I have a table called TestTable with one field, Type Varchar(5). The data in the table is:
Type A B A C A B
So there are six records in the table in a single field. Please help me in writing a query to get following result:
A B C 3 2 1
In other words I want to get the individual Type field data and their count.
There are three ways to do this.
select ( select count(*) from TestTable where Type = 'A' ) as A , ( select count(*) from TestTable where Type = 'B' ) as B , ( select count(*) from TestTable where Type = 'C' ) as C
This first query does work as is in some databases, but it's not standard SQL, because of the missing FROM clause. Add FROM DUAL (or any other 1-row table).
select sum( case when Type = 'A' then 1 else 0 end ) as A , sum( case when Type = 'B' then 1 else 0 end ) as B , sum( case when Type = 'C' then 1 else 0 end ) as C from TestTable
This second query gives you exactly what you asked for, but, like the first query, will be very bothersome to maintain if you should ever find yourself needing to add another Type.
select Type , count(*) from TestTable group by Type
This third query is the best. Not only is it drop-dead simple to understand at a glance, it also requires no maintenance when new Types are added. Furthermore, it acknowledges that presentation should not be handled in the database layer.
By the way, as your table has only one column, which has duplicate values, this column cannot be the primary key, and so the table cannot have a primary key. Some people would say it is therefore not a "real" table, but a bag.
Dig Deeper on Oracle and SQL
Related Q&A from Rudy Limeback
Read SQL expert Rudy Limeback's advice for counting combinations in a table with SQL's GROUP BY clausecontinue reading
Read an example of an SQL case expression from our SQL expert Rudy Limeback.continue reading
Read about the Mimer Validator, a tool used to verify your SQL code, in this tip from SQL expert Rudy Limeback.continue reading
Have a question for an expert?
Please add a title for your question
Get answers from a TechTarget expert on whatever's puzzling you.