如何使這個查詢更好？

Question

我正在學習SQL通過GALAXQL http://sol.gfxile.net/galaxql.html

林在第17課 - GROUP BY/HAVING

下面是這種情況：

讓我們來看看幾個我們沒有覆蓋的SELECT操作然而， 即GROUP BY和HAVING。

的語法爲這些操作是這樣的：

SELECT columns FROM table GROUP BY column HAVING expression 

在SELECT的GROUP BY命令會導致多個輸出行是 組合成單排。例如，如果我們希望將 作爲表格生成新的統計數據，這可能非常有用。

例如，找出從星爲每個 同類產品中最高的強度，我們會做：

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

的擁有操作的工作幾乎方法一樣WHERE，除了 那它在分組完成後應用。因此，我們可以計算出 每級亮度的總和，並裁掉類 其中總和比，說高，150

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

我們可以引用列未選擇在HAVING子句中， 但結果可能難以理解。您應該能夠 使用HAVING子句在聚合函數（例如， 亮度< MAX（亮度）* 0.5，但是這似乎崩潰當前 版本的SQLite。

當結合聯接，GROUP通過變得相當方便的找出 每星行星的數量，我們可以做：

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

高亮大多數軌道（組合行星和衛星）的星 （注意驗證查詢一世有點沉重，所以請耐心等待 按下「好吧，我完成了......」）。

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這裏是我的回答

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; 

這個查詢結果顯示我最oribtals的明星有170個軌道

那麼接下來：

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

我問題是我該如何做到這一點查詢更好？我不想要硬編碼170軌道，我不想創建第二個查詢來插入數據。

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

如果你用SELECT TOP(1) ...開始你的第一個查詢，那麼這將只給你第一個結果。 TOP()語法可以與一個硬編號一起使用，該編號給出了很多行或佔總數百分比的百分比。

Answer 3

只要用你的第一個查詢，並加上一句LIMIT 1只返回第一個記錄：

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 4

通過GalaXQL的第17章中，我們還沒有學會限制，我想我們可以在沒有所有LEFT OUTER JOINS的情況下（從Dale M借用（它在「Orbitals」之前缺少「AS」，儘管它仍然有效！））與我們在該部分的最後一個例子中看到的有點接近。

因此，如何：

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; 

有了限制，我們也避免LEFT OUTER JOIN的，而不必重複我們的選擇：

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; 

或使用CREATE VIEW儘量保持在不使用LIMIT的情況下加倍一些努力，我們可以：

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; 

Answer 5

星22336沒有170個軌道秒。

它有97個軌道。

檢查衛星的數量。

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 

這給出了79個月亮。

現在檢查行星的數量。

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

這給出了18顆行星。

下面的SQL計算正確計算軌道並完成任務。

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