Question

我遇到了一个问题，我无法弄清楚为什么会给我带来麻烦。

我想从带有DISTINCT语句的数据库中选择所有商品，因此每个商品ID都是唯一的。这是我在Navicat上的截图表格中获得的内容：

Screenshot http://www.dreshar.com/sql.jpg http://www.dreshar.com/sql.jpg

对于那些看不到图像的人来说，这就是选择 - 我的错误在哪里？

SELECT DISTINCT
guf_offers.id as OfferId,
guf_offers.validfrom AS OfferValidFrom,
guf_offers.validtill AS OfferValidTill,
guf_offers.days AS OfferDays,
guf_offers.active,
guf_offers.roomtype1 AS OfferRoomtype1,
guf_offers.roomprice1 AS OfferRoomprice1,
guf_countries.caption AS CountryCaption,
guf_courts.caption AS CourtCaption,
guf_hotels.caption AS HotelCaption,
guf_hotels.id AS HotelId,
guf_hotel_images.image AS HotelImage,
guf_offer_types.caption AS OffertypeCaption,
guf_regions.caption AS RegionCaption,
guf_offers.hoteloncourse AS OfferHotelOnCourse,
guf_offers.wellnessspa AS OfferWellnessSpa,
guf_offers.18hole AS Offer18Hole,
guf_offers.topangebot AS OfferTopangebot
FROM guf_offers , guf_countries , guf_courts , guf_hotels , guf_hotel_images , guf_offer_types , guf_regions
WHERE guf_offers.country_id = guf_countries.id AND guf_offers.court_id = guf_courts.id AND guf_offers.hotel_id = guf_hotels.id AND guf_hotel_images.hotel_id = guf_offers.hotel_id AND guf_offers.offer_type_id = guf_offer_types.id AND guf_offers.region_id = guf_regions.id AND guf_offers.active = 1 AND STR_TO_DATE(guf_offers.validtill, '%d.%m.%Y') > STR_TO_DATE('29.04.2013', '%d.%m.%Y')

Answer 1

DISTINCT返回任何不同的列组合。例如，图像中前两个OfferId = 89的实例在HotelImage列中显示不同的值，因此这些是DISTINCT结果。

Answer 2

如果你想要不同的ID，那么你可能想要使用group by。查询看起来像：

SELECT
guf_offers.id as OfferId,
guf_offers.validfrom AS OfferValidFrom,
guf_offers.validtill AS OfferValidTill,
guf_offers.days AS OfferDays,
guf_offers.active,
guf_offers.roomtype1 AS OfferRoomtype1,
guf_offers.roomprice1 AS OfferRoomprice1,
guf_countries.caption AS CountryCaption,
guf_courts.caption AS CourtCaption,
guf_hotels.caption AS HotelCaption,
guf_hotels.id AS HotelId,
guf_hotel_images.image AS HotelImage,
guf_offer_types.caption AS OffertypeCaption,
guf_regions.caption AS RegionCaption,
guf_offers.hoteloncourse AS OfferHotelOnCourse,
guf_offers.wellnessspa AS OfferWellnessSpa,
guf_offers.18hole AS Offer18Hole,
guf_offers.topangebot AS OfferTopangebot
FROM guf_offers , guf_countries , guf_courts , guf_hotels , guf_hotel_images , guf_offer_types , guf_regions
WHERE guf_offers.country_id = guf_countries.id AND guf_offers.court_id = guf_courts.id AND guf_offers.hotel_id = guf_hotels.id AND guf_hotel_images.hotel_id = guf_offers.hotel_id AND guf_offers.offer_type_id = guf_offer_types.id AND guf_offers.region_id = guf_regions.id AND guf_offers.active = 1 AND STR_TO_DATE(guf_offers.validtill, '%d.%m.%Y') > STR_TO_DATE('29.04.2013', '%d.%m.%Y')
group by guf_offers.id

其他列是来自不同行的任意值。这使用group by的扩展名，即使列不在聚合函数或select子句中，列也可以在group by中。

顺便说一句，您应该学习使用propoer from语法在join子句中进行联接。它使查询更具可读性和可维护性。

在MySQL中选择不同的行

2 个答案: