I am using the open_graph Explorer tool to test this out and I need to be able to get the first_name, last_name and installed from 1000 users. At the moment I group up the users in grouped of 50 for a single query and batch them into up to 20 queries in a batch request.
A single query looks as follows
?ids=100003825998801,100003825998802,547884299,100003825998803,100003825998804,100003825998805,100003825998806&fields=first_name,last_name,installed
Now, the question is, when you use multiple ids in a single query like I did, facebook seems to "force" certain fields on you, large fields, such as "metadata" and "fields". Is it possible to remove these fields from the result set of the query?
Also, any advice on how to make this more efficient would also be great.
Thank.
Have you considered using FQL instead …?
SELECT first_name, last_name, is_app_user FROM user
WHERE uid IN (100003825998801,100003825998802,547884299,100003825998803,
100003825998804,100003825998805,100003825998806)
– that gives you just the info you want, you only have to watch out for the fact that the installed field is named is_app_user in FQL.
Related
I show a model of sales that can be aggregated by different fields through a form. Products, clients, categories, etc.
view_by_choice = filter_opts.cleaned_data["view_by_choice"]
sales = sales.values(view_by_choice).annotate(........).order_by(......)
In the same form I have a string input where the user can filter the results. By "product code" for example.
input_code = filter_opts.cleaned_data["filter_code"]
sales = sales.filter(prod_code__icontains=input_code)
What I want to do is filter the queryset "sales" by the input_code, defining the field dynamically from the view_by_choice variable.
Something like:
sales = sales.filter(VARIABLE__icontains=input_code)
Is it possible to do this? Thanks in advance.
You can make use of dictionary unpacking [PEP-448] here:
sales = sales.filter(
**{'{}__icontains'.format(view_by_choice): input_code}
)
Given that view_by_choice for example contains 'foo', we thus first make a dictionary { 'foo__icontains': input_code }, and then we unpack that as named parameter with the two consecutive asterisks (**).
That being said, I strongly advice you to do some validation on the view_by_choice: ensure that the number of valid options is limited. Otherwise a user might inject malicious field names, lookups, etc. to exploit data from your database that should remain hidden.
For example if you model has a ForeignKey named owner to the User model, he/she could use owner__email, and thus start trying to find out what emails are in the database by generating a large number of queries and each time looking what values that query returned.
I am using a Django backend with postgresql.
Let's say I have a database with a table called Employees with about 20,000 records.
I need to allow multiple users to edit and verify the Area Code field for every record in Employees.
I'd prefer to allow a user to view the records, say, 30 at a time (to reduce burnout).
How can I select 30 records at a time from Employees to send to the front end UI for editing, without letting multiple users edit the same records, or re-selecting a record that has already been verified?
I don't need comments on the content of the database (these are example table and field names).
One way to do this would be to add 2 more fields to your table, say for example assigned_to and verified. You can update assigned_to, which can be a foreign key to the verifying user, when you allow the user to view that Employee. This will create a record preventing the Employee from being chosen twice. assigned_to can also double as a record of who verified this Employee for future reference.
verified could be simply a Boolean field which keeps track if the Employee has already been verified and can be updated when the user confirms the verification
The actual selects can be done like this:
employees = Employee.objects.filter(assigned_to=None, verified=False)[:30]
Then
for emp in employees:
emp.assigned_to = user
emp.save()
Note: This can still potentially cause a race condition if 2 users make this request at exactly the same time. To avoid this, another possibility could be to partition the employee tables into groups for each user with no overlap. This would ensure that no 2 users would ever have the same employees
look at the following scenario:
I have an User model and an Address model that belongs to user.
In the user index, I need to show along with user's info how many addresses does the user have, but it's generating N+1 queries as everytime I call count it executes an additional query for that user id.
How can I do that? I read about select_related but I'm trying to make it in the reverse order...
In SQL it could be translated to:
SELECT user.*,
(SELECT count(*) FROM address WHERE address.user_id = user.id) AS address_count
FROM user
Is there a way to get the above SQL with django QuerySet?
You can annotate the number of addresses, you haven't shown your models but you can use the following on your queryset
.annotate(address_count=Count('address'))
User.objects.all().annotate(address_count=Count('address')) # Im guessing you want this
This would provide an address_count property on for each result
Docs for count
Quick overview:
I have a Forecast model setup which has a workflow_state.
Now I'm trying to query the Forecast for all the forecasts that are in a certain state AND the current person logged in is_staff.
If i was writing a raw query this wouldn't be an issue because i could write something like:
SELECT * FROM forecast WHERE forecast.workflow_state_id in (1,2,3,4) AND 1 = user.is_staff
However, when trying to write this in a queryset I can't figure out how to reference a constant. I don't want to write a raw queryset and if possible want to avoid using the extra field.
Any suggestions would be appreciated. Thanks
Your constant is only a True
Forecast.objects.filter(workflow_state__in=[1,2,3,4], user__is_staf=True)
Your edit makes things rather less than clear, but you seem to be asking how to do a check on the current logged-in user, rather than on the user referenced by the model. In which case, you don't do that in a query at all; your example SQL statement wouldn't work, and neither would doing it in the ORM. You do it in Python, of course:
if request.user.is_staff:
forecasts = Forecast.objects.filter(workflow_state__in=[1,2,3,4])
Is there a way by using XPath Builder under Developer Center inside Sitecore Shell (a Fast Query interface) to select a particular attribute from an item. Example:
/sitecore/content/Home/Products/*[##templatename = 'Product Group']/#id
I would expect to see a collection of id's to be returned where id is an attribute of an item. If yes is it possible to extract an attribute with a space bar? Example:
/sitecore/content/Home/Products/*[##templatename = 'Product Group']/#more info
EDIT
The thing that I want to achieve is to get a collection of items (I have few hounded items here), not one particular item. That's why I am not interested in adding additional conditions, like specific item id or title. I want to see a collection of values of a specific attribute. As in example showed above, I want to see a collection of values that are assign to 'more info' attribute. Once again I am expecting to see few hounded different values that are set to 'more info' attribute.
EDIT2
There is a problem with a production, a critical stuff. There is no access to it other then thru Sitecore shell, but I don't have permissions to add/install additional packages. I know how to get this info by implementing custom code, or queering db directly, but I simply do not have permission to do it. Guys that will be able to grant me need credentials will wake up in 6 hours, so I was hoping to do whatever I can to analyse the situation. I would accept Maras answer if it was an answer not a comment - there is no way I can do it using fast query. thanks for help.
Try using #
/sitecore/content/Home/Products/*[##templatename = 'Product Group']/##more info#
This is the way around when selecting items with fields that contain spaces. Having said that I don't know if you would be able to get a specific result or not for your specific question but give it a try.
For example, consider this query which returns Product S1
fast:/sitecore/content/home/*[#Title = 'Item 1' and ##templatename = 'Product Group1']//*[#Title = 'Product S1' and ##id = '{787EE6C5-0885-495D-855E-1D129C643E55}']
However, if you place the special attribute (i.e. ##id) at the beginning of the condition, the query will not return a result.
fast:/sitecore/content/home/*[##templatename = 'Product Group1' and #Title = 'Product S1']//*[##id = '{787EE6C5-0885-495D-855E-1D129C643E55}' and #Title = 'Product S1']
Remember this, Sitecore Fast Query only supports the following special attributes:
##id
##name
##key
##templateid
##templatename
##templatekey
##masterid
##parentid
Let us know if this helps.