Find closest numeric value in database

Question

I need to find a select statement that will return either a record that matches my input exactly, or the closest match if an exact match is not found.

Here is my select statement so far.

SELECT * FROM [myTable] 
WHERE Name = 'Test' AND Size = 2 AND PType = 'p' 
ORDER BY Area DESC

What I need to do is find the closest match to the 'Area' field, so if my input is 1.125 and the database contains 2, 1.5, 1 and .5 the query will return the record containing 1.

My SQL skills are very limited so any help would be appreciated.

Answer 1

get the difference between the area and your input, take absolute value so always positive, then order ascending and take the first one

SELECT TOP 1 * FROM [myTable] 
WHERE Name = 'Test' and Size = 2 and PType = 'p'
ORDER BY ABS( Area - @input ) 

Answer 2

something horrible, along the lines of:

ORDER BY ABS( Area - 1.125 ) ASC LIMIT 1

Maybe?

Answer 3

If you have many rows that satisfy the equality predicates on Name , Size , and PType columns then you may want to include range predicates on the Area column in your query. If the Area column is indexed this could allow efficient index-based access.

The following query (written using Oracle syntax) uses one branch of a UNION ALL to find the record with minimal Area >= your target, while the other branch finds the record with maximal Area < your target. One of these two records will be the record that you are looking for. Then you can ORDER BY ABS(Area - ?input) to pick the winner out of those two candidates. Unfortunately the query is complex due to nested SELECTS that are needed to enforce the desired ROWNUM / ORDER BY precedence.

SELECT *
FROM
  (SELECT * FROM
    (SELECT * FROM
       (SELECT * FROM [myTable]
         WHERE Name = 'Test' AND Size = 2 AND PType = 'p' AND Area >= ?target
         ORDER BY Area)
       WHERE ROWNUM < 2
     UNION ALL
     SELECT * FROM 
       (SELECT * FROM [myTable]
         WHERE Name = 'Test' AND Size = 2 AND PType = 'p' AND Area < ?target
         ORDER BY Area DESC)
       WHERE ROWNUM < 2)
     ORDER BY ABS(Area - ?target))
WHERE rownum < 2

A good index for this query would be (Name, Size, PType, Area) , in which case the expected query execution plan would be based on two index range scans that each returned a single row.

Answer 4

How about ordering by the difference between your input and [Area], such as:

DECLARE @InputValue DECIMAL(7, 3)

SET @InputValue = 1.125

SELECT TOP 1 * FROM [myTable]
WHERE Name = 'Test' AND Size = 2 AND PType = 'p'
ORDER BY ABS(@InputValue - Area)

Answer 5

SELECT * 
  FROM [myTable] 
  WHERE Name = 'Test' AND Size = 2 AND PType = 'p' 
  ORDER BY ABS(Area - 1.125)
  LIMIT 1

-- MarkusQ

Answer 6

如果使用MySQL

SELECT * FROM [myTable] ... ORDER BY ABS(Area - SuppliedValue) LIMIT 1

Answer 7

Note that although ABS() is supported by pretty much everything, it's not technically standard (in SQL99 at least). If you must write ANSI standard SQL for some reason, you'd have to work around the problem with a CASE operator:

SELECT * FROM myTable
WHERE Name='Test' AND Size=2 AND PType='p'
ORDER BY CASE Area>1.125 WHEN 1 THEN Area-1.125 ELSE 1.125-Area END

Answer 8

选择最小值[field]> your_target_value选择最大值[field] <your_target_value