I have a two table users and orders . i want to get users with Total Orders Count. How can I do this with eloquent?
Here is my query
User::with('orders')->where('status','active')->get();
You can use withCount for this.
https://laravel.com/docs/9.x/eloquent-relationships#counting-related-models
User::withCount('orders')->where('status','active')->get();
Related
Items collection, divided on categories, a single category could be +100k documents ranging many items.
Collection of orders, an item can have many orders, either sell or bid orders.
For every item, find the lowest sell order and the highest bid order
$sort by lowest sell order and $limit results to first 20 (so a pagination is possible).
To do so, I have tried many ways via $lookup & aggregation usage, but I'm forced to fetch the lowest sell order for every item to be able to sort them all and limit the final results to the first 20, thereby the response time is huge, as this $lookup is being applied on every matching result rather than just the first 20.
How is this pattern doable without a complete category $lookup?
I cannot think of a way that doesn't apply the filter on all the results to later on, sort & limit the results to just 20.
I'm using the latest mongoose version for node.js
The title pretty much sums up my question, but for more detail, say I have a collection of houses, and each house can have multiple owners/tenants. These are just an ID reference to my owners collection.
So I want to query ALL my houses, but i want to populate all of their owners/tenants. If my query, something like housesModel.find({}) returns N number of results, but i ask it to .populate('owner'), will it perform an additional query to fetch those owners? keep in mind it can have many owners (for the sake of this question)
So if I query all my houses, and i get back 300 results, but i ask it to populate each doc's owners field, and each house has say on average 1.5 owners per house, is that 450 additional individual queries to the database? It feels like this is not very optimal, but that's what im trying to understand.
If it is actually doing N number of db queries for a parent query of N results, how does that affect the performance? Even more so when we get to the 1000s of results?
I want to create a search that filters the records like Display the name of every customer and Sum of amount of orders placed by that customer in the last thirty days. I want to apply the search on sales orders.
In the results Tabs try this:
I am adding Two tables (Transactions, Customers) to Qlikview and I need a number returned on how many customers have spent over 1000$ in a text object.
I am trying to achive this through aggregate function with no luck till now.Is this possible?or should i try an alternative root.
Num(Count( {$ < Aggr(Sum(Total),Customer) = {">1000"}>} Distinct Customer), '###.###.###')
Total is the amount spent on each transaction and customer the customer who made the transaction.
I also tried something like the below code:
count({<Customer= {"=sum(Total)> =100"} >} distinct Customer)
but still havent gotten anywhere.
If think is what you want to do. Assuming Total is the number you want to add I've used Spend to avoid confusion. This would give you the number of customers with a spend above 1000 based on the current selections.
Num(Count(if(Aggr(Sum(Spend),Customer)>1000,1)), '###.###.###')
The use of the TOTAL function inside the aggr() function will skew the results
I'm trying to create a calculated column in powerpivot to count the distinct orders per customer. The table (Sales) contains product information so has multiple rows per order and multiple orders per customer.
I want this to be a calculated column as I will use the results to build a frequency/histogram table.
I have been trying the following formula but it is very slow and it not bringing any results back.
=CALCUALTE(DISTINCTCOUNT(Sales[OrderNumber]),FILTER(Sales,Sales[Email]=EARLIER(Sales[Email])))
Does anyone have another idea how I could structure the formula to be faster? The table contains 13 million records.
Thanks
Mike