Question

我有3张桌子：发票，人和付款。我希望有一份发票清单，其中包含客户名称（来自人）以及付款总额和付款日期（来自付款）。

首先我做了这些陈述

SELECT V.id, V.datum, V.amount, P.name AS 'client', 
(SELECT SUM(B.amount) FROM payement AS B WHERE B.invoiceId = V.id) AS 'payed',
(SELECT GROUP_CONCAT(B.datum SEPARATOR ',') FROM payement AS B WHERE B.invoiceId = V.id) AS 'date payement'
FROM invoice AS V 
JOIN person AS P ON (V.clientId = P.id)
WHERE YEAR(V.datum) = '2015' 
ORDER BY V.datum;

这给了我想要的东西（例如，9月4日的1000次交易和9月10日的2400次交易），但是当我有很多发票时工作非常慢。

+------+-----------+--------+--------+-------+---------------------+
| id   |  datum    | amount | client | payed | date payement       |
+------+-----------+--------+--------+-------+---------------------+
| 75   |2015-09-10 | 3400   |Sommers | 3400  |2015-09-04,2015-09-10|
+------+-----------+--------+--------+-------+---------------------+

所以我尝试了另一个声明。

SELECT V.id, V.datum, V.amount, P.name AS 'client', B.amount AS 'payed', B.datum 'date payement'
FROM invoice AS V 
JOIN person AS P ON (V.clientId = P.id)
LEFT JOIN payement AS B ON B.invoiceId = V.id
WHERE YEAR(V.datum) = '2015' 
ORDER BY V.datum;

但是当它支付2笔交易时，这给我2张1张发票。我可以用SQL解决它，还是在我的应用程序（Java）中解决它更好？

Answer 1

如果已通过2次付款支付发票，您希望使用哪些详细信息？第一笔付款还是第二笔？

假设您需要总付款金额和最新付款日期： -

SELECT V.id, 
        V.datum, 
        V.amount, 
        P.name AS 'client', 
        SUM(B.amount) AS 'payed', 
        MAX(B.datum) AS 'date payement'
FROM invoice AS V 
JOIN person AS P ON (V.clientId = P.id)
LEFT OUTER JOIN payement AS B ON B.invoiceId = V.id
WHERE YEAR(V.datum) = '2015' 
GROUP BY V.id, 
        V.datum, 
        V.amount,
        P.name
ORDER BY V.datum

我不使用phpmyadmin。

mysql> EXPLAIN SELECT V.factuurnr, 
    ->         V.datum, 
    ->         V.somexcl, 
    ->         P.naam AS 'client', 
    ->         SUM(B.bedrag) AS 'payed', 
    ->         GROUP_CONCAT(DATE_FORMAT(B.datum,'%d/%m/%y') SEPARATOR ',') AS 'date payement'
    -> FROM verkoop AS V 
    -> JOIN persoon AS P ON (V.klantId = P.id)
    -> LEFT JOIN betaling AS B ON B.docId = V.id
    -> WHERE YEAR(V.datum) = '2015' and month(V.datum)=9
    -> GROUP BY V.factuurnr, 
    ->         V.datum, 
    ->         V.somexcl,
    ->         P.naam
    -> ORDER BY factuurnr;
+----+-------------+-------+--------+---------------+---------+---------+----------------+------+----------------------------------------------+
| id | select_type | table | type   | possible_keys | key     | key_len | ref            | rows | Extra                                        |
+----+-------------+-------+--------+---------------+---------+---------+----------------+------+----------------------------------------------+
|  1 | SIMPLE      | V     | ALL    | NULL          | NULL    | NULL    | NULL           | 1576 | Using where; Using temporary; Using filesort |
|  1 | SIMPLE      | P     | eq_ref | PRIMARY       | PRIMARY | 4       | meta.V.klantId |    1 | Using where                                  |
|  1 | SIMPLE      | B     | ALL    | NULL          | NULL    | NULL    | NULL           | 3291 |                                              |
+----+-------------+-------+--------+---------------+---------+---------+----------------+------+----------------------------------------------+
3 rows in set (0.00 sec)

mysql左连接将行连在一起

1 个答案: