Question

我有一张满是杂志的桌子，需要提取每本杂志的最新独特刊物。

我试过

    SELECT DISTINCT
    magazine
        FROM
    product p
        INNER JOIN
    (SELECT 
        title, MAX(onSale) AS Latest
    FROM
        product
    GROUP BY magazine) groupedp

返回不同的杂志，但不返回我需要的其他数据。

更新：

模式

-id----onsale----magazine
  1    1/12/12   Fishing Mag
  2    1/11/12   Fishing Mag
  3    12/03/11  Pencil Sharpening Monthly
  4    1/02/10   Pencil Sharpening Monthly
  5    16/04/09  Homes in the Sky

所以我想要的结果是：

 -id----onsale----magazine
   1    1/12/12   Fishing Mag         
   3    12/03/11  Pencil Sharpening Monthly         
   5    16/04/09  Homes in the Sky

Answer 1

SELECT id, MAX(onSale) as latest, magazine
FROM product
GROUP BY magazine
ORDER BY latest DESC

Answer 2

SELECT 
    p.*
FROM
        product p
    INNER JOIN
        ( SELECT 
              magazine, MAX(onSale) AS latest
          FROM
              product
          GROUP BY 
              magazine
        ) AS groupedp
      ON  groupedp.magazine = p.magazine
      AND groupedp.latest = p.onSale ;

Answer 3

这看起来像你需要的：

SELECT id, MAX(onsale) AS onsale, magazine FROM magazines GROUP BY magazine ORDER BY onsale DESC;

请查看：

http://sqlfiddle.com/#!2/38e78/3

<强>更新我稍微改了一下查询，返回MAX（onsale）

Answer 4

给出的答案中没有一个是正确的，因为它们返回的一组分离的数据不代表一个确切的列。 id可能不是与id值同一行的onsale。

以下将起作用：

SELECT
    id, onsale, magazine
FROM (
    SELECT
       id, onsale, magazine
    FROM
        product
    ORDER BY
        onsale DESC) AS a
GROUP BY
    magazine

Answer 5

您需要将“其余数据”放在第一个选择中：

SELECT DISTINCT magazine, "rest of data"
FROM product p 
INNER JOIN 
( SELECT title, MAX(onSale) AS Latest 
FROM product 
GROUP BY magazine ) groupedp

MySQL以最高值选择DISTINCT

5 个答案: