用于复杂逻辑的SQL查询（pl/SQL）

这个问题困扰了我一段时间。首先，我将解释表的结构

我的系统中有四个表

域

网页

插件类型（可添加到网页的插件类型） 此表列出了每个域可用的插件类型

插件：域中的每个网页都可以使用任意数量的插件

现在我想要的是，给定一个域ID，我想要所有可用网页插件的列表，这样插件就不会在多个网页中重复。 换句话说，每个插件的网页插件组合，这样插件就不会在多个网页中重复

所以对于域名1000（google.com）

我想要的结果是

我只选择了5002和5003个网页，因为它们包含了域名1000（google.com）的所有插件。

---

还有一件事，最好选择一个只有一个插件的网页。但是我对找到没有这个偏好的解决方案感兴趣，以后也许我可以改进这个解决方案。

你可以从另一个角度来看，因为你要返回每个插件类型的一行，你需要一个域上的所有插件类型，以及一个插件类型插入的示例页面，大致如下：

SELECT  PluginTypeId,
        (
        SELECT  webpageID
        FROM    plugin pi
        WHERE   pi.pluginTypeId = pd.pluginTypeId
                AND webPageID IN
                (
                SELECT  WebPageID
                FROM    WebPage
                WHERE   DomainId = 1000
                )
        ORDER BY
                (
                SELECT  COUNT(*)
                FROM    plugin pc
                WHERE   pc.webpageId = pi.webpageId
                )
        LIMIT 1
        ) AS WebPageId
FROM    Plugin pd
WHERE   WebPageId IS NOT NULL

Select PluginTypeName, 
  (select top 1 WebpageName 
    from WebPage w
      inner join Plugin p on p.WebpageID = w.WebpageID
    where p.pluginTypeID = pt.pluginTypeID) as SampleWebPage
From PluginTypes pt

这将返回如下结果：

searchButton    SearchPage.html
DropDownMenu    SearchPage.html
InteractiveForm null
loginForm       ContactUs.html
lotoutform      Welcome.html             

以上结果（斯维科提交）应该有效，我正在尝试，可能会给出重复的结果。此外，它还按域对结果进行分组。。。下面是一个修改：

Select DISTINCT DomainID, pt.PluginTypeName, 
  (select top 1 WebpageName 
    from WebPage w
      inner join Plugin p on p.WebpageID = w.WebpageID
    where p.pluginTypeID = pt.pluginTypeID) as SampleWebPage
From PluginTypes pt
GROUP BY DomainID, pt.PluginTypeName

---

您正在寻找“DISTINCT”命令吗？具体针对哪个数据库？语法不是100%相同

          5002 (Welcome.html)     8000 (SearchButton)
          5002 (Welcome.html)     8001 (DropDownMenu)
          5002 (Welcome.html)     8004 (lotoutform)
          5003 (ContactUs.htm)    8003 (loginForm)

SELECT  PluginTypeId,
        (
        SELECT  webpageID
        FROM    plugin pi
        WHERE   pi.pluginTypeId = pd.pluginTypeId
                AND webPageID IN
                (
                SELECT  WebPageID
                FROM    WebPage
                WHERE   DomainId = 1000
                )
        ORDER BY
                (
                SELECT  COUNT(*)
                FROM    plugin pc
                WHERE   pc.webpageId = pi.webpageId
                )
        LIMIT 1
        ) AS WebPageId
FROM    Plugin pd
WHERE   WebPageId IS NOT NULL

Select PluginTypeName, 
  (select top 1 WebpageName 
    from WebPage w
      inner join Plugin p on p.WebpageID = w.WebpageID
    where p.pluginTypeID = pt.pluginTypeID) as SampleWebPage
From PluginTypes pt

searchButton    SearchPage.html
DropDownMenu    SearchPage.html
InteractiveForm null
loginForm       ContactUs.html
lotoutform      Welcome.html             

Select DISTINCT DomainID, pt.PluginTypeName, 
  (select top 1 WebpageName 
    from WebPage w
      inner join Plugin p on p.WebpageID = w.WebpageID
    where p.pluginTypeID = pt.pluginTypeID) as SampleWebPage
From PluginTypes pt
GROUP BY DomainID, pt.PluginTypeName