我不太确定你的关系或外键。但从你所付出的来看，我想这也行。可能需要稍加修改，但我想你会明白的。

SELECT ___Invoices.IVC_Id, k1.KDX_Name as guest, k2.KDX_Company as company
FROM ___Invoices 
JOIN ___Bookings
ON ___Bookings.BOO_Id = ___Invoices.IVC_BookingId
LEFT JOIN ___Kardex k1 ON ___Bookings.BOO_GuestId = k1.KDX_Id
LEFT JOIN ___Kardex k2 ON ___Bookings.BOO_CompanyId = k2.KDX_Id
WHERE ___Invoices.IVC_Id='10'

请检查一下，告诉我这是不是你想要的。
http://sqlfiddle.com/#!2011年9月1日
如果你有任何疑问，尽管问。希望有帮助：）

你可以试试

SELECT IVC_Id, 
(SELECT KDX_Name FROM ___Kardex WHERE KDX_Id= ___Bookings.BOO_GuestId) AS guest,
(SELECT KDX_Company FROM ___Kardex WHERE KDX_Id= ___Bookings.BOO_CompanyId) AS company,
___Bookings.BOO_BillingId as billing
FROM ___Invoices 
JOIN ___Bookings
ON ___Bookings.BOO_Id = ___Invoices.IVC_BookingId
WHERE IVC_Id='10'

如果您关心的是性能，那么这可能不是最佳的解决方案，但是在您的模式设计中，您不应该让一个表处理多种类型的信息，这使得编写任何合法的联接都很困难。

请检查您的第二个加入查询：

LEFT JOIN ___Kardex
ON ___Kardex.KDX_Id = ___Bookings.BOO_BillingId

应该是的

LEFT JOIN ___Kardex
ON ___Kardex.KDX_Id = ___Bookings.BOO_CompanyId

sql小提琴：http://sqlfiddle.com/#!2017年9月21日

您将以0的帐单ID加入\uuu kardex。
0不等于89或90

Your code

LEFT JOIN ___Kardex
ON ___Kardex.KDX_Id = ___Bookings.BOO_BillingId

我是这样解决的：我加入到同一个表两次，每次都有不同的连接条件。然后为该联接创建了一个新别名：

CREATE TABLE `inv` (`INV_ID` VARCHAR(255) NOT NULL, `IVC_BookingID` 
VARCHAR(255) NOT NULL);

CREATE TABLE `bookings` (`BOO_id` VARCHAR(255) NOT NULL, `BOO_guestid` VARCHAR(255) NOT NULL, `BOO_CompanyID` VARCHAR(255) NOT NULL, `BOO_BillingID` VARCHAR(255) NOT NULL);

CREATE TABLE `kardex` (`KDX_ID` VARCHAR(255) NOT NULL, `KDX_TYPE` VARCHAR(255) NOT NULL, `KDX_NAME` VARCHAR(255) NOT NULL, `KDX_COMPANY` VARCHAR(255) NOT NULL);

INSERT INTO `inv` (INV_ID,IVC_BookingID) SELECT '10','31';
INSERT INTO `bookings` (BOO_id,BOO_guestid,BOO_CompanyID,BOO_BillingID) SELECT '10','89','90','0';
INSERT INTO `kardex` (KDX_ID,KDX_TYPE,KDX_NAME,KDX_COMPANY) SELECT '89','guest','Frank','';
INSERT INTO `kardex` (KDX_ID,KDX_TYPE,KDX_NAME,KDX_COMPANY) SELECT '90','company','','Google';

SELECT a.INV_ID, c.kdx_name, d.kdx_company FROM inv a 
INNER JOIN bookings b ON b.boo_id=a.INV_ID
INNER JOIN kardex c ON c.kdx_id=b.boo_guestid
INNER JOIN kardex d ON d.kdx_id=b.BOO_CompanyID
WHERE INV_ID='10';

结果

INV_ID  kdx_name    kdx_company
10       Frank      Google