3

我在这里的第一篇文章，希望我遵循正确的发布指南。如果不是，请指出。

我的带有一些示例数据的表结构（credit_notes）如下所示：

index | customer_id | user_id | doc_number | doc_date   | amount | scheme_description
+-----+-------------+---------+------------+------------+--------+-------------------
1     | 8           | 2       | IABCD001   | 2012-09-08 | 1232   | XYZ Scheme
2     | 8           | 2       | IABCD001   | 2012-09-08 | 2032   | XYZ Scheme 
3     | 8           | 2       | IABCD002   | 2012-09-09 | 2556   | XYZ Scheme 
4     | 8           | 2       | IABCD003   | 2012-09-08 | 232    | XYZ Scheme 
5     | 8           | 2       | IABCD001   | 2012-09-10 | 2556   | ABC Scheme
6     | 8           | 2       | IABCD003   | 2012-09-10 | 232    | ABC Scheme

我的要求是通过分组提取每个 scheme_description 和 doc_date 组合的总金额。这是我当前的查询：

SELECT 
    SUM( `amount` ) AS `totalamount`,
    GROUP_CONCAT( CONCAT( `amount` , ";" ) ORDER BY `index` DESC SEPARATOR '' ) AS `amounts`, 
    GROUP_CONCAT( CONCAT( `doc_number` , ";" ) ORDER BY `doc_number` DESC SEPARATOR '' ) AS `docno` , 
    `doc_date` AS `docdate` , 
    `scheme_description` AS `schemedesc` 
FROM (
        SELECT 
            `c`.`index` ,
            `c`.`doc_number` , 
            `c`.`doc_date` , 
            `c`.`amount` , 
            `c`.`scheme_description` 
        FROM `credit_notes` AS `c` 
        WHERE `c`.`customer_id` = 8 
        AND `c`.`user_id` LIKE 2
)t
GROUP BY `scheme_description` , `doc_date`

此查询返回此结构：

totalamount | amounts        | docno                       | docdate    | schemedesc
+-----------+----------------+-----------------------------+------------+-----------
3496        | 1232;2032;232; | IABCD001;IABCD001;IABCD003; | 2012-09-08 | XYZ Scheme
2556        | 2556;          | IABCD002;                   | 2012-09-09 | XYZ Scheme
23477       | 255;23222;     | IABCD007;IABCD007;          | 2012-09-10 | ABC Scheme

但是，我需要对此稍作改动，我想对每个不同docno的数量列求和，并按正确的顺序对其进行分组。因此，根据上述数据，查询应产生：

totalamount | amounts        | docno                       | docdate    | schemedesc
+-----------+----------------+-----------------------------+------------+-----------
3496        | 3264;232;      | IABCD001;IABCD003;          | 2012-09-08 | XYZ Scheme
2556        | 2556;          | IABCD002;                   | 2012-09-09 | XYZ Scheme
23477       | 23477;         | IABCD007;                   | 2012-09-10 | ABC Scheme

这可能使用 group_concat 和 distinct 和 sum。阅读我能找到的所有内容，但找不到任何指向正确方向的内容。

这可能吗？

4

3 回答 3

1

您的第一个查询可以这样写：

select
  doc_date, scheme_description,
  group_concat(doc_number),
  group_concat(amount),
  sum(amount)
from
  credit_notes
group by doc_date, scheme_description

你的第二个：

select
  doc_date, scheme_description,
  group_concat(doc_number),
  group_concat(amount),
  sum(amount)
from (
  select doc_date, scheme_description, doc_number, sum(amount) as amount
  from credit_notes
  group by doc_date, scheme_description, doc_number
) s
group by doc_date, scheme_description
于 2012-12-05T10:53:11.377 回答
0

试试这个查询

select 
    group_concat(doc_num) as DocNumm,
    group_concat(amount) as DocNumm,
    scheme_description as SchemaDesc,
    sum(amount) as TotalAmount,
    amount as Amount,
    doc_date as DocDate
from (
    SELECT * from `schema` 
) as g
group by SchemaDesc , DocDate
于 2012-12-05T11:05:10.457 回答
0

试试这个查询 -

SELECT
  SUM(amount) totalamount,
  GROUP_CONCAT(amount) amounts,
  GROUP_CONCAT(doc_number) docno,
  doc_date docdate,
  scheme_description schemedesc
FROM (
  SELECT scheme_description, doc_date, doc_number, SUM(amount) amount
    FROM credit_notes
    GROUP BY scheme_description, doc_date, doc_number) t
GROUP BY scheme_description, doc_date

如果需要，添加 WHERE、ORDER BY 和其他子句。

于 2012-12-05T10:48:59.517 回答