for example I want to update the price of products with a price of zero dollars to 100.
for price_list_items in (self.env['product.pricelist.item'].search_read([], [])):
if price_list_items.price == 0:
price_list_items.price = 100
But It's not working.
AttributeError: 'dict' object has no attribute 'price'
How can I make such a update like this ?
you can do it this way :
list_of_pricelist = self.env['product.pricelist.item'].search([])
for pricelist in list_of_pricelist:
if price_list_items.price == 0:
price_list_items.price = 100
I think you better use search() in your case.
Difference between search() and search_read() :
https://www.odoo.com/fr_FR/forum/aide-1/difference-between-search-and-search-read-in-odoo-8-83076
Search_read() returns a list of dict.
If you really want to do your code this way, i think you can do it like this, but it's not usual
for price_list_items in (self.env['product.pricelist.item'].search_read([], [])):
if price_list_items[<list_index:int>]['field_name'] == 0:
price_list_items[<list_index:int>].update({'field_name': value})
The search_read function returns a list of dicts, price_list_items is a dict of one record field values
Setting the price using the following syntax is available for recordsets:
price_list_recod.price = 100
Instead of fetching and reading all records, you can simply search for price list items with a price equal to 0.0 using the search method (which will return a recordset) and then update the price for all of them with one assignment
Example:
self.env['product.pricelist.item'].search([('price', '=', 0)]).price = 100
Related
My code:
List <bu_businesslayer> bus = ab.classid(UserAccess).Where(x => x.UserAccess == UserAccess).ToList();
return View(bus);
Now ab has count = 202, but doesn't get assigned to bus. Bus has count = 0. I don't know why. I appreciate if you help.
bu_businesslayer is class file present in class library.
Where() does not affect ab, it's copy ab and returning result, so you can check where your collection count = 0;
Check the results of ab.classid(UserAccess).ToList() what count of result? Is count of it equal 0?
If not, the purpose in Where expression. How many objects in ab satisfy the condition x => x.UserAccess == UserAccess after first operation? And what is UserAccess? If it class, so you should learn more about object equality
Hope it helps.
Try initilizing the bus first using
List <bu_businesslayer> bus=new List <bu_businesslayer>();
and then,
bus = ab.classid(UserAccess).Where(x => x.UserAccess ==
UserAccess).ToList();
and see the relults.
if this also doesnot work the where condition is not met.
I have 4 model like this
class Site(models.Model):
name = models.CharField(max_length=200)
def get_lowest_price(self, mm_date):
'''This method returns lowest product price on a site at a particular date'''
class Category(models.Model):
name = models.CharField(max_length=200)
site = models.ForeignKey(Site)
class Product(models.Model):
name = models.CharField(max_length=200)
category = models.ForeignKey(Category)
class Price(models.Model):
date = models.DateField()
price = models.IntegerField()
product = models.ForeignKey(Product)
Here every have many category, every category have many product. Now product price can change every day so price model will hold the product price and date.
My problem is I want list of site filter by price range. This price range will depends on the get_lowest_price method and can be sort Min to Max and Max to Min. Already I've used lambda expression to do that but I think it's not appropriate
sorted(Site.objects.all(), key=lambda x: x.get_lowest_price(the_date))
Also I can get all site within a price range by running a loop but this is also not a good idea. Please help my someone to do the query in right manner.
If you still need more clear view of the question please see the first comment from "Ishtiaque Khan", his assumption is 100% right.
*In these models writing frequency is low and reading frequency is high.
1. Using query
If you just wanna query using a specific date. Here is how:
q = Site.objects.filter(category__product__price__date=mm_date) \
.annotate(min_price=Min('category__product__price__price')) \
.filter(min_price__gte=min_price, min_price__lte=max_price)
It will return a list of Site with lowest price on mm_date fall within range of min_price - max_price. You can also query for multiple date using query like so:
q = Site.objects.values('name', 'category__product__price__date') \
.annotate(min_price=Min('category__product__price__price')) \
.filter(min_price__gte=min_price, min_price__lte=max_price)
2. Eager/pre-calculation, you can use post_save signal. Since the write frequency is low this will not be expensive
Create another Table to hold lowest prices per date. Like this:
class LowestPrice(models.Model):
date = models.DateField()
site = models.ForeignKey(Site)
lowest_price = models.IntegerField(default=0)
Use post_save signal to calculate and update this every time there. Sample code (not tested)
from django.db.models.signals import post_save
from django.dispatch import receiver
#receiver(post_save, sender=Price)
def update_price(sender, instance, **kwargs):
cur_price = LowestPrice.objects.filter(site=instance.product.category.site, date=instance.date).first()
if not cur_price:
new_price = LowestPrice()
new_price.site = instance.product.category.site
new_price.date = instance.date
else:
new_price = cur_price
# update price only if needed
if instance.price<new_price.lowest_price:
new_price.lowest_price = instance.price
new_price.save()
Then just query directly from this table when needed:
LowestPrice.objects.filter(date=mm_date, lowest_price__gte=min_price, lowest_price__lte=max_price)
Solution:
from django.db.models import Min
Site.objects.annotate(
price_min=Min('categories__products__prices__price')
).filter(
categories__products__prices__date=the_date,
).distinct().order_by('price_min') # prefix '-' for descending order
For this to work, you need to modify the models by adding a related_name attribute to the ForeignKey fields.
