Question

我有一个包含这些示例行的表：

CouponID ShopID ShopLongitude ShopLatitude
365      1881   55,5574105    9,9613295
365      23550  55,5510846    9,9936818
365      33550  55,6220936    10,0663895
365      33551  55,5573436    9,9611765
366      1881   55,5574105    9,9613295
366      23550  55,5510846    9,9936818
367      1881   55,5574105    9,9613295
533      1881   55,5574105    9,9613295
533      23550  55,5510846    9,9936818
533      33550  55,6220936    10,0663895
533      33551  55,5573436    9,9611765
354      1881   55,5574105    9,9613295
354      23550  55.5510846    9,9936818
354      33550  55,6220936    10,0663895
354      33551  55,5573436    9,9611765

我希望得到每张CouponID

最近的ShopID

我已经有了查询：

SELECT CouponID, MIN (dbo.DistanceBetween (53.54613,9.98537,ShopLatitude,ShopLongitude)) as distance
FROM Table
GROUP BY CouponID
ORDER BY CouponID, distance ASC

为每张优惠券输出CouponID和ShopID的最小距离：

CouponID distance
354     0,778524633472375
365     0,778524633472375
366     0,778524633472375
367     2,02548179145764

在LINQ中语句如下：

var coupon = (from c in dbContext.table
                      group c by c.CouponID into cgrp
                      select new
                      {
                        CouponID = cgrp.Key,
                        Distance = cgrp.Min(c => DistanceBetween(latitude, longitude, c.ShopLatitude, c.ShopLongitude))
                      })
                      .OrderBy(c => c.CouponID).ThenBy(c => DistanceBetween(latitude, longitude, c.ShopLatitude, c.ShopLongitude));

如何从相关的最小距离取回相关的ShopID 查询和Linq声明？

Answer 1

SELECT c.CouponID, c.ShopID, nearest.distance
FROM
Coupons C JOIN
(
    SELECT CouponID, MIN (dbo.DistanceBetween (53.54613,9.98537,ShopLat,ShopLon)) as distance
    FROM Coupons
    GROUP BY CouponID   
) nearest on c.CouponID = nearest.CouponID
          AND dbo.DistanceBetween (53.54613,9.98537,ShopLat,ShopLon) = nearest.distance
ORDER BY CouponID, distance ASC

Answer 2

只需修改您的SQL查询即可执行分组并加入您想要的地方。

创建一个表来保存距离，然后选择couponid和shopid匹配的最小值。

SQL / Linq Group By具有最小值和从属列

2 个答案: