下载包括下载时间,下载时间ID和buno ID.故障由故障代码,下载时间ID,状态和类型组成.下载可能有许多错误,可以在下载时ID上加入.
给定一组故障代码,结果必须包含每个具有相应故障计数的故障代码.如果在下载中未找到故障代码,则必须返回故障代码,故障计数为零.
问题似乎需要一个OUTER JOIN,但是没有看到它在Postgres上按预期工作,因为它似乎没有从LEFT表返回带有空值的集合.
查询如下,为简洁起见,省略了一些细节:
SELECT f.faultcode, f.downloadtimeid, d.downloadtime, count(*) as faultcount
FROM download_time d
LEFT OUTER JOIN fs_fault f ON f.downloadtimeid = d.id
AND f.faultcode IN (1000,1100)
AND f.statusid IN(2, 4)
WHERE (d.downloadtime BETWEEN '04/11/2011' AND '05/01/2012')
AND d.bunoid = 166501
GROUP BY d.bunoid, f.downloadtimeid, d.downloadtime, f.faultcode
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第二天,我编辑了以显示答案.所有答案都很接近,并有各种各样的帮助.然而,JayC的答案是最接近的. 这是最终的SQL,只有作为WHERE子句采用故障代码IN语句的更改:
SELECT f.faultcode, f.downloadtimeid, d.downloadtime, count(*) as faultcount
FROM download_time d
RIGHT OUTER JOIN fs_fault f ON f.downloadtimeid = d.id
AND f.statusid IN(2, 4)
AND d.downloadtime BETWEEN '04/11/2011' AND '05/01/2012'
AND d.bunoid = 166501
WHERE f.faultcode IN (1000,1100)
GROUP BY d.bunoid, f.downloadtimeid, d.downloadtime, f.faultcode
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谢谢,一切都是为了你的帮助!喜欢这个网站!
Jay*_*ayC 22
我给出了答案,因为我对其他答案有很大疑问.你必须小心过滤器的要求.请记住,where子句在您加入后运行.因此,如果where子句中有任何过滤器要求引用非外连接表,则(在许多情况下)使外连接无效.所以拿你的sql,似乎最简单的解决方案是使用正确的连接或适当地移动表名,然后将过滤条件移出where子句并进入join子句.
SELECT f.faultcode, f.downloadtimeid, d.downloadtime, count(*) as faultcount
FROM download_time d
RIGHT OUTER JOIN fs_fault f ON
f.downloadtimeid = d.id
AND f.faultcode IN (1000,1100)
AND f.statusid IN(2, 4)
AND d.downloadtime BETWEEN '04/11/2011' AND '05/01/2012')
AND d.bunoid = 166501
GROUP BY d.bunoid, f.downloadtimeid, d.downloadtime, f.faultcode
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我认为应该相同的另一种方式是
SELECT f.faultcode, f.downloadtimeid, d.downloadtime, count(*) as faultcount
FROM download_time d
RIGHT OUTER JOIN fs_fault f ON
f.downloadtimeid = d.id
AND d.downloadtime BETWEEN '04/11/2011' AND '05/01/2012')
AND d.bunoid = 166501
WHERE
f.faultcode IN (1000,1100)
AND f.statusid IN(2, 4)
GROUP BY d.bunoid, f.downloadtimeid, d.downloadtime, f.faultcode
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因为fs_fault的过滤器要求在哪里并不严格.(而且你的SQL引擎无论如何都要改变这一点).
编辑:这是一个SQLFiddle,演示了对join子句和where子句的过滤.
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