三个表的结构。
预订
CREATE TABLE `booking` (
`bookingID` int(11) NOT NULL AUTO_INCREMENT,
`receipt_no` int(11) NOT NULL,
`client` varchar(32) NOT NULL,
`operator` varchar(32) NOT NULL,
`discount` int(11) NOT NULL,
`total_amount` int(64) NOT NULL,
`amount_paid` int(32) NOT NULL,
`balance` int(32) NOT NULL,
`date` date NOT NULL,
PRIMARY KEY (`bookingID`)
ENGINE=InnoDB AUTO_INCREMENT=209 DEFAULT CHARSET=latin1
JOBTITLE
CREATE TABLE `jobtitle` (
`jobtitleID` int(11) NOT NULL AUTO_INCREMENT,
`jobtitle` varchar(255) NOT NULL,
`quantity` int(11) NOT NULL,
`amount` varchar(255) NOT NULL,
`jobtypeID` int(11) NOT NULL,
`bookingID` int(11) NOT NULL,
PRIMARY KEY (`jobtitleID`)
ENGINE=InnoDB AUTO_INCREMENT=463 DEFAULT CHARSET=latin1
First_graphics_debtors
CREATE TABLE `first_graphics_debtors`
`id` int(11) NOT NULL AUTO_INCREMENT,
`receipt_no` int(11) NOT NULL,
`date_paid` date NOT NULL,
`old_balance` int(32) NOT NULL,
`debtor_amount_paid` int(32) NOT NULL,
`new_balance` int(32) NOT NULL,
PRIMARY KEY (`id`)
ENGINE=InnoDB AUTO_INCREMENT=16 DEFAULT CHARSET=latin1
插入少量数据后
预订
+-----------+------------+----------+----------+----------+--------------+-------------+---------+------------+
| bookingID | receipt_no | client | operator | discount | total_amount | amount_paid | balance | date |
+-----------+------------+----------+----------+----------+--------------+-------------+---------+------------+
| 205 | 156872940 | Osaro | Obi | 10 | 156380 | 135000 | 5742 | 2012-05-15 |
| 206 | 227349168 | Amaka | Stephen | 4 | 73250 | 70320 | 0 | 2012-05-15 |
| 207 | 155732278 | Aghahowa | Ibori | 0 | 116836 | 15000 | 101836 | 2012-05-15 |
| 208 | 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 |
+-----------+------------+----------+----------+----------+--------------+-------------+---------+------------+
JOBTITLE
+------------+---------------------------+----------+--------+-----------+-----------+
| jobtitleID | jobtitle | quantity | amount | jobtypeID | bookingID |
+------------+---------------------------+----------+--------+-----------+-----------+
| 454 | A1 Full Colour | 10 | 4334 | 1 | 205 |
| 455 | Complementry Card | 20 | 5652 | 2 | 205 |
| 456 | A4 Printout (graphics)B/W | 25 | 2930 | 4 | 206 |
| 457 | Scan | 2 | 4334 | 5 | 207 |
| 458 | A4 Full Colour | 199 | 500 | 3 | 207 |
| 459 | ID Card | 2 | 4334 | 2 | 207 |
| 460 | A3 Full Colour | 10 | 4334 | 3 | 208 |
| 461 | Flex Banner | 20 | 2930 | 2 | 208 |
| 462 | A2 Single Colour | 199 | 650 | 1 | 208 |
+------------+---------------------------+----------+--------+-----------+-----------+
First_graphics_debtors
+----+------------+------------+-------------+--------------------+-------------+
| id | receipt_no | date_paid | old_balance | debtor_amount_paid | new_balance |
+----+------------+------------+-------------+--------------------+-------------+
| 7 | 156872940 | 2012-05-15 | 5742 | 5000 | 742 |
| 8 | 156872940 | 2012-05-15 | 5742 | 5742 | 0 |
| 9 | 753263343 | 2012-05-15 | 152161 | 152161 | 0 |
| 13 | 753263343 | 2012-05-15 | 152161 | 14524 | 137637 |
| 14 | 753263343 | 2012-05-15 | 152161 | 2000 | 150161 |
| 15 | 753263343 | 2012-05-15 | 152161 | 1000 | 151161 |
+----+------------+------------+-------------+--------------------+-------------+
当我运行此查询时：
