如何正确设计 SQL 查询以实现最优行 output

Question

How to properly design a SQL query code to achieve most optimal row output:

select (top 3) C1 from Table where...... (for given input c2 to c9)

Suppose there is one table which has 9 (data is percentages values) columns

Imagine Table has millions of rows.

     C1   C2   C3   C4   C5   C6   C7   C8   C9
     ==========================================
     0.1  0.9  0.4  0.7  0.7  0.3  0.5  0.8  0.9  R1
     0.6  0.7  0.3  0.5  0.8  0.9  0.1  0.9  0.4  R2
     0.1  0.9  0.4  0.7  0.7  0.3  0.5  0.8  0.9  R3
     0.1  0.7  0.3  0.5  0.8  0.9  0.3  0.5  0.8  R4
     0.7  0.7  0.9  0.1  0.9  0.4  0.2  0.2  0.2  R5
     0.7  0.7  0.3  0.5  0.8  0.9  0.4  0.7  0.2  R6

Result would be top 3 C1 values where the sum of absolute difference for each column value is minimal for give input of C2... to C9.

For example if input is C2=0.7, C3=0.3, C4=0.6, C5=0.8 C6=0.9 C7=0.3 C8=0.9 C9=0.2 then

Top 3 C1 values returned by SQL query should be:

     Result1 is 0.7 (as result was 0.4 from R6)
     Result2 is 0.6 (as result was 0.5 from R2)
     Result3 is 0.7 (as result was 2.5 from R5) 

Calculations are explained below:

           C2=0.7    C3=0.3    C4=0.6    C5=0.8    C6=0.9    C7=0.3    C8=0.9    C9=0.2
     R6 =  (0.7-0.7)+(0.3-0.3)+(0.6-0.5)+(0.8-0.8)+(0.9-0.9)+(0.3-0.4)+(0.9-0.7)+(0.2-0.2)
     absolute  0    +   0     +   0.1   +   0     +   0     +   0.1     +   0.2 +   0
     result = 0.4


     R5 =  (0.7-0.7)+(0.3-0.9)+(0.6-0.1)+(0.8-0.9)+(0.9-0.4)+(0.3-0.2)+(0.9-0.2)+(0.2-0.2)
     absolute  0.0  +   0.6   +   0.5   +   0.1   +   0.5   +   0.1   +   0.7   +   0.0
     result = 2.5

     R2 =  (0.7-0.7)+(0.3-0.3)+(0.6-0.5)+(0.8-0.8)+(0.9-0.9)+(0.3-0.1)+(0.9-0.9)+(0.2-0.4)
     absolute  0.0  +   0.0   +   0.1   +   0.0   +   0.0   +   0.2   +   0.0   +   0.2
     result = 0.5

Answer 1

This should give you C1 values ordered ascending by your formula:

SELECT C1
FROM PERCENTAGES
ORDER BY ABS(C2 - 0.7) + ABS(C3 - 0.3) + ABS(C4 - 0.6) + ABS(C5 - 0.8) + 
         ABS(C6 - 0.9) + ABS(C7 - 0.3) + ABS(C8 - 0.9) + ABS(C9 - 0.2);

Tested with Oracle... but it's plain SQL that should work anywhere. To get the top 3 values, you'll have to apply a DB-specific clause.

For MySQL it should be easy. Just add a LIMIT 3 to the query.

I recommend to replace the constant input with variables and use a prepared statement to actually run the query. But that's a topic for another question.