Question

我有一个与Restaurent加入的SQL Server表Product：

SELECT 
    rm.RestaurantId, rm.RestaurantImage, rm.RestaurantName, 
    rm.RestaurantLocation,
    pm.ProductName, rm.IsActive,
    (6371 * acos(cos(radians(@userlat)) * cos(radians(RestaurantLatitude)) * cos(radians(RestaurantLongitude) - radians(@userlong)) + sin(radians(@userlat)) * sin(radians(RestaurantLatitude)))) AS distance 
FROM 
    HomeMadeFood.dbo.RestaurentMaster AS rm 
INNER JOIN
    ProductMaster AS pm ON pm.RestaurentId = rm.RestaurantId
GROUP BY 
    rm.RestaurantId
ORDER BY 
    distance ASC

此查询返回如下输出：

RestID     RestName      Distance    ProductName
-------------------------------------------------
  1         A             16           Demo1
  2         B             20           Demo2
  1         A             16           Demo3
  3         C             50           Demo4

我的预期输出应该是这样的

RestID     RestName      Distance    ProductName
-------------------------------------------------
   1         A             16           Demo1
   2         B             20           Demo2
   3         C             50           Demo4

如果我在group by上使用RestId，则会在第二列（图片）上引发错误

Answer 1

以下内容通过将所有非聚合列移动到group by子句来修复您看到的错误：

SELECT rm.RestaurantId, rm.RestaurantImage, rm.RestaurantName, rm.RestaurantLocation, pm.ProductName, rm.IsActive,
       ( 6371 * acos( cos( radians(@userlat) ) * cos( radians(RestaurantLatitude) ) * cos( radians(RestaurantLongitude) - radians(@userlong) ) + sin( radians(@userlat) ) * sin( radians(RestaurantLatitude) ) ) ) AS distance 
from HomeMadeFood.dbo.RestaurentMaster rm inner join
     ProductMaster pm 
     on pm.RestaurentId = rm.RestaurantId
group by rm.RestaurantId, rm.RestaurantImage, rm.RestaurantName, rm.RestaurantLocation, pm.ProductName, rm.IsActive,
         rm.RestaurantLongitude, rm.RestaurantLatitude
order by distance asc;

您想要的结果要简单得多，所以我建议：

SELECT rm.RestaurantId, rm.RestaurantName, rm.IsActive,
       ( 6371 * acos( cos( radians(@userlat) ) * cos( radians(RestaurantLatitude) ) * cos( radians(RestaurantLongitude) - radians(@userlong) ) + sin( radians(@userlat) ) * sin( radians(RestaurantLatitude) ) ) ) AS distance 
from HomeMadeFood.dbo.RestaurentMaster rm inner join
     ProductMaster pm 
     on pm.RestaurentId = rm.RestaurantId
group by rm.RestaurantId, rm.RestaurantName, rm.IsActive,
         rm.RestaurantLongitude, rm.RestaurantLatitude
order by distance asc;

Answer 2

根据我的假设，你需要的是具有特定id的第一行，其中距离最小。试试这个。

WITH MyNewTable AS(
SELECT rm.RestaurantId,rm.RestaurantImage,rm.RestaurantName,rm.RestaurantLocation,pm.ProductName,rm.IsActive,( 6371 * acos( cos( radians(@userlat) ) * cos( radians(RestaurantLatitude) ) * cos( radians(RestaurantLongitude) - radians(@userlong) ) + sin( radians(@userlat) ) * sin( radians(RestaurantLatitude) ) ) ) AS distance,
ROW_NUMBER() OVER(PARTITION BY pm.RestaurentId ) AS rk 
FROM HomeMadeFood.dbo.RestaurentMaster AS rm 
INNER JOIN ProductMaster AS pm 
on pm.RestaurentId = rm.RestaurantId
ORDER BY distance ASC
) 
SELECT
*
FROM
MyNewTable
WHERE rk = 1

如果它不起作用，请告诉我。

如何只选择一行特定的id？没有明确解决

2 个答案: