Mysql 计算各门店每月创建的平均潜在客户数_Mysql_Sql

Mysql 计算各门店每月创建的平均潜在客户数

mysql sql

Mysql 计算各门店每月创建的平均潜在客户数,mysql,sql,Mysql,Sql,我想计算每个商店每月创建的潜在客户（行）的平均数量模式和输入： CREATE TABLE leads (`id` int, `store_id` int, `created_at` datetime) ; INSERT INTO leads (`id`, `store_id`, `created_at`) VALUES (5211, 1, '2019-09-13 23:29:29'), (5212, 1, '2019-08-13 21:29:29'),

我想计算每个商店每月创建的潜在客户（行）的平均数量

模式和输入：

CREATE TABLE leads
    (`id` int, `store_id` int, `created_at` datetime)
;

INSERT INTO leads
    (`id`, `store_id`, `created_at`)
VALUES
    (5211, 1, '2019-09-13 23:29:29'),
    (5212, 1, '2019-08-13 21:29:29'),
    (5781, 1, '2019-08-16 21:29:29'),
    (3349, 5, '2019-10-16 23:29:29'),
    (3344, 5, '2019-10-16 23:29:29'),
    (6291, 8, '2019-08-14 21:29:29'),
    (6292, 8, '2019-08-14 22:29:29'),
    (6299, 8, '2019-08-14 11:29:29'),
    (7799, 8, '2019-10-16 23:29:29'),
    (9898, 8, '2019-08-13 23:29:29'),
    (7791, 8, '2019-10-16 23:29:29'),
    (7792, 8, '2019-10-16 23:29:29'),
    (7793, 8, '2019-10-16 23:29:29'),
    (7794, 8, '2019-10-16 23:29:29'),
    (7795, 8, '2019-10-16 23:29:29')
;

average_leads_per_month_per_store, month_name
3, 2018-08
1, 2018-09
4, 2018-10

SELECT COUNT(id) AS leads_count, DATE_FORMAT(created_at, "%Y-%m") month_name, store_id, created_at
FROM leads
GROUP BY month_name, store_id
ORDER BY month_name, store_id;

| leads_count | month_name | store_id |           created_at |
|-------------|------------|----------|----------------------|
|           2 |    2019-08 |        1 | 2019-08-13T21:29:29Z |
|           4 |    2019-08 |        8 | 2019-08-14T21:29:29Z |
|           1 |    2019-09 |        1 | 2019-09-13T23:29:29Z |
|           2 |    2019-10 |        5 | 2019-10-16T23:29:29Z |
|           6 |    2019-10 |        8 | 2019-10-16T23:29:29Z |

所需输出：

CREATE TABLE leads
    (`id` int, `store_id` int, `created_at` datetime)
;

INSERT INTO leads
    (`id`, `store_id`, `created_at`)
VALUES
    (5211, 1, '2019-09-13 23:29:29'),
    (5212, 1, '2019-08-13 21:29:29'),
    (5781, 1, '2019-08-16 21:29:29'),
    (3349, 5, '2019-10-16 23:29:29'),
    (3344, 5, '2019-10-16 23:29:29'),
    (6291, 8, '2019-08-14 21:29:29'),
    (6292, 8, '2019-08-14 22:29:29'),
    (6299, 8, '2019-08-14 11:29:29'),
    (7799, 8, '2019-10-16 23:29:29'),
    (9898, 8, '2019-08-13 23:29:29'),
    (7791, 8, '2019-10-16 23:29:29'),
    (7792, 8, '2019-10-16 23:29:29'),
    (7793, 8, '2019-10-16 23:29:29'),
    (7794, 8, '2019-10-16 23:29:29'),
    (7795, 8, '2019-10-16 23:29:29')
;

average_leads_per_month_per_store, month_name
3, 2018-08
1, 2018-09
4, 2018-10

SELECT COUNT(id) AS leads_count, DATE_FORMAT(created_at, "%Y-%m") month_name, store_id, created_at
FROM leads
GROUP BY month_name, store_id
ORDER BY month_name, store_id;

| leads_count | month_name | store_id |           created_at |
|-------------|------------|----------|----------------------|
|           2 |    2019-08 |        1 | 2019-08-13T21:29:29Z |
|           4 |    2019-08 |        8 | 2019-08-14T21:29:29Z |
|           1 |    2019-09 |        1 | 2019-09-13T23:29:29Z |
|           2 |    2019-10 |        5 | 2019-10-16T23:29:29Z |
|           6 |    2019-10 |        8 | 2019-10-16T23:29:29Z |

