我有一个包含rowID,经度,纬度,businessName,url,标题的表格。这可能看起来像:
rowID | long | lat | businessName | url | caption
1 20 -20 Pizza Hut yum.com null
如何删除所有重复项,但只保留具有URL(第一优先级)的副本,或者如果另一个没有URL(第二优先级)则保留具有标题的副本并删除休息?
答案 0 :(得分:4)
这个解决方案是通过“我在Stack Overflow上学到的东西”在上周提供的:
DELETE restaurant
WHERE rowID in
(SELECT rowID
FROM restaurant
EXCEPT
SELECT rowID
FROM (
SELECT rowID, Rank() over (Partition BY BusinessName, lat, long ORDER BY url DESC, caption DESC ) AS Rank
FROM restaurant
) rs WHERE Rank = 1)
警告:我没有在真实数据库上测试过这个
答案 1 :(得分:3)
这是我的循环技术。这可能会因为不成为主流而被否决 - 而且我对此感到很满意。
DECLARE @LoopVar int
DECLARE
@long int,
@lat int,
@businessname varchar(30),
@winner int
SET @LoopVar = (SELECT MIN(rowID) FROM Locations)
WHILE @LoopVar is not null
BEGIN
--initialize the variables.
SELECT
@long = null,
@lat = null,
@businessname = null,
@winner = null
-- load data from the known good row.
SELECT
@long = long,
@lat = lat,
@businessname = businessname
FROM Locations
WHERE rowID = @LoopVar
--find the winning row with that data
SELECT top 1 @Winner = rowID
FROM Locations
WHERE @long = long
AND @lat = lat
AND @businessname = businessname
ORDER BY
CASE WHEN URL is not null THEN 1 ELSE 2 END,
CASE WHEN Caption is not null THEN 1 ELSE 2 END,
RowId
--delete any losers.
DELETE FROM Locations
WHERE @long = long
AND @lat = lat
AND @businessname = businessname
AND @winner != rowID
-- prep the next loop value.
SET @LoopVar = (SELECT MIN(rowID) FROM Locations WHERE @LoopVar < rowID)
END
答案 2 :(得分:1)
基于设置的解决方案:
delete from T as t1
where /* delete if there is a "better" row
with same long, lat and businessName */
exists(
select * from T as t2 where
t1.rowID <> t2.rowID
and t1.long = t2.long
and t1.lat = t2.lat
and t1.businessName = t2.businessName
and
case when t1.url is null then 0 else 4 end
/* 4 points for non-null url */
+ case when t1.businessName is null then 0 else 2 end
/* 2 points for non-null businessName */
+ case when t1.rowID > t2.rowId then 0 else 1 end
/* 1 point for having smaller rowId */
<
case when t2.url is null then 0 else 4 end
+ case when t2.businessName is null then 0 else 2 end
)
答案 3 :(得分:1)
delete MyTable
from MyTable
left outer join (
select min(rowID) as rowID, long, lat, businessName
from MyTable
where url is not null
group by long, lat, businessName
) as HasUrl
on MyTable.long = HasUrl.long
and MyTable.lat = HasUrl.lat
and MyTable.businessName = HasUrl.businessName
left outer join (
select min(rowID) as rowID, long, lat, businessName
from MyTable
where caption is not null
group by long, lat, businessName
) HasCaption
on MyTable.long = HasCaption.long
and MyTable.lat = HasCaption.lat
and MyTable.businessName = HasCaption.businessName
left outer join (
select min(rowID) as rowID, long, lat, businessName
from MyTable
where url is null
and caption is null
group by long, lat, businessName
) HasNone
on MyTable.long = HasNone.long
and MyTable.lat = HasNone.lat
and MyTable.businessName = HasNone.businessName
where MyTable.rowID <>
coalesce(HasUrl.rowID, HasCaption.rowID, HasNone.rowID)
答案 4 :(得分:1)
与其他答案类似,但您希望根据行号而不是排名进行删除。混合使用常见的表表达式:
;WITH GroupedRows AS
( SELECT rowID, Row_Number() OVER (Partition BY BusinessName, lat, long ORDER BY url DESC, caption DESC) rowNum
FROM restaurant
)
DELETE r
FROM restaurant r
JOIN GroupedRows gr ON r.rowID = gr.rowID
WHERE gr.rowNum > 1
答案 5 :(得分:0)
如果可能,你可以均质化,然后删除重复吗?
第1步:
UPDATE BusinessLocations
SET BusinessLocations.url = LocationsWithUrl.url
FROM BusinessLocations
INNER JOIN (
SELECT long, lat, businessName, url, caption
FROM BusinessLocations
WHERE url IS NOT NULL) LocationsWithUrl
ON BusinessLocations.long = LocationsWithUrl.long
AND BusinessLocations.lat = LocationsWithUrl.lat
AND BusinessLocations.businessName = LocationsWithUrl.businessName
UPDATE BusinessLocations
SET BusinessLocations.caption = LocationsWithCaption.caption
FROM BusinessLocations
INNER JOIN (
SELECT long, lat, businessName, url, caption
FROM BusinessLocations
WHERE caption IS NOT NULL) LocationsWithCaption
ON BusinessLocations.long = LocationsWithCaption.long
AND BusinessLocations.lat = LocationsWithCaption.lat
AND BusinessLocations.businessName = LocationsWithCaption.businessName
第2步: 删除重复项。