Question

我不知何故需要以下方法：

用户位于某个博客页面现在我想让他有能力移动：

到下一个更受欢迎的页面或下一个不太受欢迎的页面。意思是这样的：

[PAGE 5] | [PAGE 3 (User is here)] | [PAGE 2]

（第1页=最受欢迎，第3页=最不受欢迎）。我的MySQL表看起来像这样：

[ID] [VIEWS]
[1] [1000]
[2] [2560]
[3] [3200]
[4] [200]
[5] [4000]

我的问题是具体的查询。在这种情况下，唯一给定的变量是ID：2。也许你可以帮助我。只要告诉我你是否需要进一步的信息。（我只需要两个邻居。不需要等等。）

编辑：@Trevor不，对不起，我不能。我更清楚地改变了这个例子。

Answer 1

注意：正如各种评论所述，@mellamokb's answer比我的好。我会删除这个答案但是因为它被接受了所以我无法接受。

/* Next Page */
SELECT ID
    FROM YourTable
    WHERE VIEWS <= (SELECT VIEWS FROM YourTable WHERE ID = $CurrentPageId)
        AND ID < $CurrentPageId
    ORDER BY VIEWS DESC, ID DESC LIMIT 1

/* Previous Page */
SELECT ID
    FROM YourTable
    WHERE VIEWS >= (SELECT VIEWS FROM YourTable WHERE ID = $CurrentPageId)
        AND ID > $CurrentPageId
    ORDER BY VIEWS, ID LIMIT 1

Answer 2

  select ID as PreviousId
    from PageViews
   where Views > (select Views From PageViews Where ID = @Id)
      or (
             Views = (select Views From PageViews Where ID = @Id)
             and ID > @id
         )
order by Views ASC, ID DESC
   limit 1

  select ID as PreviousId
    from PageViews
   where Views < (select Views From PageViews Where ID = @Id)
      or (
             Views = (select Views From PageViews Where ID = @Id)
             and ID < @id
         )
order by Views DESC, ID DESC
   limit 1

Answer 3

三个问题：

SELECT @MyID:=ID,@MyViews:=Views FROM Table WHERE ID=2;

使用本地“视图”：

SELECT ID,Views FROM Table WHERE Views <= @MyViews AND ID != @MyID ORDER BY Views DESC LIMIT 1;
SELECT ID,Views FROM Table WHERE Views >= @MyViews AND ID != @MyID ORDER BY Views ASC LIMIT 1;

并且，请确保您有关于视图的索引。

Answer 4

或

    /* Previous Page */
SELECT ID
    FROM YourTable
    WHERE VIEWS < $CurrentPageViews 
    ORDER BY VIEWS DESC LIMIT 1

/* Next Page */
SELECT ID
    FROM YourTable
    WHERE VIEWS > $CurrentPageViews
    ORDER BY VIEWS LIMIT 1

PHP / MYSQL查询：获取上下结果！

4 个答案: