Count distinct value pairs in multiple columns in SQL
To get a count of the number of unique combinations of id, name and address: SELECT Count(*) FROM ( SELECT DISTINCT id , name , address FROM your_table ) As distinctified
distinct
LINQ Select Distinct Count in Lambda form
Use this: items.Select(i => i.Value).Distinct().Count()
MySQL Counting Distinct Values on Multiple Columns
Just list multiple columns in the GROUP BY clause. SELECT computer, user, count(*) AS count FROM login GROUP BY computer, user
How to make a “distinct” join with MySQL
Use: SELECT p.upc, p.name, ph.price, ph.date FROM PRODUCT p LEFT JOIN PRICE_H ph ON ph.product_id = p.id JOIN (SELECT a.product_id, MAX(a.date) AS max_date FROM PRICE_H a GROUP BY a.product_id) x ON x.product_id = ph.product_id AND x.max_date = ph.date
Select top distinct results ordered by frequency
I haven’t tested it, so the syntax might not be perfect, but what about something like this : select name, count(*) as frequency from your_table group by name order by count(*) desc Should give you unique names and the corresponding number of times each name appears in the table, ordered by that number.
SqlAlchemy: count of distinct over multiple columns
The exact query can be produced using the tuple_() construct: session.query( func.count(distinct(tuple_(Hit.ip_address, Hit.user_agent)))).scalar()
Counting Using Group By Linq
var noticesGrouped = notices.GroupBy(n => n.Name). Select(group => new { NoticeName = group.Key, Notices = group.ToList(), Count = group.Count() });
Select distinct values from multiple columns in same table
It’s better to include code in your question, rather than ambiguous text data, so that we are all working with the same data. Here is the sample schema and data I have assumed: CREATE TABLE tbl_data ( id INT NOT NULL, code_1 CHAR(2), code_2 CHAR(2) ); INSERT INTO tbl_data ( id, code_1, code_2 ) VALUES … Read more
MYSQL UNION DISTINCT
No. You cannot specify which exact field you need to distinct with. It only works with the whole row. As of your problem – just make your query a subquery and in outer one GROUP BY user_id SELECT * FROM (SELECT a.user_id,a.updatecontents as city,b.country FROM userprofiletemp AS a LEFT JOIN userattributes AS b ON a.user_id=b.user_id … Read more
MongoDB: How to get distinct list of sub-document field values?
You can just do: db.collection.distinct(“children.child_name”); In your case it returns: [ “John”, “Anna”, “Kevin” ]