SELECT `booking`.`receipt_no`, `client`, `operator`, `discount`, `total_amount`,
`amount_paid`, `balance`, `date`, `jobtitle`, `quantity`, `amount`,
`date_paid`, `old_balance`, `debtor_amount_paid`, `new_balance`
FROM (`booking`)
JOIN `jobtitle` ON `jobtitle`.`bookingID` = `booking`.`bookingID`
JOIN `first_graphics_debtors`
ON `first_graphics_debtors`.`receipt_no` = `booking`.`receipt_no`
WHERE `booking`.`receipt_no` = '753263343'
AND `first_graphics_debtors`.`receipt_no` = '753263343'
GROUP BY `jobtitle`.`quantity`
我得到了这个输出：
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
| receipt_no | client | operator | discount | total_amount | amount_paid | balance | date | jobtitle | quantity | amount | date_paid | old_balance | debtor_amount_paid | new_balance |
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | A3 Full Colour | 10 | 4334 | 2012-05-15 | 152161 | 152161 | 0 |
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | Flex Banner | 20 | 2930 | 2012-05-15 | 152161 | 152161 | 0 |
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | A2 Single Colour | 199 | 650 | 2012-05-15 | 152161 | 152161 | 0 |
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
以下列的数据重复三次，而不是从与receipt_no相关的四行中获取数据
date_paid, old_balance, debtor_amount_paid, new_balance
预期结果
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
| receipt_no | client | operator | discount | total_amount | amount_paid | balance | date | jobtitle | quantity | amount | date_paid | old_balance | debtor_amount_paid | new_balance |
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | A3 Full Colour | 10 | 4334 | 2012-05-15 | 152161 | 152161 | 0 |
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | Flex Banner | 20 | 2930 | 2012-05-15 | 152161 | 14524 | 137637 |
| 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | 2012-05-15 | A2 Single Colour | 199 | 650 | 2012-05-15 | 152161 | 2000 | 150161 |
+------------+--------+----------+----------+--------------+-------------+---------+------------+------------------+----------+--------+------------+-------------+--------------------+-------------+
尽管如此，预期的输出尚未完成，但也许您可以从debtor_amount_paid和new_balance上的数据中获取图片
答案 0 :(得分：1)
我很确定你的问题是因为你的GROUP BY条款。不确定你使用它的目的是什么，但尝试没有它。如果您获得重复的行，请尝试使用SELECT DISTINCT。如果您的目标是根据该列订购结果，请使用ORDER BY。
此外，在receipt_no子句中指定WHERE两次是多余的。这两个表已经被该列连接，因此您需要在一个表中对其进行过滤。并且除了一些特殊的异常（例如列名中的空格或逗号）或者与保留字命名相同的列（也可能是其他几个）之外，反引号并不是必需的。您的唯一一列看起来需要反引号是date列，但即使在该列上排除它们，您仍应该没问题。我发现到处都是反复使得查询更长，更难阅读。他们不会因为在那里而伤害任何东西，所以如果这是你的偏好，你可以留下他们，但就个人而言，我不是粉丝。
我用上面提到的修改重写了你的查询，而且我给表别名以进一步缩短它。这不会提高性能或任何东西，只是让它更容易阅读IMO：
SELECT DISTINCT
b.receipt_no, client, operator, discount, total_amount,
amount_paid, balance, `date`, jobtitle, quantity,
amount, date_paid, old_balance, debtor_amount_paid, new_balance
FROM
booking b
INNER JOIN jobtitle jt ON jt.bookingID = b.bookingID
INNER JOIN first_graphics_debtors fgd ON fgd.receipt_no = b.receipt_no
WHERE
b.receipt_no = '753263343'
ORDER BY
jt.quantity
答案 1 :(得分：1)
您可能希望使用FULL JOIN来对齐这些连接的行，但由于MySQL上没有FULL JOIN，您必须与UNION竞争：
压缩代码以避免滚动条：
select bx.*, null as sepy, booking_particulars.*
from booking bx
join (
select same_n,
max(Receipt_No) as Receipt_No, max(Quantity) as Quantity,
max(Amount) as Amount,
null as sepx, -- separator
max(Date_Paid) as Date_Paid, max(Old_Balance) as Old_Balance,
