Question

不确定如何做到这一点，但我试图从MYSQL数据库中获取所有包含多年拖欠的包裹。现在，我可以使用以下查询计算多年来我有多少次违约：

SELECT 
    parcel, COUNT(*), due
FROM 
    `deliquent_property`
WHERE year IN ('2013', '2012', '2011', '2010', '2009') AND
      CAST(replace(replace(ifnull(due,0),',',''),'$','') AS decimal(10,2)) > 0
GROUP BY parcel
HAVING COUNT(*) > 2;

所以我得到这样的东西

'XXX-XXX-XXX1'    '3'  '167.00'
'XXX-XXX-XXX2'    '4'  '190'

其中金额是数据库中该宗地的最后一条记录。问题是我真正想要的是每年显示过期金额的包裹清单，类似这样（跳过那些没有至少3次拖欠的包裹）：

'XXX-XXX-XXX1'    '2013'  '1267.78'
'XXX-XXX-XXX1'    '2012'  '1000.38'
'XXX-XXX-XXX1'    '2011'  '167.00'
'XXX-XXX-XXX2'    '2013'  '1000.00'
'XXX-XXX-XXX2'    '2012'  '500.00'
'XXX-XXX-XXX2'    '2011'  '100.00'
'XXX-XXX-XXX2'    '2010'  '190.00'

我已经到了一半，但我迷失了最后一部分。

Answer 1

您可以将表格加入列表。这是一种方式：

select dp.*
from delinquent_property dp join
     (SELECT parcel, COUNT(*)
      FROM delinquent_property
      WHERE year IN ('2013', '2012', '2011', '2010', '2009') AND
           CAST(replace(replace(ifnull(due,0),',',''),'$','') AS decimal(10,2)) > 0
     GROUP BY parcel
     HAVING COUNT(*) > 2
    ) p3
    on dp.parcel = p3.parcel
where CAST(replace(replace(ifnull(due,0),',',''),'$','') AS decimal(10,2)) > 0
order by parcel;

MYSQL查询 - 抓取符合条件3次或更多次的所有记录

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