Question

我在德比中形成了以下查询来计算所有棒球运动员的击打百分比......

 SELECT p1.nameFirst as "First", 
 p1.nameLast as "Last", 
 (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0) as "SLUGGING" 
FROM Batting b1 join Players p1 on b1.playerID = p1.playerID
WHERE AB >= 400;

它生产所有击球时有400或以上的击球手，并计算他们的击打率。我的问题是我只需要选择一个具有最大段塞百分比的玩家。我尝试用SELECT MAX包围数学块，但无济于事。我想要的输出是......

Jose Bautista 0.7426900584795321637426900

另外，有没有办法将该十进制格式化为一定数量的数字？任何建议都表示赞赏。

Answer 1

将ORDER BY SLUGGING DESC LIMIT 1添加到您的查询中，您应该获得最佳结果。

编辑：似乎Derby不支持限制返回的行数。那么最好的方法可能就是创建一个带有最大段塞结果的子选择，并加入它以使playerId打印出来。

Answer 2

我开始怀疑自己是在帮助你为MBA写点什么，还是做计算机科学作业。： - ）

SELECT
    p1.nameFirst as "First", 
    p1.nameLast as "Last",
    b.s AS "SLUGGING"
FROM
(
    SELECT
        MAX(
            (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0)
        ) AS s
    FROM
        Batting b1
    WHERE
        b1.AB >= 400
) m1
JOIN
    Batting b
        ON 
                m1.s = (b.H + (2 * b.doubles) + (3 * b.triples) + (4 * b.HR)) / (b.AB * 1.0)
JOIN
    Players p1
        ON
            b1.playerID = p1.playerID
WHERE
    b1.AB >= 400

OR

*不确定是否支持

SELECT
    TOP 1
    p1.nameFirst as "First"
    , p1.nameLast as "Last"
    , (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0) AS "SLUGGING" 
FROM
    Batting b1
JOIN
    Players p1 on b1.playerID = p1.playerID
WHERE
    b1.AB >= 400
ORDER BY
    (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0) DESC

OR

*我认为这个会起作用

SELECT * FROM (
SELECT
    ROW_NUMBER() OVER(
        ORDER BY
            (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0) DESC
    ) as rownum
    , p1.nameFirst as "First"
    , p1.nameLast as "Last"
    , (b1.H + (2 * b1.doubles) + (3 * b1.triples) + (4 * b1.HR)) / (b1.AB * 1.0) AS "SLUGGING" 
FROM
    Batting b1
JOIN
    Players p1 on b1.playerID = p1.playerID
WHERE
    b1.AB >= 400
) AS r
WHERE r.rownum = 1

如何选择MAX SQL

2 个答案: