我有两个表: invoices和invoiceitems。
invoiceitems包含每张发票上的项目
例如:
invoices
----------------------------------
| id |status| net | tax | total |
----------------------------------
| 72 |paid | 100 | 120 | 220 |
| 73 |unpaid| 50 | 5 | 55 |
| 74 |paid | 400 | 45 | 445 |
| 75 |paid | 250 | 67 | 317 |
invoiceitems
-------------------------------
| invoiceid |itemdescription |
-------------------------------
| 72 | apples |
| 72 | pears |
| 72 | oranges |
| 73 | lemons |
| 73 | oranges |如您所见,在本例中,发票号72有3个项目
我想搜索我的发票中的某些东西,并显示某些字段的计数。
但我的问题是,sum值似乎乘以了第二个表中的字段数。
$sql = "SELECT COUNT(DISTINCT invoices.id) AS num,
SUM(CASE invoices.status WHEN 'Paid' THEN 1 ELSE 0 END) AS numpaid,
SUM(CASE invoices.status WHEN 'Paid' THEN invoices.total ELSE 0 END) AS sumtotal,
FROM invoices
LEFT JOIN invoiceitems ON invoices.id=invoiceitems.invoiceid
WHERE invoices.id LIKE :invoiceid
AND IFNULL(opcinvoiceitems.itemdescription, '') LIKE :itemdescription
AND invoices.net LIKE :net
AND invoices.tax LIKE :tax
AND invoices.total LIKE :total
AND ......" 因此,使用上面的方法,发票72的总和将乘以3
我真的很抱歉,我知道这真的解释得很糟糕,但我无法用其他方式解释它,我已经寻找了很长时间,但没有找到解决方案。希望有人能帮上忙。谢谢
发布于 2013-09-10 08:48:48
执行联接时,通过匹配原始表中的记录来生成所创建的记录。因此,您将有3条记录用于发票#72,每条记录都是通过将#72的单个发票记录与#72的每个发票项进行匹配而创建的。每个合并的记录将具有相同的总数(在本例中为220),因此总和将是该总数的3倍。
听起来你只是想要总数;你可以直接使用total,或者你可以把你的sum除以count (你似乎也在计算)。
发布于 2013-09-10 08:51:28
完成所需操作的一种方法是在连接之前预先聚合invoiceItems表:
SELECT COUNT(i.id) AS num,
SUM(CASE i.status WHEN 'Paid' THEN 1 ELSE 0 END) AS numpaid,
SUM(CASE i.status WHEN 'Paid' THEN i.total ELSE 0 END) AS sumtotal,
FROM invoices i LEFT JOIN
(select ii.invoiceid, sum(. . .) as . . .
from invoiceitems ii
where IFNULL(ii.itemdescription, '') LIKE :itemdescription AND
group by ii.invoiceid
) ii
ON i.id = ii.invoiceid
WHERE i.id LIKE :invoiceid AND
i.net LIKE :net AND
i.tax LIKE :tax AND
i.total LIKE :total AND .....您的查询实际上并未在from子句中使用invoiceitems,因此很难提供更详细的示例。
https://stackoverflow.com/questions/18709104
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