mysql左连接将行放在一起
我有三张表:发票、人员和付款。 我想有一个发票清单,上面有客户名称(来自个人)和付款总额以及付款日期(来自付款) 首先,我发表了这些声明mysql左连接将行放在一起,mysql,join,Mysql,Join,我有三张表:发票、人员和付款。 我想有一个发票清单,上面有客户名称(来自个人)和付款总额以及付款日期(来自付款) 首先,我发表了这些声明 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
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(即9月4日的1000笔交易和9月10日的2400笔交易中的一笔),但当我有很多发票时,工作非常缓慢
+------+-----------+--------+--------+-------+---------------------+
| id | datum | amount | client | payed | date payement |
+------+-----------+--------+--------+-------+---------------------+
| 75 |2015-09-10 | 3400 |Sommers | 3400 |2015-09-04,2015-09-10|
+------+-----------+--------+--------+-------+---------------------+
所以我尝试了另一种说法
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;
但这给了我2行1发票,当它是支付2交易。
我可以用SQL来解决它,还是在我的应用程序(Java)中解决它更好 当发票已支付两次付款时,您希望使用哪些详细信息?第一次付款还是第二次付款 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)
谢谢,这是可行的,但是它仍然很慢,但速度是我的第一个解决方案的两倍。作为“日期付款”,我不需要日期,而是不同日期的串联。因此我使用了:GROUP_CONCAT(B.datum)分隔符','),而不是MAX(B.datum)。索引可能需要调整以提高性能。此外,使用YEAR(V.datum)需要在每一行上使用一个函数,而不能使用索引,但是在'2015-01-01'和'2014-12-31'之间使用WHERE V.datum可能更快。据我所知,是'LEFT'连接需要时间。请发布表定义。还可以尝试对查询进行解释,然后发布。没有LEFT需要0.03秒,LEFT需要0.66秒。
但是我不会得到没有付款的发票。
如果我省略了LEFT,并且使用相同的语句进行联合,而没有JOIN payment,则需要0.08秒,但是我有多个副本(正如我所说,distinct不适用于1列)。