i have one model userprofile in my django project and mobile_number is one for the modelfiled in that model. i want to get all the userprofiles where length of mobile_number is less than 10.
i know the solution when field is CharField
qs = UserProfile.objects.annotate(text_len=Length('mobile_number')).filter(text_len__lt=10)
but here mobile_number is BigIntegerField.
thanks for your help.
oh i did it using regex..Thanks
qs = UserProfile.objects.filter(mobile_number__regex = r'^\d{1,9}$')
Related
I have a Work model with a video_id, a user_id and some other simple fields. I need to display the last 12 works on the page, but only take 1 per user. Currently I'm trying to do it like this:
def self.latest_works_one_per_user(video_id=nil)
scope = self.includes(:user, :video)
scope = video_id ? scope.where(video_id: video_id) : scope.where.not(video_id: nil)
scope = scope.order(created_at: :desc)
user_ids = works = []
scope.each do |work|
next if user_ids.include? work.user_id
user_ids << work.user_id
works << work
break if works.size == 12
end
works
end
But I'm damn sure there is a more elegant and faster way of doing it especially when the number of works gets bigger.
Here's a solution that should work for any SQL database with minimal adjustment. Whether one thinks it's elegant or not depends on how much you enjoy SQL.
def self.latest_works_one_per_user(video_id=nil)
scope = includes(:user, :video)
scope = video_id ? scope.where(video_id: video_id) : scope.where.not(video_id: nil)
scope.
joins("join (select user_id, max(created_at) created_at
from works group by created at) most_recent
on works.user_id = most_recent.user_id and
works.created_at = most_recent.created_at").
order(created_at: :desc).limit(12)
end
It only works if the combination of user_id and created_at is unique, however. If that combination isn't unique you'll get more than 12 rows.
It can be done more simply in MySQL. The MySQL solution doesn't work in Postgres, and I don't know a better solution in Postgres, although I'm sure there is one.
I am using flask-sqlalchemy together with a sqlite database. I try to get all votes below date1
sub_query = models.VoteList.query.filter(models.VoteList.vote_datetime < date1)
sub_query = sub_query.filter(models.VoteList.group_id == selected_group.id)
sub_query = sub_query.filter(models.VoteList.user_id == g.user.id)
sub_query = sub_query.subquery()
old_votes = models.Papers.query.join(sub_query, sub_query.c.arxiv_id == models.Papers.arxiv_id).paginate(1, 4, False)
where the database model for VoteList looks like this
class VoteList(db.Model):
id = db.Column(db.Integer, primary_key=True)
user_id = db.Column(db.Integer, db.ForeignKey('user.id'))
group_id = db.Column(db.Integer, db.ForeignKey('groups.id'))
arxiv_id = db.Column(db.String(1000), db.ForeignKey('papers.arxiv_id'))
vote_datetime = db.Column(db.DateTime)
group = db.relationship("Groups", backref=db.backref('vote_list', lazy='dynamic'))
user = db.relationship("User", backref=db.backref('votes', lazy='dynamic'), foreign_keys=[user_id])
def __repr__(self):
return '<VoteList %r>' % (self.id)
I made sure that the 'old_votes' selection above has 20 elements. If I use .all() instead of .paginate() I get the expected 20 result?
Since I used a max results value of 4 in the example above I would expect that old_votes.items has 4 elements. But it has only 2? If I increase the max results value the number of elements also increases, but it is always below the max result value? Paginate seems to mess up something here?
any ideas?
thanks
carl
EDIT
I noticed that it works fine if I apply the paginate() function on add_columns(). So if I add (for no good reason) a column with
old_votes = models.Papers.query.join(sub_query, sub_query.c.arxiv_id == models.Papers.arxiv_id)
old_votes = old_votes.add_columns(sub_query.c.vote_datetime).paginate(page, VOTES_PER_PAGE, False)
it works fine? But since I don't need that column it would still be interesting to know what goes wrong with my example above?
Looks to me that for the 4 rows returned (and filtered) by the query, there are 4 rows representing 4 different rows of the VoteList table, but they refer/link/belong to only 2 different Papers models. When model instances are created, duplicates are filtered out, and therefore you get less rows. When you add a column from a subquery, the results are tuples of (Papers, vote_datetime), and in this case no duplicates are removed.
I encountered the same issue and I applied van's answer but it did not work. However I agree with van's explanation so I added .distinct() to the query like this:
old_votes = models.Papers.query.distinct().join(sub_query, sub_query.c.arxiv_id == models.Papers.arxiv_id).paginate(1, 4, False)
It worked as I expected.
I am new to entity framework and learning making queries. Can anyone please help me how can I convert the following SQL query to run in entity framework?
select max(isnull(TInvoice.InvoiceNr, 0)) + 1
from TInvoice inner join TOrders
on TInvoice.OrderId = TOrders.OrderId
where TOrders.ClientFirmId = 1
As comments have said, without the data model it is hard to be exact.
Would really need to see how you have defined your relations in your data model.
I guess from first read my first impression is something along the lines of:
int max = context.TInvoice.Where(x => x.TOrders.ClientFirmId == 1).Max(x => x.InvoiceNr);
I'm really struggling on this one.
I need to be able to sort my user by the number of positive vote received on their comment.
I have a table userprofile, a table comment and a table likeComment.
The table comment has a foreign key to its user creator and the table likeComment has a foreign key to the comment liked.
To get the number of positive vote a user received I do :
LikeComment.objects.filter(Q(type = 1), Q(comment__user=user)).count()
Now I want to be able to get all the users sorted by the ones that have the most positive votes. How do I do that ? I tried to use extra and JOIN but this didn't go anywhere.
Thank you
It sounds like you want to perform a filter on an annotation:
class User(models.Model):
pass
class Comment(models.Model):
user = models.ForeignKey(User, related_name="comments")
class Like(models.Model):
comment = models.ForeignKey(Comment, related_name="likes")
type = models.IntegerField()
users = User \
.objects \
.all()
.extra(select = {
"positive_likes" : """
SELECT COUNT(*) FROM app_like
JOIN app_comment on app_like.comment_id = app_comment.id
WHERE app_comment.user_id = app_user.id AND app_like.type = 1 """})
.order_by("positive_likes")
models.py
class UserProfile(models.Model):
.........
def like_count(self):
LikeComment.objects.filter(comment__user=self.user, type=1).count()
views.py
def getRanking( anObject ):
return anObject.like_count()
def myview(request):
users = list(UserProfile.objects.filter())
users.sort(key=getRanking, reverse=True)
return render(request,'page.html',{'users': users})
Timmy's suggestion to use a subquery is probably the simplest way to solve this kind of problem, but subqueries almost never perform as well as joins, so if you have a lot of users you may find that you need better performance.
So, re-using Timmy's models:
class User(models.Model):
pass
class Comment(models.Model):
user = models.ForeignKey(User, related_name="comments")
class Like(models.Model):
comment = models.ForeignKey(Comment, related_name="likes")
type = models.IntegerField()
the query you want looks like this in SQL:
SELECT app_user.id, COUNT(app_like.id) AS total_likes
FROM app_user
LEFT OUTER JOIN app_comment
ON app_user.id = app_comment.user_id
LEFT OUTER JOIN app_like
ON app_comment.id = app_like.comment_id AND app_like.type = 1
GROUP BY app_user.id
ORDER BY total_likes DESCENDING
(If your actual User model has more fields than just id, then you'll need to include them all in the SELECT and GROUP BY clauses.)
Django's object-relational mapping system doesn't provide a way to express this query. (As far as I know—and I'd be very happy to be told otherwise!—it only supports aggregation across one join, not across two joins as here.) But when the ORM isn't quite up to the job, you can always run a raw SQL query, like this:
sql = '''
SELECT app_user.id, COUNT(app_like.id) AS total_likes
# etc (as above)
'''
for user in User.objects.raw(sql):
print user.id, user.total_likes
I believe this can be achieved with Django's queryset:
User.objects.filter(comments__likes__type=1)\
.annotate(lks=Count('comments__likes'))\
.order_by('-lks')
The only problem here is that this query will miss users with 0 likes. Code from #gareth-rees, #timmy-omahony and #Catherine will include also 0-ranked users.
I have to tables, one is profile and the other is rating.
rating has a field profileid which is primary key to profile.id and a field with rating values. Now I want to find the field with highest rating and display the corresponding profile. Since I'm new to YII framework I'm having troubles with it. Please help me getting out of it. What I'm doing is described below.
$topRage=new CDbCriteria();
$topRage->select="*";
$topRage->alias="t1";
$topRage->order="rateing DESC";
$topRage->join="JOIN `ratings` ON `profile`.`id` = `t1`.`profileId`";
$topRage->limit="1";
Try this :
join='JOIN profile ON profile.id = t1.profileId';
If you are doing this: Ratings::model()->findAll($topRage) , then ratings table is already being queried, so you need to join with profile table.
Edit:
for echo you'll need to do this:
$echo "Rating id: ".$rating->id."| Profile Id: ".$rating->profile->id."| Profile Name: ".$rating->profile->name."| Rating: ".$rating->ratingvalue;
Don't forget to pass $rating from the controller though.
You could also use find($topRage) instead of the findAll($topRage) and remove the limit, but that 's just another way of doing the same thing.
just query without join .
$topRage=new CDbCriteria();
$topRage->select="*";
$topRage->alias="t1";
$topRage->order="rateing DESC";
$topRage->limit="1";
$rating=Ratings::model()->findAll($topRage);
$profile=Profile::model()->findByPk($rating->profileId);