Question

我试图通过选择2列的不同组合来获取行（“column_a”和“column_b”，如下例所示），并在MySQL中使用一个sql查询获取每个区别的结果的最新日期时间行。最好是看样本数据而不是阅读我的英语。

sample_table（简化了“created_datetime”列）

SELECT SUM(a) as total_a, SUM(b) as total_b, (sum(a) + sum(b)) as grand_total
FROM table

我理想的结果

id column_a column_b column_c created_datetime
-----------------------------------------
1 100 200 aaaa t
2 200 100 bbbb t+20
3 100 300 cccc t+40
4 100 200 zzzzzzz t+60

以下是我尝试的一些模式：

A：

id column_a column_b column_c created_datetime
-----------------------------------------
3 100 300 cccc t+40
4 100 200 zzzzzzz t+60

结果将是

SELECT DISTINCT
    column_a,
    column_b,
    MAX(created_datetime)
FROM sample_table
GROUP BY column_a, column_b;

B：

100 200 t+60
100 300 t+40
200 100 t+20

结果将是

SELECT DISTINCT 
    least(column_a, column_b) as column_a,
    greatest(column_a, column_b) as column_b
FROM sample_table
GROUP BY column_a, column_b;

C：

100 200
100 300

结果将是

SELECT DISTINCT
least(column_a, column_b) as column_a,
    greatest(column_a, column_b) as column_b,
    MAX(created_datetime)
FROM sample_table
GROUP BY column_a, column_b;

我试图在没有任何其他语言（如JavaScript或Python）的结果获得数据后进行调整。是不可能用一个SQL查询来实现这一点，还是我需要采取进一步的MySQL步骤？提前谢谢。

Answer 1

SELECT sample_table.column_a,
       sample_table.column_b,
       sample_table.column_c,
       sample_table.created_datetime
FROM sample_table
     INNER JOIN (SELECT least(column_a, column_b) as column_a,
                        greatest(column_a, column_b) as column_b,
                        MAX(created_datetime) as created_datetime
                 FROM sample_table
                 GROUP BY least(column_a, column_b),
                          greatest(column_a, column_b)) t
     ON t.column_a = least(sample_table.column_a, sample_table.column_b)
     AND t.column_b = greatest(sample_table.column_a, sample_table.column_b)
     AND t.created_datetime = sample_table.created_datetime

选择2列的不同组合，并从每个区别中获取最新的日期时间行

1 个答案: