從多個表中選擇與交叉引用和自引用[brain cracker]

Question

任何人都達到SQL挑戰？因爲我迄今爲止所做的所有努力幾乎不足以簡化問題並將其置於一個問題中......

在這裏。在下面的例子中，我們需要包括：

• 一個國家（薪或無薪）所有支付航班
• 機票如果一個人做出了另一架飛機到付費城市在該國

這已經很棘手，但還有更多。

• 如果一個人對飛行沒有報名費一個城市，但它位於一個 國家，確實有一定的費用，該航班仍然被認爲是支付 ，必須包括在內。

編輯：我增加航班110，這將有助於揭示不必要的添加免費fligts。

下面是結果集將要來的SQL查詢出來：

+--------------------------------------------------------------+ 
| Desired result set           | 
+--------------------------------------------------------------+ 
| FlightNumber | ID | Name | LocationID | location.Name  | 
+--------------------------------------------------------------+ 
| 102   | 2 | Tom | 500  | NL - NoFee   | -> because Tom has a paid flight to Amsterdam 
| 103   | 2 | Tom | 501  | Amsterdam (NL) - Fee | -> because Amsterdam is a paid location 
| 105   | 4 | Bob | 501  | Amsterdam (NL) - Fee | -> because Amsterdam is a paid location 
| 107   | 6 | Bill | 503  | ITA - Fee   | -> because ITA is a paid location 
| 108   | 7 | Ryan | 503  | ITA - Fee   | -> because ITA is a paid location 
| 109   | 7 | Ryan | 505  | Venice (ITA) - NoFee | -> because Venice is located inside ITA 
+--------------------------------------------------------------+ 

有誰知道如何得到這個甜蜜的結果與SQL設置？

一個良好的開端：

SELECT flights.FlightNumber, people.ID, people.Name, flights.LocationID, locations.Name 
FROM flights 
INNER JOIN people ON (people.ID = flights.ID) 
INNER JOIN locations ON (locations.LocationID = flights.LocationID) 

創建/插入

CREATE TABLE `people` (
    `ID` INT NOT NULL, 
    `Name` VARCHAR(45) NULL, 
    PRIMARY KEY (`ID`)); 

CREATE TABLE `locations` (
    `LocationID` INT NOT NULL, 
    `Name` VARCHAR(45) NULL, 
    `EntryFee` TINYINT(1) NULL, 
    `ParentLocationID` INT NULL, 
    PRIMARY KEY (`LocationID`)); 

CREATE TABLE `flights` (
    `FlightNumber` INT NOT NULL, 
    `ID` INT NULL, 
    `LocationID` INT NULL, 
    PRIMARY KEY (`FlightNumber`) , 
    INDEX `fk_purchases_buyers_idx` (`LocationID` ASC) , 
    INDEX `fk_flights_people1_idx` (`ID` ASC) , 
    CONSTRAINT `fk_purchases_buyers` 
    FOREIGN KEY (`LocationID`) 
    REFERENCES `locations` (`LocationID`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    CONSTRAINT `fk_flights_people1` 
    FOREIGN KEY (`ID`) 
    REFERENCES `people` (`ID`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION); 

INSERT INTO `people` (`ID`, `Name`) VALUES 
(1, 'John'), 
(2, 'Tom'), 
(3, 'Kate'), 
(4, 'Bob'), 
(5, 'Mike'), 
(6, 'Bill'), 
(7, 'Ryan'); 

INSERT INTO `locations` (`LocationID`, `Name`, `EntryFee`, `ParentLocationID`) VALUES 
(500, 'NL - NoFee', 0, NULL), 
(501, 'Amsterdam (NL) - Fee', 1, 500), 
(502, 'Rotterdam (NL) - NoFee', 0, 500), 
(503, 'ITA - Fee', 1, NULL), 
(504, 'Rome (ITA) - Fee', 1, 503), 
(505, 'Venice (ITA) - NoFee', 0, 503); 

INSERT INTO `flights` VALUES 
(100, 1, 500), 
(101, 1, 502), 
(102, 2, 500), 
(103, 2, 501), 
(104, 3, 500), 
(105, 4, 501), 
(106, 5, 502), 
(107, 6, 503), 
(108, 7, 503), 
(109, 7, 505), 
(110, 6, 502); 

不重要注：我知道這個例子是不是在存儲國家意識完全合乎邏輯和在同一張桌子上的城市，有一個國家和一個城市的航班記錄。但這只是一個例子。至少它比t1.col1更具可讀性。

Answer 1

下面應該工作：

SELECT F.FlightNumber, P.ID, P.Name, F.LocationID, LOC.Name AS Loc_Name 
FROM flights F 
INNER JOIN people P ON P.ID = F.ID 
INNER JOIN (SELECT L1.LocationID, L1.`Name`, L1.`ParentLocationID` 
      FROM locations L1 
      LEFT JOIN locations L2 ON L1.`ParentLocationID` = L2.LocationID 
      WHERE L1.`EntryFee` = 1 OR L2.`EntryFee` = 1) AS LOC ON LOC.LocationID = F.LocationID 
UNION 
SELECT F.FlightNumber, PAID_FL.ID, PAID_FL.Name, F.LocationID, PAID_FL.Loc_Name 
FROM flights F 
INNER JOIN (SELECT F.FlightNumber, P.ID, P.Name, LOC.Name AS Loc_Name,LOC.`ParentLocationID` AS LocationID 
      FROM flights F 
      INNER JOIN people P ON P.ID = F.ID 
      INNER JOIN (SELECT L1.LocationID, L1.`ParentLocationID`, L2.`Name` 
         FROM locations L1 
         LEFT JOIN locations L2 ON L1.`ParentLocationID` = L2.LocationID 
         WHERE L1.`EntryFee` = 1 OR L2.`EntryFee` = 1) AS LOC ON LOC.LocationID = F.LocationID) PAID_FL ON F.ID = PAID_FL.ID AND F.LocationID = PAID_FL.LocationID 

Answer 2

SELECT x.* 
    , COALESCE(y.entryfee,x.entryfee,0) fee 
    FROM locations x 
    LEFT 
    JOIN locations y 
    ON y.parentlocationid = x.locationid; 

Answer 3

我沒有做任何測試，但我認爲這將工作：

SELECT F.FlightNumber, P.ID, P.Name, L.LocationID, L.Name 
FROM flights F 
INNER JOIN people P ON P.ID = F.ID 
INNER JOIN locations L ON L.LocationID = F.LocationID 
WHERE P.ID IN (SELECT F.ID FROM flights F INNER JOIN locations L ON L.LocationID = F.LocationID WHERE L.EntryFee = 1) 

See it at SQLFiddle.com.