Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
269 views
in Technique[技术] by (71.8m points)

sql - Fastest way to determine if record exists

As the title suggests... I'm trying to figure out the fastest way with the least overhead to determine if a record exists in a table or not.

Sample query:

SELECT COUNT(*) FROM products WHERE products.id = ?;

    vs

SELECT COUNT(products.id) FROM products WHERE products.id = ?;

    vs

SELECT products.id FROM products WHERE products.id = ?;

Say the ? is swapped with 'TB100'... both the first and second queries will return the exact same result (say... 1 for this conversation). The last query will return 'TB100' as expected, or nothing if the id is not present in the table.

The purpose is to figure out if the id is in the table or not. If not, the program will next insert the record, if it is, the program will skip it or perform an UPDATE query based on other program logic outside the scope of this question.

Which is faster and has less overhead? (This will be repeated tens of thousands of times per program run, and will be run many times a day).

(Running this query against M$ SQL Server from Java via the M$ provided JDBC driver)

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

EXISTS (or NOT EXISTS) is specially designed for checking if something exists and therefore should be (and is) the best option. It will halt on the first row that matches so it does not require a TOP clause and it does not actually select any data so there is no overhead in size of columns. You can safely use SELECT * here - no different than SELECT 1, SELECT NULL or SELECT AnyColumn... (you can even use an invalid expression like SELECT 1/0 and it will not break).

IF EXISTS (SELECT * FROM Products WHERE id = ?)
BEGIN
--do what you need if exists
END
ELSE
BEGIN
--do what needs to be done if not
END

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...