sql嵌套连接内部和左连接

Question

嗨，我想使用带有跟踪列表的MySQL过滤日志。

每个日志都属于一台服务器，

每个跟踪都属于服务器，并且具有0..N个模式

每个模式都属于一个跟踪

我有3张桌子：

logs     : |  id  |      ip     | url     | server_id | ...
tracking : |  id  | server_id   | name    | other fields...
pattern  : |  id  | tracking_id | pattern |

我想计算与特定服务器的跟踪匹配的日志，我的问题是我的查询混淆了有模式和无模式的跟踪。

SQL小提琴： http ://sqlfiddle.com/#!2/f11b1/2

SELECT COUNT(DISTINCT logs.ip), tr.name
FROM `logs`
INNER JOIN `trackings` as tr ON 
        ( tr.server_id = logs.server_id )
        AND -- OTHER conditions between log and tracking
    LEFT JOIN `patterns` as pt ON 
        ( pt.tracking_id = tr.id )
        AND (logs.url LIKE pt.pattern )
GROUP BY tr.id

我的问题是在第二次INNER JOIN patterns as pt ON ，如果我将INNER JOIN patterns as pt ON我将获得正确的结果，但仅在具有某些模式的跟踪上才能获得结果;如果我将LEFT JOIN patterns as pt ON我将获得所有跟踪，但计数错误（ SELECT COUNT(DISTINCT logs.ip) FROM logs获取SELECT COUNT(DISTINCT logs.ip) FROM logs的结果

编辑我可以在跟踪字段中获得正确的结果，该字段指示跟踪是否具有模式和UNION：

(
SELECT COUNT(DISTINCT lg.ip), tr.name
FROM `logs` as lg
INNER JOIN `trackings` as tr ON 
        ( tr.server_id = lg.server_id )
        AND (tr.hasPatterns = 1)
        AND -- Other conditions
    INNER JOIN `patterns` as pt ON 
        ( pt.tracking_id = tr.id )
        AND (lg.url LIKE pt.pattern )
WHERE
GROUP BY tr.id
)
UNION
(
SELECT COUNT(DISTINCT lg.ip), tr.name
FROM `logs` as lg
INNER JOIN `trackings` as tr ON 
        ( tr.server_id = lg.server_id )
        AND (tr.hasPatterns = 0)
WHERE
GROUP BY tr.id, lg.date
)

但我想有一种方法可以不使用Union ...

Answer 1

您可以添加一个有条件的内部count ，因此我认为以下内容可以满足您的需求：

SELECT COUNT(DISTINCT (case when tr.hasPatterns = 1 and pt.tracking_id is not null
                            then lg.ip
                            when tr.hasPatterns = 0
                            then lg.ip
                       end)), tr.name
FROM `logs` as lg
INNER JOIN `trackings` as tr ON 
        ( tr.server_id = lg.server_id )
        AND -- Other conditions
    LEFT JOIN `patterns` as pt ON 
        ( pt.tracking_id = tr.id )
        AND (lg.url LIKE pt.pattern )
WHERE
GROUP BY tr.id

编辑：

这将返回您想要的：

SELECT COUNT(DISTINCT (case when tr.size = 0 and pt.tracking_id is not null
                            then lg.ip
                            when tr.size > 0 and lg.size > tr.size
                            then lg.ip
                       end)), tr.name
FROM `logs` as lg
INNER JOIN `trackings` as tr ON 
        ( tr.server_id = lg.server_id )
    LEFT JOIN `patterns` as pt ON 
        ( pt.tracking_id = tr.id )
        AND (lg.url LIKE pt.pattern )
GROUP BY tr.id;

您的SQL Fiddle具有附加条件lg.size > tr.size ，该条件lg.size > tr.size原始问题中。