaviour queryset then you can refer to the last two points.
If you are thinking to do this, Model.objects.all().last() to retrieve last and Model.objects.all().first() to retrieve first element in a queryset or using filters without a second thought. Then see some caveats below.
The important part to note here is that if you haven't included any ordering in your model the data can be in any order and you will have a random last or first element which was not expected.
Eg. Let's say you have a model named Model1 which has 2 columns id and item_count with 10 rows having id 1 to 10.[There's no ordering defined]
If you fetch Model.objects.all().last() like this, You can get any element from the list of 10 elements. Yes, It is random as there is no default ordering.
So what can be done?
You can define ordering based on any field or fields on your model. It has performance issues as well, Please check that also. Ref: Here
OR you can use order_by while fetching.
Like this: Model.objects.order_by('item_count').last()
If using django 1.6 and up, its much easier now as the new api been introduced -
Model.object.earliest()
It will give latest() with reverse direction.
p.s. - I know its old question, I posting as if going forward someone land on this question, they get to know this new feature and not end up using old method.
In a Django template I had to do something like this to get it to work with a reverse queryset:
thread.forumpost_set.all.last
Hope this helps someone looking around on this topic.
MyModel.objects.order_by('-id')[:1]
If you use ids with your models, this is the way to go to get the latest one from a qs.
obj = Foo.objects.latest('id')
You can try this:
MyModel.objects.order_by('-id')[:1]
The simplest way, without having to worry about the current ordering, is to convert the QuerySet to a list so that you can use Python's normal negative indexing. Like so:
list(User.objects.all())[-1]
I have some codes like this:
cats = Category.objects.filter(is_featured=True)
for cat in cats:
entries = Entry.objects.filter(score>=10, category=cat).order_by("-pub_date")[:10]
But, the results just show the last item of cats and also have problems with where ">=" in filter. Help me solve these problems. Thanks so much!
You may want to start by reading the django docs on this subject. However, just to get you started, the filter() method is just like any other method, in that it only takes arguments and keyword args, not expressions. So, you can't say foo <= bar, just foo=bar. Django gets around this limitation by allowing keyword names to indicate the relationship to the value you pass in. In your case, you would want to use:
Entry.objects.filter(score__gte=10)
The __gte appended to the field name indicates the comparison to be performed (score >= 10).
Your not appending to entries on each iteration of the for loop, therefore you only get the results of the last category. Try this:
entries = Entry.objects.filter(score__gte=10, category__is_featured=True).order_by("-pub_date")[:10]
Recently i have implemented django-sphinx search on my website.
It is working fine of each separate model.
But now my client requirement has changed.
To implement that functionality i need field name to whom search is made.
suppose my query is:
"select id, name,description from table1"
and search keyword is matched with value in field "name". So i need to return that field also.
Is it possible to get field name or any method provided by django-sphinx which return field name.
Please help me...
As far as I know, this isn't possible. You might look at the contents of _sphinx though.
Well from django-sphinx it might not be possible. But there is a solution -
Make different indexes, each index specifying the field that you need to search.
In your django-sphinx models while searching do this -
search1 = SphinxSearch(index='index1')
search2 = SphinxSearch(index='index2')
...
After getting all the search results, you aggregate them & you have the info of from where they have come.