我尝试过的：

CREATE TABLE leads
    (`id` int, `store_id` int, `created_at` datetime)
;

INSERT INTO leads
    (`id`, `store_id`, `created_at`)
VALUES
    (5211, 1, '2019-09-13 23:29:29'),
    (5212, 1, '2019-08-13 21:29:29'),
    (5781, 1, '2019-08-16 21:29:29'),
    (3349, 5, '2019-10-16 23:29:29'),
    (3344, 5, '2019-10-16 23:29:29'),
    (6291, 8, '2019-08-14 21:29:29'),
    (6292, 8, '2019-08-14 22:29:29'),
    (6299, 8, '2019-08-14 11:29:29'),
    (7799, 8, '2019-10-16 23:29:29'),
    (9898, 8, '2019-08-13 23:29:29'),
    (7791, 8, '2019-10-16 23:29:29'),
    (7792, 8, '2019-10-16 23:29:29'),
    (7793, 8, '2019-10-16 23:29:29'),
    (7794, 8, '2019-10-16 23:29:29'),
    (7795, 8, '2019-10-16 23:29:29')
;

average_leads_per_month_per_store, month_name
3, 2018-08
1, 2018-09
4, 2018-10

SELECT COUNT(id) AS leads_count, DATE_FORMAT(created_at, "%Y-%m") month_name, store_id, created_at
FROM leads
GROUP BY month_name, store_id
ORDER BY month_name, store_id;

| leads_count | month_name | store_id |           created_at |
|-------------|------------|----------|----------------------|
|           2 |    2019-08 |        1 | 2019-08-13T21:29:29Z |
|           4 |    2019-08 |        8 | 2019-08-14T21:29:29Z |
|           1 |    2019-09 |        1 | 2019-09-13T23:29:29Z |
|           2 |    2019-10 |        5 | 2019-10-16T23:29:29Z |
|           6 |    2019-10 |        8 | 2019-10-16T23:29:29Z |

这是按年度、月份和门店分组的，现在我需要计算各个门店每月的平均值

据我所知，这将为您提供平均值

SELECT COUNT(id)*1.0/(1.0*count(distinct store_id)) AS leads_count, DATE_FORMAT(created_at, "%Y-%m") month_name
FROM leads T
GROUP BY month_name
ORDER BY month_name;

根据你的结果,

在

2019-08

月，您有

2+4个Lead\u count

每个

1和8

store\u id，因此，您有

（6/2=3）

在2019-09年

month，每个1
store\u id您有1个Lead\u count
，因此，您有（1/1=1）

在2019-10
月，您有2+6个Lead\u count
每个5和8
store\u id，因此，您有（8/2=4）

因此，您需要计算Lead\u count
然后划分store\u id
的总数
你可以通过这种方式实现它
SELECT COUNT(id) / Count(Distinct (store_id)) AS average_leads_per_month_per_store, DATE_FORMAT(created_at, "%Y-%m") month_name
FROM leads T
GROUP BY month_name
ORDER BY month_name;

实时演示。
MySQL中最简单的语法似乎是：
SELECT DATE_FORMAT(created_at, '%Y-%m') as month_name,
       COUNT(*) / COUNT(DISTINCT store_id) as avg_per_month
FROM leads T
GROUP BY month_name
ORDER BY month_name;

编辑：
如果您需要统计没有潜在客户的门店，那么您需要所有门店的一些来源。您大概有一个存储
表：
SELECT DATE_FORMAT(l.created_at, '%Y-%m') as month_name,
       COUNT(*) / s.num_stores as avg_per_month
FROM leads l. CROSS JOIN
     (SELECT COUNT(*) as num_stores FROM stores) s
GROUP BY l.month_name, s.num_stores
ORDER BY month_name;

这回答了你的问题吗？如果您需要帮助，请告诉我@Andrew.MastonNote，DISTINCT不是一个功能，如果某个商店在某个特定月份没有潜在客户，该怎么办？它是被忽略还是被计为0
？它应该被计为@GordonLinoff。