Question

我正在通过GALAXQL http://sol.gfxile.net/galaxql.html

学习SQL

我在第17课 - GROUP BY / HAVING

以下是该方案：

让我们看看我们尚未涉及的几个SELECT操作，   即GROUP BY和HAVING。

这些操作的语法如下所示：

SELECT columns FROM table GROUP BY column HAVING expression

SELECT中的GROUP BY命令会导致多个输出行   合并成一排。例如，如果这个非常有用   我们希望以表格的形式生成新的统计数据。

例如，从星星中找出最高强度   上课，我们会这样做：

Select Class, Max(Intensity) As Brightness From Stars Group By Class Order By Brightness Desc

HAVING操作符与WHERE的工作方式几乎相同，除外在分组完成后应用它。因此，我们可以计算每个类的亮度之和，并裁剪出类其中总和高于150。

SELECT class, SUM(intensity) AS brightness FROM stars GROUP BY class HAVING brightness < 150 ORDER BY brightness DESC

我们可以引用未在HAVING子句中选择的列，   但结果可能难以理解。你应该可以   使用HAVING子句中的聚合函数（例如，   亮度＆lt; MAX（亮度）* 0.5，但这似乎会使电流崩溃   SQLite的版本。

与联接结合使用时，GROUP BY变得非常方便。要找出答案   每颗恒星的行星数量，我们可以做到：

SELECT stars.starid AS starid, COUNT(planets.planetid) AS planet_count FROM planets, stars WHERE stars.starid=planets.starid GROUP BY stars.starid

Hilight是大多数轨道的恒星（行星和卫星的组合）。（请注意，验证查询有点重，所以请耐心等待按“好的，我已经完成......”）。

这是我的回答

SELECT stars.starid AS HighStar, 
(COUNT(planets.planetid) + COUNT(moons.moonid)) AS OrbitalsTotal 
FROM stars
LEFT OUTER JOIN planets
ON stars.starid = planets.starid
LEFT OUTER JOIN moons
ON planets.planetid = moons.planetid
GROUP BY stars.starid
ORDER BY OrbitalsTotal DESC;

此查询向我显示具有最多oribtals的恒星有170个轨道

那么：

INSERT INTO hilight SELECT result.HighStar
FROM result
INNER JOIN stars
ON result.HighStar = stars.starid
WHERE result.OrbitalsTotal = 170

我的问题是如何让这个查询更好？我不想硬编码170轨道，我不想创建第二个查询来插入数据。

Answer 1

SELECT stars.starid AS HighStar, 
       (COUNT(planets.planetid) + COUNT(moons.moonid)) AS OrbitalsTotal 
FROM stars
     LEFT OUTER JOIN
     planets ON stars.starid = planets.starid
     LEFT OUTER JOIN
     moons ON planets.planetid = moons.planetid
GROUP BY stars.starid
HAVING OrbitalsTotal = (SELECT MAX(Orbitals)
                        FROM (SELECT (COUNT(planets.planetid) + COUNT(moons.moonid)) Orbitals
                              FROM stars
                                   LEFT OUTER JOIN
                                   planets ON stars.starid = planets.starid
                                   LEFT OUTER JOIN
                                   moons ON planets.planetid = moons.planetid
                              GROUP BY stars.starid))

Answer 2

只需使用您的第一个查询，并添加条款LIMIT 1以仅返回第一条记录：

INSERT INTO hilight
SELECT stars.starid AS HighStar
FROM stars
LEFT OUTER JOIN planets
ON stars.starid = planets.starid
LEFT OUTER JOIN moons
ON planets.planetid = moons.planetid
GROUP BY stars.starid
ORDER BY COUNT(planets.planetid) + COUNT(moons.moonid) DESC
LIMIT 1

Answer 3

Star 22336没有170个轨道。

它有97个轨道。

检查卫星数量。

SELECT COUNT(moons.moonid) as moon_count
FROM stars JOIN planets
ON stars.starid=planets.starid
JOIN moons
ON planets.planetid=moons.planetid
WHERE stars.starid=22336

这给了79个卫星。

现在检查一下行星的数量。

SELECT COUNT(planets.planetid) AS planet_count
FROM stars JOIN planets
ON stars.starid=planets.starid
WHERE stars.starid=22336

这给了18个行星。

下面的SQL正确计算轨道计数并完成任务。

INSERT INTO hilight
SELECT pc.starid FROM

(SELECT p.starid, COUNT(p.planetid) AS p_count
FROM planets AS p
GROUP BY p.starid) AS pc

JOIN

(SELECT p.starid, COUNT(m.moonid) AS m_count
FROM planets AS p JOIN moons AS m
ON p.planetid=m.planetid
GROUP BY p.starid) AS mc

ON pc.starid=mc.starid

ORDER BY p_count+m_count DESC
LIMIT 1

Answer 4

如果您使用SELECT TOP(1) ...开始第一个查询，那么这将只为您提供第一个结果。 TOP()语法可以与硬编号一起使用，该编号可以提供许多行或百分比，从而提供总数的百分比。

Answer 5

通过GalaXQL的第17章，我们还没有学到LIMIT，而且我认为我们可以在没有所有LEFT OUTER JOINS的情况下做到这一点（借用Dale M＆＃39;其中缺少＆＃34; AS＆＃ 34;之前＆＃34; Orbitals＆＃34;虽然它仍然有效？！））与我们在该部分的最后一个例子中看到的更接近。

那怎么样：

DELETE FROM hilight;
INSERT INTO hilight 
SELECT HighStar FROM (
    SELECT stars.starid AS HighStar, 
        (COUNT(planets.planetid) + COUNT(moons.moonid)) AS OrbitalsTotal 
    FROM stars, planets, moons 
    WHERE stars.starid = planets.starid
        AND planets.planetid = moons.planetid
    GROUP BY stars.starid
    HAVING OrbitalsTotal = (
        SELECT MAX(Orbitals) FROM (
            SELECT (COUNT(planets.planetid) + COUNT(moons.moonid)) AS Orbitals
            FROM stars, planets, moons
            WHERE stars.starid = planets.starid
                AND planets.planetid = moons.planetid
            GROUP BY stars.starid))
);
SELECT * FROM hilight;

使用LIMIT，我们也可以避免LEFT OUTER JOIN而不必复制我们的选择：

DELETE FROM hilight;
INSERT INTO hilight
SELECT bodygroup FROM (
    SELECT stars.starid as bodygroup, 
        (COUNT(planets.planetid)+COUNT(moons.moonid)) AS bodycount
    FROM stars, planets, moons 
    WHERE stars.starid=planets.starid 
        AND planets.planetid=moons.planetid 
    GROUP BY stars.starid
    ORDER BY bodycount DESC) 
LIMIT 1;
SELECT * FROM hilight;

或者使用CREATE VIEW试图避免在不使用LIMIT的情况下加倍努力，我们可以：

DELETE FROM hilight;    
CREATE VIEW bodyview AS
    SELECT stars.starid as bodygroup,
        (COUNT(planets.planetid)+COUNT(moons.moonid)) AS bodycount
    FROM stars, planets, moons
    WHERE stars.starid=planets.starid
       AND planets.planetid=moons.planetid
    GROUP BY stars.starid
    ORDER BY bodycount DESC;
INSERT INTO hilight SELECT bodygroup FROM bodyview  
    WHERE bodycount = (SELECT MAX(bodycount) FROM bodyview );
DROP VIEW bodyview;
SELECT * FROM hilight;

如何更好地进行此查询？

5 个答案: