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Extend queryset in django python

I am looking for way how to add new objects to existing queryset, or how to implement what I want by other way.
contact = watson.filter(contacts, searchline)
This line returns queryset, which I later use to iterate.
Then I want to do this to add more objects, which watson couldn't find
contact_in_iteration = Contact.objects.get(id = fild.f_for)
contact.append(contact_in_iteration)
And sorry for my poor english
Did this
contacts = Contact.objects.filter(crm_id=request.session['crm_id'])
query = Q(contacts,searchline)
contact = watson.filter(query)
and get 'filter() missing 1 required positional argument: 'search_text'' error

You can use | and Q lookups. See the docs.
I'm not sure I've fully understood your initial query, but I think that in your case you would want to do:
query = Q(contacts='Foo', searchline='Bar')
contact = watson.filter(query)
Then later:
contact = watson.filter(query | Q(id=field.f_for))
Strictly speaking it won't append to the queryset, but will return a new queryset. But that's okay, because that's what .filter() does anyway.

You should look at a queryset as a sql query that will be executed later. When constructing a queryset and save the result in a variable, you can later filter it even more, but you can not expand it. If you need a query that has more particular rules (like, you need an OR operation) you should state that when you are constructing the query. One way of doing that is indeed using the Q object.
But it looks like you are confused about what querysets really are and how they are used. First of all:
Contact.objects.get(id = fild.f_for)
will never return a queryset, but an instance, because you use get and thus ask for a single particular record. You need to use filter() if you want to get a quersyet. So if you had an existing queryset say active_contacts and you wanted to filter it down so you only get the contacts that have a first_name of 'John' you would do:
active_contacts = Contact.objects.filter(active=True)
active_contacts_named_John = active_contacts.filter(first_name='John')
Of course you could do this in one line too, but I'm assuming you do the first queryset construction somewhere else in your code.
Second remark:
If in your example watson is a queryset, your user of filter() is unclear. This doesn't really make sense:
contact = watson.filter(contacts, searchline)
As stated earlier, filtering a queryset returns another queryset. So you should use a plurar as your variable name e.g. contacts. Then the correct use of filter would be:
contacts = watson.filter(first_name=searchline)
I'm assuming searchline here is a variable that contains a user inputted search term. So maybe here you should name your variable searchterm or similar. This will return all contacts that are filtered by whatever watson is filtering out already and whose first_name matches searchline exactly. You could also use a more liberate method and filter out results that 'contains' the searching term, like so:
contacts = watson.filter(first_name__contains=searchline)
Hope this helps you get on the right path.

Django OR query using Extra and Filter

I am trying to use Django's ORM to generate a query using both extra and filter methods. Something like this:
Model.objects.filter(clauseA).extra(clauseB).all()
This generates a query, but the issue is that everything in the filter clause is AND'd with everything in the extra clause, so the sql looks like:
SELECT * FROM model WHERE clauseA AND clauseB.
My question is, is there a way to change the default combination operator for a query in Django such that the query generated will be:
SELECT * FROM model WHERE clauseA OR clauseB.

Try Q object
Model.objects.filter(Q(clauseA) | ~Q(clauseB))
EDIT
try this
Model.objects.filter(clauseA) | Model.objects.extra(clauseB)

It might be easier if you just get rid of the filter clause, and include that filter directly into extra OR'd with your Postgres specific function. I think it is already a limitation of the Django ORM.
You can attempt to create your own Func expression though. Once you have created one for your Postgres specific function, you might be able to use a combination of Func(), F(), and Q() objects to get rid of that nasty .extra() function and chain them nicely.

Variable interpolation in python/django, django query filters [duplicate]

Given a class:
from django.db import models
class Person(models.Model):
name = models.CharField(max_length=20)
Is it possible, and if so how, to have a QuerySet that filters based on dynamic arguments? For example:
# Instead of:
Person.objects.filter(name__startswith='B')
# ... and:
Person.objects.filter(name__endswith='B')
# ... is there some way, given:
filter_by = '{0}__{1}'.format('name', 'startswith')
filter_value = 'B'
# ... that you can run the equivalent of this?
Person.objects.filter(filter_by=filter_value)
# ... which will throw an exception, since `filter_by` is not
# an attribute of `Person`.

Python's argument expansion may be used to solve this problem:
kwargs = {
'{0}__{1}'.format('name', 'startswith'): 'A',
'{0}__{1}'.format('name', 'endswith'): 'Z'
}
Person.objects.filter(**kwargs)
This is a very common and useful Python idiom.

A simplified example:
In a Django survey app, I wanted an HTML select list showing registered users. But because we have 5000 registered users, I needed a way to filter that list based on query criteria (such as just people who completed a certain workshop). In order for the survey element to be re-usable, I needed for the person creating the survey question to be able to attach those criteria to that question (don't want to hard-code the query into the app).
The solution I came up with isn't 100% user friendly (requires help from a tech person to create the query) but it does solve the problem. When creating the question, the editor can enter a dictionary into a custom field, e.g.:
{'is_staff':True,'last_name__startswith':'A',}
That string is stored in the database. In the view code, it comes back in as self.question.custom_query . The value of that is a string that looks like a dictionary. We turn it back into a real dictionary with eval() and then stuff it into the queryset with **kwargs:
kwargs = eval(self.question.custom_query)
user_list = User.objects.filter(**kwargs).order_by("last_name")

Additionally to extend on previous answer that made some requests for further code elements I am adding some working code that I am using
in my code with Q. Let's say that I in my request it is possible to have or not filter on fields like:
publisher_id
date_from
date_until
Those fields can appear in query but they may also be missed.
This is how I am building filters based on those fields on an aggregated query that cannot be further filtered after the initial queryset execution:
# prepare filters to apply to queryset
filters = {}
if publisher_id:
filters['publisher_id'] = publisher_id
if date_from:
filters['metric_date__gte'] = date_from
if date_until:
filters['metric_date__lte'] = date_until
filter_q = Q(**filters)
queryset = Something.objects.filter(filter_q)...
Hope this helps since I've spent quite some time to dig this up.
Edit:
As an additional benefit, you can use lists too. For previous example, if instead of publisher_id you have a list called publisher_ids, than you could use this piece of code:
if publisher_ids:
filters['publisher_id__in'] = publisher_ids

Django.db.models.Q is exactly what you want in a Django way.

This looks much more understandable to me:
kwargs = {
'name__startswith': 'A',
'name__endswith': 'Z',
***(Add more filters here)***
}
Person.objects.filter(**kwargs)

A really complex search forms usually indicates that a simpler model is trying to dig it's way out.
How, exactly, do you expect to get the values for the column name and operation?
Where do you get the values of 'name' an 'startswith'?
filter_by = '%s__%s' % ('name', 'startswith')
A "search" form? You're going to -- what? -- pick the name from a list of names? Pick the operation from a list of operations? While open-ended, most people find this confusing and hard-to-use.
How many columns have such filters? 6? 12? 18?
A few? A complex pick-list doesn't make sense. A few fields and a few if-statements make sense.
A large number? Your model doesn't sound right. It sounds like the "field" is actually a key to a row in another table, not a column.
Specific filter buttons. Wait... That's the way the Django admin works. Specific filters are turned into buttons. And the same analysis as above applies. A few filters make sense. A large number of filters usually means a kind of first normal form violation.
A lot of similar fields often means there should have been more rows and fewer fields.

Django - Following ForeignKey relationships "backward" for entire QuerySet

is it possible to follow ForeignKey relationships backward for entire querySet?
i mean something like this:
x = table1.objects.select_related().filter(name='foo')
x.table2.all()
when table1 hase ForeignKey to table2.
in
https://docs.djangoproject.com/en/1.2/topics/db/queries/#following-relationships-backward
i can see that it works only with get() and not filter()
Thanks

You basically want to get QuerySet of different type from data you start with.
class Kid(models.Model):
mom = models.ForeignKey('Mom')
name = models.CharField…
class Mom(models.Model):
name = models.CharField…
Let's say you want to get all moms having any son named Johnny.
Mom.objects.filter(kid__name='Johnny')
Let's say you want to get all kids of any Lucy.
Kid.objects.filter(mom__name='Lucy')

You should be able to use something like:
for y in x:
y.table2.all()
But you could also use get() for a list of the unique values (which will be id, unless you have a different specified), after finding them using a query.
So,
x = table1.objects.select_related().filter(name='foo')
for y in x:
z=table1.objects.select_related().get(y.id)
z.table2.all()
Should also work.

You can also use values() to fetch specific values of a foreign key reference. With values the select query on the DB will be reduced to fetch only those values and the appropriate joins will be done.
To re-use the example from Krzysztof Szularz:
jonny_moms = Kid.objects.filter(name='Jonny').values('mom__id', 'mom__name').distinct()
This will return a dictionary of Mom attributes by using the Kid QueryManager.

Filtering with joined tables

I'm trying to get some query performance improved, but the generated query does not look the way I expect it to.
The results are retrieved using:
query = session.query(SomeModel).
options(joinedload_all('foo.bar')).
options(joinedload_all('foo.baz')).
options(joinedload('quux.other'))
What I want to do is filter on the table joined via 'first', but this way doesn't work:
query = query.filter(FooModel.address == '1.2.3.4')
It results in a clause like this attached to the query:
WHERE foos.address = '1.2.3.4'
Which doesn't do the filtering in a proper way, since the generated joins attach tables foos_1 and foos_2. If I try that query manually but change the filtering clause to:
WHERE foos_1.address = '1.2.3.4' AND foos_2.address = '1.2.3.4'
It works fine. The question is of course - how can I achieve this with sqlalchemy itself?

If you want to filter on joins, you use join():
session.query(SomeModel).join(SomeModel.foos).filter(Foo.something=='bar')
joinedload() and joinedload_all() are used only as a means to load related collections in one pass, not used for filtering/ordering!. Please read:
http://docs.sqlalchemy.org/en/latest/orm/tutorial.html#joined-load - the note on "joinedload() is not a replacement for join()", as well as :
http://docs.sqlalchemy.org/en/latest/orm/loading.html#the-zen-of-eager-loading