Question

我试图在SQLFiddle上运行一个复杂的连接查询，但一直遇到错误。想要在select语句中打印所有主要属性。过去几天卡住了。伸出援助之手将非常感激。

//location of ship
var location1 = 3;
var location2 = 4;
var location3 = 5;

//number of guesses 
var guess;
var guesses = 0;
var hits = 0;

//is the ship sunk?
var isSunk = false;

while (isSunk == false) {
    guess = prompt("Ready, aim, FIRE! Enter a number 0 - 9 to hit the ship!");

    if (guess > 6 || guess < 0) {
        alert("Please enter a valid number fool!");
    } else {
        guesses = guesses + 1;
    }
if (guess == location1 || guess == location2 || guess == location3) {
    hits = hits + 1;
    if (hits == 3) {
        alert("You sank my B ship hombre");
    }
    }
    }
}
var stats = "You took " + guesses + " guesses to sink the battleship. Your accuracy is " + (3/guesses);

Answer 1

您试图在实际SalesInvoice之前引用join中的join表格，因此它并不存在。您只需要更改联接的顺序：

SELECT Customer.CustomerID, FirstName, LastName, StreetAddress, 
      Apartment, City, State, 
      ZipCode, HomePhone, MobilePhone, 
      OtherPhone, Donut.DonutID, Name, Description, 
      UnitPrice, DonutOrder.DonutOrderID, DonutOrder.DonutID, 
      Quantity, SalesInvoice.DonutOrderID, 
      SalesInvoice.CustomerID 
FROM donut 
  JOIN DonutOrder ON Donut.DonutID = DonutOrder.DonutID 
  JOIN SalesInvoice ON DonutOrder.DonutID = SalesInvoice.DonutOrderID
  JOIN Customer ON SalesInvoice.CustomerID = Customer.CustomerID

SQL Fiddle Demo

Answer 2

简化为运行查询，如下所示

SELECT  c.*, d.*, dor.*, si.*
    FROM donut d
    left outer JOIN DonutOrder dor ON dor.DonutID = d.DonutID
    left outer JOIN SalesInvoice si ON si.DonutOrderID = si.DonutOrderID
    left outer JOIN Customer c on c.CustomerID = si.CustomerID

复杂连接查询给出错误

2 个答案: