opsqert.blogg.se - Mysql count distinct group by

#MYSQL COUNT DISTINCT GROUP BY HOW TO#

SELECTĬode language: SQL (Structured Query Language) ( sql ) F) Oracle COUNT() with HAVING clause example The following examples get all category names and the number of products in each category by joining the product_categories with the products table and using the COUNT() function with the GROUP BY clause.

Second, the COUNT(*) function returns the number of products for each group.

First, the GROUP BY clause divides the products into groups based on the product category ( category_id).

To find the number of products in each product category, you use the following statement: SELECT If you want to find the number of products in the category id 1, you can add a WHERE clause to the query above: SELECT COUNT(*)Ĭode language: SQL (Structured Query Language) ( sql ) D) Oracle COUNT() with GROUP BY clause example The following example returns the number of rows in the products table: SELECT COUNT(*)Ĭode language: SQL (Structured Query Language) ( sql ) C) Oracle COUNT() with WHERE clause example SELECT COUNT( ALL val )Ĭode language: SQL (Structured Query Language) ( sql ) B) Simple Oracle COUNT() example The following statement uses the COUNT(ALL val) function to return the number of non-null rows in the items table, considering duplicates. The following statement uses the COUNT(DISTINCT val) to return only the number of distinct and non-null rows from the items table: SELECT COUNT( DISTINCT val ) The following statement uses the COUNT(*) function to return the number of rows in the items table including NULL and duplicate values: SELECT COUNT(*) Let’s create a table named items that consists of a val column and insert some sample data into the table for the demonstration. Let’s take some examples of using the COUNT() function. Note that, unlike other aggregate functions such as AVG() and SUM(), the COUNT(*) function does not ignore NULL values.

If you don’t explicitly specify DISTINCT or ALL, the COUNT() function uses the ALL by default. COUNT(ALL expression) evaluates the expression and returns the number of non-null items in a group, including duplicate values.COUNT(DISTINCT expression) function returns the number of unique and non-null items in a group.COUNT(*) function returns the number of items in a group, including NULL and duplicate values.The COUNT() function accepts a clause which can be either ALL, DISTINCT, or *: The syntax of the COUNT() function is as follows: COUNT( expression)Ĭode language: SQL (Structured Query Language) ( sql ) The Oracle COUNT() function is an aggregate function that returns the number of items in a group.

#MYSQL COUNT DISTINCT GROUP BY HOW TO#

However, using the above solution had virtually no increase on the query time (comparing with using simply the SUM), and should be completely reliable! It should be able to help others in a similar situation so I'm posting it here.Summary: in this tutorial, you will learn how to use the Oracle COUNT() function to get the number of items in a group. The CHECKSUM solution was also far too slow in its conversion, particularly as a result of the various data types, and I couldn't risk its unreliability. And due to the complexity, statistics simply wasn't a viable option. The table is very large, and using a sub-query dramatically increased the query time. My specific problem required me to divide a SUM by the COUNT of the distinct combination of various foreign keys and a date field, grouping by another foreign key and occasionally filtering by certain values or keys. SELECT COUNT(DISTINCT CAST(DocumentId as binary(4)) + CAST(DocumentSessionId as binary(4))) Assuming DocumentId and DocumentSessionId are both ints, and are therefore 4 bytes long. If you're working with datatypes of fixed length, you can cast to binary to do this very easily and very quickly. Obviously the chosen separator must be a character, or set of characters, which can never appear in either column. Given the following data the concatenating solution provided above will miscount: col1 col2 SQL> select count(distinct concat(deptno,job)) from emp

I went down a blind alley with analytics but the answer was depressingly obvious. SQL> select distinct deptno, job from emp It certainly works as you might expect in Oracle. What is it about your existing query that you don't like? If you are concerned that DISTINCT across two columns does not return just the unique permutations why not try it?