Question

我一直在处理因使用Join-Ons而不是逗号连接而导致的问题。我的SQL目前看起来像这样：

SELECT islandID AS parentIslandID, islandName, island.longDesc, 
imageLocation, COUNT(resort.resortID) AS totalResorts, resort.resortID 
FROM island, resort, images 
join resort as r1 
on island.islandID = resort.parentIslandID 
where 
r1.resortID IN ( 
59,62,65,69,71,72,74,75,76,82,86,89,91,93,95,105, 
106,116,117,118,120,121,122,123,124,125,126,127, 
131,145,146,150,157,159,160,167,170,174,176,185,188,189,193, 
194,198,199,200,203,205,213,217 
) 
&& resort.active = '-1' 
GROUP BY resort.parentIslandID 
ORDER BY totalResorts DESC

执行时，我收到以下错误：

#1054 - Unknown column 'island.islandID' in 'on clause'

我做了一些研究并了解错误的起源但是我试图通过为“孤岛”表创建别名来纠正这个问题。当我这样做时，像“island.longDesc”这样的列就是“未知”。

如果有人能够纠正似乎是一个小的语法问题，我将非常感激。我更习惯使用逗号连接这么简单。谢谢你的帮助 -

-Aaron

编辑：

Images Structure:
CREATE TABLE `images` (
  `imageID` int(11) NOT NULL auto_increment,
  `imageType` int(11) NOT NULL COMMENT 'used to tell if its for an artist, header image, etc.',
  `parentObjectID` int(11) NOT NULL COMMENT 'used to tell what island/resort the image applies to',
  `imageLocation` text NOT NULL,
  `largeImageLocation` text NOT NULL,
  `imageLinkLabel` text NOT NULL,
  `imageURL` text NOT NULL,
  PRIMARY KEY  (`imageID`)
)

Island Structure:
CREATE TABLE `island` (
  `islandID` int(11) NOT NULL auto_increment,
  `islandName` text NOT NULL,
  `shortDesc` text NOT NULL,
  `longDesc` text NOT NULL,
  `getTo` text NOT NULL,
  `getAround` text NOT NULL,
  `photoInfo` text NOT NULL,
  `flowerInfo` text NOT NULL,
  `musicInfo` text NOT NULL,
  `cakeInfo` text NOT NULL,
  `activityInfo` text NOT NULL,
  `wedCoord` text NOT NULL,
  `regs` text NOT NULL,
  `climate` text NOT NULL,
  `languageID` int(11) NOT NULL,
  `currencyID` int(11) NOT NULL,
  `wideAccept` int(11) NOT NULL,
  `passportReq` int(11) NOT NULL,
  `picture` text NOT NULL,
  `daysSearchable` int(11) NOT NULL,
  `active` tinyint(1) NOT NULL,
  PRIMARY KEY  (`islandID`)
)

Resort Structure:
CREATE TABLE `resort` (
  `resortID` int(11) NOT NULL auto_increment,
  `resortName` text NOT NULL,
  `parentIslandID` int(11) NOT NULL,
  `longDesc` text NOT NULL,
  `website` text NOT NULL,
  `genBooking` text NOT NULL,
  `eventCoord` text NOT NULL,
  `amenInfo` text NOT NULL,
  `roomInfo` text NOT NULL,
  `coordInfo` text NOT NULL,
  `localeInfo` text NOT NULL,
  `spaInfo` text NOT NULL,
  `roomPrice` text NOT NULL,
  `maxGuests` text NOT NULL,
  `picture` text NOT NULL,
  `search_Inclusive` int(11) NOT NULL,
  `search_resortType` int(11) NOT NULL,
  `search_onBeach` int(11) NOT NULL,
  `search_wedCoord` int(11) NOT NULL,
  `search_roomRate` int(11) NOT NULL,
  `search_airportDist` int(11) NOT NULL,
  `search_HotelSuite` tinyint(1) NOT NULL,
  `search_VillaCondo` tinyint(1) NOT NULL,
  `search_Amenities` text NOT NULL,
  `active` tinyint(1) NOT NULL,
  PRIMARY KEY  (`resortID`)
)

Answer 1

您正在混合两种不同类型的JOIN语法 - FROM中列出表的隐式类型和显式JOIN类型。相反，尝试：

SELECT
   islandID AS parentIslandID,
   islandName, 
   island.longDesc, 
   imageLocation,
   COUNT(r1.resortID) AS totalResorts, 
   r1.resortID 
FROM island
  JOIN resort r1 ON island.islandID = r1.parentIslandID 
  JOIN images ON island.islandID = images.parentObjectID 
WHERE 
  r1.resortID IN ( 
    59,62,65,69,71,72,74,75,76,82,86,89,91,93,95,105, 
    106,116,117,118,120,121,122,123,124,125,126,127, 
    131,145,146,150,157,159,160,167,170,174,176,185,188,189,193, 
    194,198,199,200,203,205,213,217 
  ) 
AND resort.active = '-1' 
GROUP BY r1.parentIslandID 
ORDER BY totalResorts DESC

**在发布表格结构后编辑包含岛JOIN。

此外：

MySQL使用AND而不是&&来表示布尔AND
表别名不需要AS关键字（JOIN resort r1）
请务必在选择列表中使用resort别名r1（r1.resortID）

使用连接时的未知列

1 个答案: