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Home/ Questions/Q 8588991
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T22:56:12+00:00 2026-06-11T22:56:12+00:00

Given the following table info: A B C D E 1 786788 123.45 0

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Given the following table info:

A       B           C           D       E

1       786788      123.45      0       617
2       786788      654.21      0       3271
3       786788      111.11      0       556
4       786788      224.45      0       1122
5       786788      0.12        12      0   
6       786788      3.21        321     0   
7       786788      0.22        22      0   
8       786788      0.33        33      0   
9       786788      757.57      0       3788
10      786788      545.41      0       2727
11      786788      0.07        7       0   
12      786788      0.05        5       0   
13      786790      50          0       125 
14      786790      50          0       125 
15      786790      50          0       125 
16      786790      50          0       125 
17      786790      50          0       125 
18      786788      1.23        0       6   
19      786788      0.12        0       1   
20      786788      0.12        0       1   
21      786788      0.05        0       0   
22      786788      0.55        0       3   
23      786788      0.45        45      0   
24      786788      1.47        0       7   
25      786788      0.41        0       2   
26      786788      0.74        74      0   
27      786788      100         0       500 
36      786788      5           0       25  
37      786788      24          0       120 
38      786788      23          0       115 
39      786788      1.47        0       7   
40      786788      12          0       60  
41      786788      14          0       70  
43      786222      7.8         0       39  
44      786222      0.12        12      0   
45      786788      47.47       0       237 
46      786788      9.89        989     0   
47      786788      1.23        0       6   
48      786788      1.23        0       6   

I’m trying to get the following info in one query:

  1. DISTINCT(column B)
  2. COUNT(DISTINCT(column B))
  3. SUM(column C) WHERE column E != 0
  4. SUM(column D)
  5. SUM(column E)

If all goes well, the result should look like this example (I didn’t “SUM” everything for the example):

786788      32      SUM(column C) WHERE column E != 0       SUM(column D)       SUM(column E)
786790      5       SUM(column C) WHERE column E != 0       SUM(column D)       SUM(column E)
786222      2       SUM(column C) WHERE column E != 0       SUM(column D)       SUM(column E)
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T22:56:13+00:00Added an answer on June 11, 2026 at 10:56 pm

    Unless I’m mistaken, your proposed field-list is pretty close to what you’d actually need in a query. I haven’t tested this, but give this one a shot (it uses MySQL’s GROUP BY to do the DISTINCT/grouping you require):

    SELECT
        B,
        COUNT(*),
        SUM((CASE WHEN E != 0 THEN C ELSE 0 END)),
        SUM(D),
        SUM(E)
    
    FROM
        your_table
    
    GROUP BY B
    
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