Question

我有3个表，定义如下。（文件夹和地址簿条目，连接表中的连接为1：多表。地址簿可以在多个文件夹中）

我正在尝试获取文件夹列表，这很容易。我添加了COUNT（总数）文件夹中有多少条目，这也很容易。

但是我还需要添加另外一个COUNT，其中COUNT（总数）中有多少人的addressbook_lastchecked为0

所以或多或少我需要计算该文件夹中的总名称数，然后再计算其中有多少总数未经检查...

**addressbook**
    addressbook_id
    addressbook_lastchecked
    addressbook_name

**folders**
    folder_id
    folder_name

**folders_connections**
    connection_id
    folder_id
    addressbook_id

到目前为止我的努力..我需要选择不选择吗？

SELECT f . * , c.lastchecked, COUNT( c.lastchecked ) AS entries
FROM folders AS f
LEFT JOIN (

SELECT fc.folder_id, a.lastchecked
FROM folder_connections AS fc
LEFT JOIN addressbook AS a
USING ( addressbook_id ) 
WHERE a.lastchecked =0
) AS c
USING ( folder_id )

我的额外努力是（这是不正确的，因为它有重复。）

SELECT f.*, COUNT(c.connection_id) AS entries, COUNT(csub.connection_id) AS unchecked  
FROM folders AS f 
LEFT JOIN folder_connections AS c USING(folder_id)
LEFT JOIN (SELECT fc.*, ab.lastchecked FROM folder_connections AS fc 
     LEFT JOIN addressbook AS ab USING(addressbook_id) WHERE ab.lastchecked=0) AS csub USING(folder_id)
GROUP BY c.folder_id ORDER BY folder_name Asc LIMIT 0, 15

通过混合两件事来解决。

SELECT f.folder_id, f.folder_name, SUM( 
CASE c.lastchecked
WHEN 0 
THEN 1 
ELSE 0 
END ) AS unchecked, COUNT( c.lastchecked ) AS entries
FROM folders AS f
LEFT JOIN (

SELECT fc.folder_id, a.lastchecked
FROM folder_connections AS fc
INNER JOIN addressbook AS a
USING ( addressbook_id )
)c
USING ( folder_id ) 
GROUP BY f.folder_id, f.folder_name
LIMIT 0 , 30

Answer 1

在我理解的范围内是解决方案：

首先选择addressbook_lastchecked is 0

所在的企业

SELECT fc.folder_id, a.lastchecked
FROM folder_connections AS fc
Inner JOIN addressbook AS a // Inner join is a best option as you get all folders
USING ( addressbook_id ) 
WHERE a.lastchecked =0

//现在您拥有所有上次检查的文件夹0

现在通过组合两个查询来计算文件夹数和最后检查数。

SELECT f . folder_id,f.folder_name
 , count(c.lastchecked) checkedcount, COUNT( c.lastchecked ) AS entries
FROM folders AS f left join (    SELECT fc.folder_id, a.lastchecked
    FROM folder_connections AS fc
    Inner JOIN addressbook AS a // Inner join is a best option as you get all folders
    USING ( addressbook_id ) 
    WHERE a.lastchecked =0) c
USING ( folder_id ) 
group by f . folder_id,f.folder_name

现在你有文件夹ID和相应的计数和检查计数。

如果您不是在寻找，请告诉我。

mysql - 从不同的表计数补充来自subresult的另一个计数

1 个答案: