Question

arrayA[3]==null?"":arrayA[3];

goods
id     name     rent_price
1     Toyota     400000
2     Honda      500000
3     Mazda      600000
4     Suzuki     700000
5     KIA        800000

customers
id     name
1     Ali
2     Budi
3     Candra
4     Doni
5     Eka

history_transactions
id    customer_id      date
1      1             2016-03-01 15:00:00
2      4             2016-03-02 15:00:00
3      4             2016-03-02 17:00:00
4      2             2016-03-03 09:00:00
5      4             2016-03-05 10:00:00
6      1             2016-04-06 10:00:00
7      5             2016-05-07 10:00:00
8      2             2016-06-08 10:00:00
9      1             2016-07-09 10:00:00
10     2             2016-08-10 10:00:00

我有很多数据历史交易的165.000行客户18.000行

我需要数天的客户为每个客户进行交易

history_rents
id    history_transaction_id         good_id
1                1                      1
2                2                      2
3                2                      5
4                3                      3
5                4                      1
6                4                      2
7                4                      3
8                4                      4
9                4                      5
10               5                      5
11               6                      3
12               7                      4
13               8                      3
14               9                      1
15               10                     1

我需要计算每个客户租用的物品

customers
id     name       day_trans
1     Ali            3
2     Budi           3
3     Candra         0
4     Doni           2 (history_transaction_id 2 & 3 count as 1)
5     Eka            1

请帮助查询以及我应该为索引

指定哪个字段

Answer 1

如果您希望通过两个不同的查询来输出两个输出：

输出＃1：

customers
id     name       day_trans
1     Ali            3
2     Budi           3
3     Candra         0
4     Doni           2 (history_transaction_id 2 & 3 count as 1)
5     Eka            1

查询＃1：

SELECT 
C.id,
C.name,
COUNT(DISTINCT DATE(HT.`date`)) AS day_trans
FROM customers C
INNER JOIN history_transactions HT 
ON C.id = HT.customer_id
GROUP BY C.id 
ORDER BY C.id;

输出＃2：

customers
id     name       rent_count
1     Ali            3
2     Budi           7
3     Candra         0
4     Doni           4
5     Eka            1

查询＃2：

SELECT 
    C.id,
    C.name,
    COALESCE(t.total,0) AS rent_count
FROM customers C
LEFT JOIN 
(
    SELECT 
    HT.customer_id,
    COUNT(*) total
    FROM history_rents HR 
    INNER JOIN history_transactions HT 
    ON HR.history_transaction_id = HT.id 
    GROUP BY HT.customer_id
) AS t
ON C.id = t.customer_id
ORDER BY C.id;

如果你想让他们通过一个：

SELECT 
    C.id,
    C.name,
    COALESCE(t.day_trans,0) AS day_trans
    COALESCE(t.total_rent,0) AS rent_count
FROM customers C
LEFT JOIN 
(
    SELECT 
    HT.customer_id,
    COUNT(DISTINCT DATE(HT.`date`)) AS day_trans,
    COUNT(*) total_rent
    FROM history_rents HR 
    INNER JOIN history_transactions HT 
    ON HR.history_transaction_id = HT.id 
    GROUP BY HT.customer_id
) AS t
ON C.id = t.customer_id

Answer 2

您可以在同一个查询中执行这两项操作：

SELECT c.id, c.name, count(distinct date(ht.date)) as day_trans, count(*) as rent_count
FROM customers c
LEFT JOIN history_transactions ht on c.id=ht.customer_id 
LEFT JOIN history_rents hr on ht.id=hr.history_transaction_id
GROUP by c.id

请求MySQL查询

2 个答案: