0

嗨，我想使用带有跟踪列表的 MySQL 过滤日志。

每个日志都属于一个服务器，

每个跟踪都属于一个服务器并且有 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没有模式的跟踪。

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加入时，如果我使用INNER JOIN patterns as pt ON我得到正确的结果，但只在有一些模式的跟踪上，如果我使用LEFT JOIN patterns as pt ON我得到所有跟踪但计数错误（我得到的结果SELECT COUNT(DISTINCT logs.ip) FROM logs）

编辑 我可以通过跟踪中的字段获得正确的结果，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在不使用 Union 的情况下做到这一点......

4

1 回答 1

0

你可以在里面放一个条件count，所以我认为以下是你想要的：

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：

这是返回你想要的：

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lg.size > tr.size原始问题中没有的附加条件。

于 2013-08-14T17:04:05.147 回答