Like this -
class Category(models.Model):
# rest of the fields
site = models.ForeignKey(Site, related_name='categories')
Similary, for Product and Price models, add related_name as products and prices in the ForeignKey fields.
Explanation:
Starting with related_name, it describes the reverse relation from one model to another.
After the reverse relationship is setup, you can use them to inner join the tables.
You can use the reverse relationships to get the price of a product of a category on a site and annotate the min price, filtered by the_date. I have used the annotated value to order by min price of the product, in ascending order. You can use '-' as a prefix character to do in descending order.
Do it with django queryset operations
Price.objects.all().order_by('price') #add [0] for only the first object
or
Price.objects.all().order_by('-price') #add [0] for only the first object
or
Price.objects.filter(date= ... ).order_by('price') #add [0] for only the first object
or
Price.objects.filter(date= ... ).order_by('-price') #add [0] for only the first object
or
Price.objects.filter(date= ... , price__gte=lower_limit, price__lte=upper_limit ).order_by('price') #add [0] for only the first object
or
Price.objects.filter(date= ... , price__gte=lower_limit, price__lte=upper_limit ).order_by('-price') #add [0] for only the first object
I think this ORM query could do the job ...
from django.db.models import Min
sites = Site.objects.annotate(price_min= Min('category__product__price'))
.filter(category__product__price=mm_date).unique().order_by('price_min')
or /and for reversing the order :
sites = Site.objects.annotate(price_min= Min('category__product__price'))
.filter(category__product__price=mm_date).unique().order_by('-price_min')
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.
Let's say I do a simple query:
$products =
Yii::app()->db->createCommand()->setFetchMode(PDO::FETCH_OBJ)
->select('*')
->from('products')
->limit(9)
->queryAll();
Let's say there are 500 products in the database. Is there a way I can get YII to automatically return the total number (count) of products if the "limit" was included? Perhaps return an object like this:
$products->products = array( ... products ... )
$products->totalProducts = 500;
The problem is, if LIMIT is included, it will return items, and the count will therefore be 9. I want a solution whereby it will return the 9 items, but also the count of say 200 items if there were 200 items.
Why not an easy:
$сount = Yii::app()->db->createCommand('select count(*) from table')->queryScalar();
echo $count;
You'll either have to run two queries (a count(*) query without the limit and then the limited query) or you can send you can retrieve your products using CSqlDataProvider and let it do it for you. But it generally takes two queries.
Note: one of the nifty features in Yii 1.1.13 is that you can send your query builder command in to the CSqlDataProvider if you're going to be using a dataprovider. More information at on this pull request that fixed it. That way, you can both use the power of query builder while also being able to shift your data into a dataprovider. Previously you had to build your SQL statement manually or grab the queryText of the command.
Yii::app()->db->createCommand('select count(*) from tbl_table')->queryScalar();
Try to use execute() instead of query() because execute returns rows count.
example:
$rowCount = $command->execute();
You could try using COUNT like this:
$dbCommand = Yii::app()->db->createCommand("
SELECT COUNT(*) as count FROM `products`");
$data = $dbCommand->queryAll();
Hope that helps!
EDIT: You might find this useful too: CDataProvider
Try this -
$sql = Yii::app()->db->createCommand('select * from tbl_table')->queryAll(); //It's return the Array
echo count($sql); //Now using count() method we can count the array.
I have a need to get a Count of Documents in a particular collection :
There is an existing index Raven/DocumentCollections that stores the Count and Name of the collection paired with the actual documents belonging to the collection. I'd like to pick up the count from this index if possible.
Here is the Map-Reduce of the Raven/DocumentCollections index :
from doc in docs
let Name = doc["#metadata"]["Raven-Entity-Name"]
where Name != null
select new { Name , Count = 1}
from result in results
group result by result.Name into g
select new { Name = g.Key, Count = g.Sum(x=>x.Count) }
On a side note, var Count = DocumentSession.Query<Post>().Count(); always returns 0 as the result for me, even though clearly there are 500 odd documents in my DB atleast 50 of them have in their metadata "Raven-Entity-Name" as "Posts". I have absolutely no idea why this Count query keeps returning 0 as the answer - Raven logs show this when Count is done
Request # 106: GET - 0 ms - TestStore - 200 - /indexes/dynamic/Posts?query=&start=0&pageSize=1&aggregation=None
For anyone still looking for the answer (this question was posted in 2011), the appropriate way to do this now is:
var numPosts = session.Query<Post>().Count();
To get the results from the index, you can use:
session.Query<Collection>("Raven/DocumentCollections")
.Where(x=>x.Name == "Posts")
.FirstOrDefault();
That will give you the result you want.