2014-01-30 50 views
0

我知道我错过了一些东西,我尝试了几种不同的方法,但我忽略了(或者太努力)。有人可以告诉我在这个SQL中我错了吗?SQL JOIN计数和SUM问题

SELECT id, 
    COUNT(id) AS dupBlocks 
FROM tbl_duplicates8 INNER JOIN (
     tbl_accounts8, 
     tbl_delaccounts, 
     tbl_bad_bots, 
     tbl_log, 
     tbl_ipban, 
     tbl_ipban8 
    ) ON (
     tbl_accounts8.SUM(num_attacks) AND 
     tbl_delaccounts.SUM(noattacks) AND 
     tbl_bad_bots.COUNT(id) AND 
     tbl_log.COUNT(id) AND 
     tbl_ipban.COUNT(txt_ip) AND 
     tbl_ipban8.COUNT(ip) 
    ); 

我没有注意到这个MySQL Join two tables count and sum from second table,但它返回时返回null。

任何帮助,将不胜感激。

为了进一步为更好的回答这个问题,这是我目前在做什么:

$statsresults['newIPBan'] = $db->query("SELECT COUNT(ip) AS newIPBan FROM tbl_ipban8;"); 
$statsresults['oldIPBan'] = $db->query("SELECT COUNT(txt_ip) AS oldIPBan FROM tbl_ipban;"); 
$statsresults['log_blocks'] = $db->query("SELECT COUNT(id) AS logBlocks FROM tbl_log;"); 
$statsresults['badbots'] = $db->query("SELECT COUNT(id) AS badBots FROM tbl_bad_bots;"); 
$statsresults['del_num_attacks'] = $db->query("SELECT SUM(noattacks) AS deltotalattacks FROM tbl_delaccounts;"); 
$statsresults['num_attacks'] = $db->query("SELECT SUM(num_attacks) AS totalattacks FROM tbl_accounts8;"); 
$statsresults['dup_blocks'] = $db->query("SELECT COUNT(id) AS dupBlocks FROM tbl_duplicates8;"); 

将返回此:

| ['newIPBan0newIPBan'] = String(6) "289033" 
| ['oldIPBan0oldIPBan'] = String(6) "125723" 
| ['log_blocks0logBlocks'] = String(4) "6481" 
| ['badbots0badBots'] = String(5) "15310" 
| ['del_num_attacks0deltotalattacks'] = String(9) "119494860" 
| ['num_attacks0totalattacks'] = String(8) "25286478" 
| ['dup_blocks0dupBlocks'] = String(6) "179916" 

所以现在它被调用数据库7倍获得每笔金额或数量。我希望将其改为1个数据库调用并将其全部返回。

+4

能否请您提供更多的信息对你正在尝试做的?例如'JOIN''tbl_duplicates8'到6个其他表,然后返回特定字段的SUM和COUNT?就目前而言,Sql查询没有多大意义。 – StuartLC

+0

请提供样本数据和期望的结果。 SQL的问题在于它不是SQL,很难(不可能)弄清楚你正在做什么。 –

+0

我更新了这个问题,举了一个我目前成功做的例子,虽然需要7次数据库调用才能实现。我希望我能在1次通话中完成这一切,并返回总数。 谢谢 – SecureLive

回答

2

这是你可以将它们结合起来的一种方法:

select (newIPBan + oldIPBan + logBlocks + badBots + deltotalattacks + totalattacks + dupBlocks 
     ) as NumIPs 
from (SELECT COUNT(ip) AS oldIPBan FROM tbl_ipban8) ipb8 cross join 
    (SELECT COUNT(txt_ip) AS newIPBan FROM tbl_ipban) ipb cross join 
    (SELECT COUNT(id) AS logBlocks FROM tbl_log) l cross join 
    (SELECT COUNT(id) AS badBots FROM tbl_bad_bots) bb cross join; 
    (SELECT coalesce(SUM(noattacks), 0) AS deltotalattacks FROM tbl_delaccounts) da cross join 
    (SELECT coalesce(SUM(num_attacks), 0) AS totalattacks FROM tbl_accounts8) ta cross join 
    (SELECT COUNT(id) AS dupBlocks FROM tbl_duplicates8) d; 
+0

我可以看到这是如何工作的,虽然它仍然返回null。我会玩这个几个,看看我能否得到这个工作。谢谢 – SecureLive

+0

@SecureLive。 。 。它由于sum()而返回'NULL'。那些可能返回NULL,导致'select'中的全部和为NULL。用'coalesce()'固定。 –

+0

另外,我注意到newIPBan + newIPBan,将第二个更改为oldIPBan,这两个地方解决了这个问题......将添加coalesce以及...谢谢!这真是太棒了,让我给你买一只熊,你可以把你的贝宝地址寄给我,并且我会发一些钱给你。再次感谢! – SecureLive

0

编辑

我会亲自创造的各种度量view,大意如下:

CREATE VIEW vwMetrics AS 
    SELECT 'newIPBan' AS Metric, COUNT(ip) AS Value FROM tbl_ipban8 
    UNION 
    SELECT 'oldIPBan', COUNT(txt_ip) FROM tbl_ipban 
    UNION 
    SELECT 'logBlocks', COUNT(id) FROM tbl_log 
    UNION 
    SELECT 'badBots', COUNT(id) AS badBots FROM tbl_bad_bots 
    UNION 
    SELECT 'deltotalattacks', SUM(noattacks) FROM tbl_delaccounts 
    UNION 
    SELECT 'totalattacks', SUM(num_attacks) FROM tbl_accounts8 
    UNION 
    SELECT 'dupBlocks', COUNT(id) FROM tbl_duplicates8; 

然后你就可以聚集部分:

SELECT SUM(Value) AS TotalOfEverything 
FROM vwMetrics; 

的该观点的好处在于,您可以深入查看组件以进行详细/调试,而不是ar在一个神奇的总计上徘徊 - 可能会有一些重用,以便在系统中的其他位置使用该视图。

Prototype Fiddle here

(+感谢澄清的问题)