With DENSE_RANK it will never ‘skip’ a number, but will still assign the same number to identical entries. Incidentally, SQL Server (unsure about similar functionality in Teradata) also provides the function DENSE_RANK(). The assignment of rank to rows always starts with 1 for a new partition. The rank is assigned to rows in a sequential manner. The ranking functions always assign a rank on the basis of the ORDER BY clause. ,ROW_NUMBER() OVER (PARTITION BY Region ORDER BY sales DESC) AS SalesRank These functions have to be used with the OVER() clause. This function works in exactly the same way, but always numbers each record uniquely. But what if it’s not? What if we always want to see 1-4? We can use ROW_NUMBER(). Use a CTE that will return the distinct rows for each userid and dateaccessed filtered for the dates that you want and then use ROWNUMBER() to get the rank. Sometimes, this is fine and is what we want. The next store in the list will then skip to the next one. Using Rank, if you have multiple entries with the same value (eg 2 stores with the same sales value), rank will give them the same rank number. ,RANK() OVER (PARTITION BY Region ORDER BY sales DESC) AS SalesRank You ‘Partiton’ your data, and ‘Qualify’ the results Now, lets say we want to show the ranking by Region, and perhaps only want to see the top 4 stores in each region. This will list your stores, their sales and the sales rank for each store. ,RANK() OVER (ORDER BY sales DESC) AS SalesRank RANK() is pretty simple to use, and works pretty much the same way in Teradata & SQL Server I will cover both SQL Server and Teradata. ![]() In this post, I want to cover off a little bit of RANK() usage in SQL, but then focus on how to use ranking to get a unique rank number for each row. There are other options to achieve the same result, but after trying them the query plan generated was way more complex.Aug2 comments Article Intermediate, SQL Server, Teradata This query could benefit from additional indexes and improve the JOIN, but besides that, the plan seems quite ok. The problem is the rank alias you gave to the call to ROWNUMBER.In MySQL 8+, RANK becomes a reserved keyword referring to the analytic function RANK().Note that using an alias of rank on MySQL 5.7 and earlier was perfectly fine. It should be 1 as first three columns used in Partition By has same data. Even without order by used in the last line, the RN number varies. The execution plan generated by this query is not too bad as we could imagine. because you Order by the result, it will disorder the Partition part, try to Order by TW.EMPLID,TW.HRDEPTID,TW.TransferStartdate,TW.RN or remove the Order by. ColorPerCategory: It becomes a SUM, adding all the distinct count results of each category.CategoriesPerOrder: It becomes a SUM to achieve the result we would like.However, there are some different calculations: Some of them are the same of the 2nd query, aggregating more the rows. Once again, the calculations are based on the previous queries. Sum ( colorpercategory ) ColorPerCategory Sum ( categoriesperorder ) CategoriesPerOrder , On the 3 rd step we reduce the aggregation, achieving our final result, the aggregation by SalesOrderId ColorPerCategory: After making the DENSE_RANK, now we need to extract the MAX value to have the same effect as the COUNT DISTINCT.ColorPerOrder: We make a SUM over the already existing COUNT from the previous query.CategoriesPerOrder: It can have a fixed number of 1, since we are still aggregating per category.Total: A simple SUM over the SUM already made.ItemsPerOrder: We make a SUM on the results of the COUNT, this will aggregate the different counts. ![]() The calculations on the 2 nd query are defined by how the aggregations were made on the first query: Max ( colorpercategory ) ColorPerCategory ![]() Partition BY salesorderid, productcategoryid The following query makes an example of the difference: RANK: After a tie, the count jumps the number of tied items, leaving a hole.ĭENSE_RANK: No jump after a tie, the count continues sequentially The difference is how they deal with ties. There are two ranking functions: RANK and DENSE_RANK. This works in a similar way as the distinct count because all the ties, the records with the same value, receive the same rank value, so the biggest value will be the same as the distinct count. We are counting the rows, so we can use DENSE_RANK to achieve the same result, extracting the last value in the end, we can use a MAX for that. Count Distinct is not supported by window partitioning, we need to find a different way to achieve the same result. Partition BY productcategoryid ) ColorPerCategory ![]() Count ( DISTINCT productcategoryid ) CategoriesPerOrder ,
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