Question

我坚持这个问题。有人可以帮忙吗？

为符合以下条件的所有投保人写一个查询，将2015年所有投资总额的总和（TIV_2012）打印到2位小数的比例：

1）与一个或多个其他保单持有人具有相同的TIV_2011价值。

2）与其他保单持有人不在同一个城市（即（纬度，经度）属性对必须是唯一的，

输入格式是这样的，表是

保险表描述如下：

列名称类型 PID INTEGER TIV_2011 NUMERIC TIV_2012 NUMERIC LAT NUMERIC LON NUMERIC

其中PID是保单持有人的政策ID，TIV_2011是2011年的总投资，TIV_2012是2012年的总投资，LAT是保单持有人城市的纬度，而LON是经度保单持有人的城市。

例如，如果数据是 PID，TIV_2011，TIV_2012，lat，lon

1,300,400.5,60,70
2,300,500.7,70,80
3,400,400,60,90
4,500,600,80,80
5,400,300.1,6,6

答案是1601.30。总和（300.1,400,500.7,400.5）

所以，到目前为止，我已经提出了这个

SELECT SUM（TIV_2012）FROM Insurance WHERE NOT UNIQUE（从保险中选择TIV_2011）;

这不起作用，我收到错误。有人帮忙。

Answer 1

SELECT SUM(t1.TIV_2012)
FROM Insurance t1
INNER JOIN
(
    SELECT TIV_2011
    FROM Insurance
    GROUP BY TIV_2011
    HAVING COUNT(*) > 1
) t2
    ON t1.TIV_2011 = t2.TIV_2011
INNER JOIN
(
    SELECT lat, lon
    FROM Insurance
    GROUP BY lat, lon
    HAVING COUNT(*) = 1
) t3
    ON t1.lat = t3.lat AND
       t1.lon = t3.lon

Answer 2

SELECT CAST(SUM(t1.TIV_2012) as DECIMAL(11,2)) 
FROM Insurance t1
INNER JOIN (
  SELECT TIV_2011
  FROM Insurance
  GROUP BY TIV_2011 HAVING COUNT(*) > 1 ) t2 ON t1.TIV_2011 = t2.TIV_2011
INNER JOIN ( 
  SELECT lat, lon 
  FROM Insurance 
  GROUP BY lat, lon HAVING COUNT(*) = 1 ) t3 ON t1.lat = t3.lat AND t1.lon = t3.lon

Answer 3

Round(x,2) -> to scale to 2 decimal digits
inner join 1 -> for condition 1 (finds all the repeating TIV_2011)
inner join 2 -> for condition 2 (finds LAT, LON that are distinct as a pair)

Select ROUND(SUM(i1.TIV_2012),2) 
from Insurance i1
inner join 
    (Select TIV_2011 
    from Insurance 
    group by TIV_2011 
    having count(*) > 1) i2 
on
    i1.TIV_2011 = i2.TIV_2011
inner join 
    (Select LAT, LON 
    from Insurance 
    group by LAT, LON 
    having count(*) = 1) i3 
on 
    i1.LAT = i3.LAT 
and  
    i1.LON = i3.LON

Answer 4

我相信你需要编写的查询有两个内连接，如下所示：

SELECT SUM(ins1.column_of_interest) as value_needed
FROM Insurance ins1
INNER JOIN Insurance ins2 ON (ins1.id = ins2.id AND <<conditions to get same TIV_2011 applied to ins2 >> )
INNER JOIN Insurance ins3 ON (ins1.id = ins3.id AND << conditions to get unique latitude, longitude on ins3 >> )
WHERE << other conditions you may apply to ins1 >>

Answer 5

这里的所有答案都采用一种方法。我尝试了另一种方法，所以提交它。

SELECT CAST(SUM(i.TIV_2012) AS NUMERIC(18,2))
FROM Insurance i
WHERE i.PID IN (SELECT DISTINCT(i1.PID)
                FROM Insurance i1
                INNER JOIN Insurance i2 ON (i1.TIV_2011 = i2.TIV_2011 AND i1.PID <> i2.PID)
                WHERE NOT EXISTS (SELECT i3.PID 
                    FROM Insurance i3 
                    WHERE i1.PID <> i3.PID
                    AND i1.LAT = i3.LAT
                    AND i1.LON = i3.LON));

Answer 6

SELECT SUM(i1.TIV_2012) FROM (
  SELECT * ,COUNT(1) OVER (PARTITION BY i.TIV_2011 ORDER BY i.TIV_2011) R2 ,COUNT(1) OVER (PARTITION BY i.Lat, i.Lon ORDER BY i.Lat, i.Lon) R1 FROM 
    #Insurance i) i1 
WHERE R2 > 1 AND R1 = 1

Answer 7

从以下位置选择总和（d.tiv_2012）作为Total_2012_TIV

（选择a.TIV_2011，a.TIV_2012，a.lat，a.lon 来自保险加入（SELECT TIV_2011，count（）count 来自保险按TIV_2011分组拥有COUNT（）> 1 ）b 在a.TIV_2011 = b.TIV_2011 加入（选择纬度，计数（纬度）纬度计数来自保险按拉特分组具有count（lat）= 1 ）C 在a.lat = c.lat）d

具有多个条件的SQL查询

7 个答案: