Sql Postgres：字符串聚合和连接

如果我有这样一个问题：

SELECT 
    u.client_code, 
    max(d.created_at) as last_document_created_at, 
    u.brand_id, 
    t.name as template_name, 
    count(d)
FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
GROUP BY 1, 3, 4

client_code last_document_created_at    brand_id template_name  count
---------------------------------------------------------------------
client1     2017-12-06 10:03:47 +1100   39       newsletter_r   1
client1     2017-12-05 15:23:24 +1100   39       Other media    5
client2     2017-12-21 17:07:11 +1100   39       newsletter_r   4
client3     2018-01-11 12:10:43 +1100   39       newsletter_r   2
client3     2017-12-06 11:45:21 +1100   39       Other media    1

SELECT u.client_code, 
       max(d.created_at) as last_document_created_at, 
       u.brand_id, 
       string_agg(t.name, ',') as template_name, 
       count(distinct d.id)
FROM users u INNER JOIN
     documents d
     ON u.id = d.user_id INNER JOIN
     templates t
     ON t.id = d.template_id
GROUP BY 1, 3;

SELECT client_code, STRING_AGG(template_name || template_count, ',') 
FROM (
    SELECT 
        u.client_code, 
        MAX(d.created_at) AS last_document_created_at, 
        u.brand_id, 
        t.name AS template_name, 
        COUNT(d) AS template_count
    FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
    GROUP BY 1, 3, 4
) src
GROUP BY client_code

返回如下信息：

SELECT 
    u.client_code, 
    max(d.created_at) as last_document_created_at, 
    u.brand_id, 
    t.name as template_name, 
    count(d)
FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
GROUP BY 1, 3, 4

client_code last_document_created_at    brand_id template_name  count
---------------------------------------------------------------------
client1     2017-12-06 10:03:47 +1100   39       newsletter_r   1
client1     2017-12-05 15:23:24 +1100   39       Other media    5
client2     2017-12-21 17:07:11 +1100   39       newsletter_r   4
client3     2018-01-11 12:10:43 +1100   39       newsletter_r   2
client3     2017-12-06 11:45:21 +1100   39       Other media    1

SELECT u.client_code, 
       max(d.created_at) as last_document_created_at, 
       u.brand_id, 
       string_agg(t.name, ',') as template_name, 
       count(distinct d.id)
FROM users u INNER JOIN
     documents d
     ON u.id = d.user_id INNER JOIN
     templates t
     ON t.id = d.template_id
GROUP BY 1, 3;

SELECT client_code, STRING_AGG(template_name || template_count, ',') 
FROM (
    SELECT 
        u.client_code, 
        MAX(d.created_at) AS last_document_created_at, 
        u.brand_id, 
        t.name AS template_name, 
        COUNT(d) AS template_count
    FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
    GROUP BY 1, 3, 4
) src
GROUP BY client_code

我有哪些选项可以连接

template\u name

和

count

字段，以便每个用户（在

u.client\u code

中表示）都在一行上？我知道我可以像这样调用

string\u agg

列：

...
string_agg(distinct t.name, ', ') as template_name, 
...

但这当然破坏了各自的计算：

newsletter_r, Other media   6

更新

我可以这样做：

string_agg(concat_ws(': ', t.name::text, count(d)::text), ', ') as template_count

但这给了我一个错误：

aggregate function calls cannot be nested LINE 5: string_agg(concat_ws(': ', t.name::text, count(d)::text)... ^ : SELECT u.client_code,

---

我想你想要这样的东西：

SELECT 
    u.client_code, 
    max(d.created_at) as last_document_created_at, 
    u.brand_id, 
    t.name as template_name, 
    count(d)
FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
GROUP BY 1, 3, 4

client_code last_document_created_at    brand_id template_name  count
---------------------------------------------------------------------
client1     2017-12-06 10:03:47 +1100   39       newsletter_r   1
client1     2017-12-05 15:23:24 +1100   39       Other media    5
client2     2017-12-21 17:07:11 +1100   39       newsletter_r   4
client3     2018-01-11 12:10:43 +1100   39       newsletter_r   2
client3     2017-12-06 11:45:21 +1100   39       Other media    1

SELECT u.client_code, 
       max(d.created_at) as last_document_created_at, 
       u.brand_id, 
       string_agg(t.name, ',') as template_name, 
       count(distinct d.id)
FROM users u INNER JOIN
     documents d
     ON u.id = d.user_id INNER JOIN
     templates t
     ON t.id = d.template_id
GROUP BY 1, 3;

SELECT client_code, STRING_AGG(template_name || template_count, ',') 
FROM (
    SELECT 
        u.client_code, 
        MAX(d.created_at) AS last_document_created_at, 
        u.brand_id, 
        t.name AS template_name, 
        COUNT(d) AS template_count
    FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
    GROUP BY 1, 3, 4
) src
GROUP BY client_code

---

不确定要如何设置连接字段的格式，但是否尝试将原始查询放入子查询并对其应用

string\u agg

？大概是这样的：

SELECT 
    u.client_code, 
    max(d.created_at) as last_document_created_at, 
    u.brand_id, 
    t.name as template_name, 
    count(d)
FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
GROUP BY 1, 3, 4

client_code last_document_created_at    brand_id template_name  count
---------------------------------------------------------------------
client1     2017-12-06 10:03:47 +1100   39       newsletter_r   1
client1     2017-12-05 15:23:24 +1100   39       Other media    5
client2     2017-12-21 17:07:11 +1100   39       newsletter_r   4
client3     2018-01-11 12:10:43 +1100   39       newsletter_r   2
client3     2017-12-06 11:45:21 +1100   39       Other media    1

SELECT u.client_code, 
       max(d.created_at) as last_document_created_at, 
       u.brand_id, 
       string_agg(t.name, ',') as template_name, 
       count(distinct d.id)
FROM users u INNER JOIN
     documents d
     ON u.id = d.user_id INNER JOIN
     templates t
     ON t.id = d.template_id
GROUP BY 1, 3;

SELECT client_code, STRING_AGG(template_name || template_count, ',') 
FROM (
    SELECT 
        u.client_code, 
        MAX(d.created_at) AS last_document_created_at, 
        u.brand_id, 
        t.name AS template_name, 
        COUNT(d) AS template_count
    FROM users u
    INNER JOIN documents d ON u.id = d.user_id
    INNER JOIN templates t ON t.id = d.template_id
    GROUP BY 1, 3, 4
) src
GROUP BY client_code

---

我还没有测试它，所以您可能有一些语法错误。让我知道这是否有效。

您的

字符串\u agg（）

中有一个错误（需要两个参数）。而且，当这个问题得到解决时，它不会产生任何与我的查询不同的结果。效果非常好！但是，我不知道如何通过

last\u document\u created\u at

列，因为我被迫将

u.client\u code

拆分为多行的列进行分组：

column“src.last\u document\u created\u at”必须出现在group by子句中或用于聚合函数中，如何确定每个
客户机代码在
值处创建的最后一个
文档要返回哪个
？或者您只是将其包含在
字符串\u agg
中？我刚刚做了另一个
agg
：
（数组\u agg（上次创建的文档））[1]作为上次创建的
。完成了（我想？）。非常感谢！呜呼！很高兴它成功了：）另一个选择是应用MAX--
MAX（上次创建文档时）
。噢，天哪。再次感谢！