max(Debtor_Amount_Paid) as Debtor_Amount_Paid,
max(New_Balance) as New_Balance
from (
select *
from (
select b.Receipt_No, j.Quantity, j.Amount,
null as Date_Paid, null as Old_Balance, null as Debtor_Amount_Paid,
null as New_Balance, @j_rn := @j_rn + 1 as same_n
from (booking b, (select @j_rn := 0) as vars)
join jobtitle j ON j.bookingID = b.bookingID
where b.receipt_no = '753263343'
order by j.JobTitleId
) as jx
union all
select * from (
select f.Receipt_No, null as Quantity, null as Amount,
f.Date_Paid, f.Old_Balance, f.Debtor_Amount_Paid,
f.New_Balance, @f_rn := @f_rn + 1 as same_n
from (first_graphics_debtors f, (select @f_rn := 0) as vars)
where f.receipt_no = '753263343'
order by f.id
) as fx
) as z
group by same_n order by same_n
) as booking_particulars on booking_particulars.receipt_no = bx.receipt_no
输出：
| BOOKINGID | RECEIPT_NO | CLIENT | OPERATOR | DISCOUNT | TOTAL_AMOUNT | AMOUNT_PAID | BALANCE | DATE | SEPY | SAME_N | QUANTITY | AMOUNT | SEPX | DATE_PAID | OLD_BALANCE | DEBTOR_AMOUNT_PAID | NEW_BALANCE |
|-----------|------------|--------|----------|----------|--------------|-------------|---------|----------------------------|--------|--------|----------|--------|--------|----------------------------|-------------|--------------------|-------------|
| 208 | 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | May, 15 2012 08:00:00-0700 | (null) | 1 | 10 | 4334 | (null) | May, 15 2012 08:00:00-0700 | 152161 | 152161 | 0 |
| 208 | 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | May, 15 2012 08:00:00-0700 | (null) | 2 | 20 | 2930 | (null) | May, 15 2012 08:00:00-0700 | 152161 | 14524 | 137637 |
| 208 | 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | May, 15 2012 08:00:00-0700 | (null) | 3 | 199 | 650 | (null) | May, 15 2012 08:00:00-0700 | 152161 | 2000 | 150161 |
| 208 | 753263343 | Chaka | Stephen | 10 | 231290 | 56000 | 152161 | May, 15 2012 08:00:00-0700 | (null) | 4 | (null) | (null) | (null) | May, 15 2012 08:00:00-0700 | 152161 | 1000 | 151161 |
在此处进行实时测试和查询进展：http://www.sqlfiddle.com/#!2/d8d3b/45
答案 2 :(得分：0)
我完全同意Travesty3关于表别名和反引号的内容。
我认为，您的问题是，您正试图将“应该”和“应该”的内容结合起来。是两个单独的查询到一个。对于任何给定的预订，您可以拥有任意数量的职位条目，并且完全独立于职位信息，您可以拥有任意数量的First Graphics Debtors条目。在样品收据753263343的样本数据中，您有一个预订，三个职称条目和4个第一图形债务人条目。
默认情况下（没有GROUP BY子句），您将得到12个条目，对应于三组行的交叉积（1 x 3 x 4）。
MySQL中的GROUP BY与大多数其他DBMS不同;标准SQL不允许您省略GROUP BY子句中的非聚合列（并且所有列都是非聚合的）。
我认为您之后的输出结果集类似于（使用FGD作为First Graphics Debtors的缩写）：
+-------------------+---------------------+---------------+
| Data from Booking | Data from Job Title | Data from FGD |
+-------------------+---------------------+---------------+
| Booking ID 208... | Job Title ID 460... | FGD ID 9 |
| Booking ID 208... | Job Title ID 461... | FGD ID 13 |
| Booking ID 208... | Job Title ID 462... | FGD ID 14 |
| Booking ID 208... | ...?????????????... | FGD ID 15 |
+-------------------+---------------------+---------------+
问号的最合理的值集可能是一组NULL。
但是，您可能会更好，使用两个单独的查询，一个用于获取职位信息，一个用于获取FGD信息（两个中只有一个需要返回预订信息）。但是，您必须在应用程序代码中进行一些演示工作（这是应该进行演示工作的地方，但这是一个不同的讨论）。
因此，对于胆小的人，我建议使用两个查询。对于勇敢者，请继续阅读...
更新： TDQD - 包括测试的材料有点错误。
首先，让我说明一下：如果这是你所追求的，那么这是相对艰苦的工作，尽管可以做到。其次，它在DBMS中使用OLAP函数更简单。使用MySQL（以及其他一些DBMS），您可以使用临时表来保存中间结果。
您在预订ID上加入预订和职位。您在收据编号上加入Booking和FGD。假设我们创建了两个中间结果集：
SequencedJobTitles(Booking ID, Job_Seq_No)
SequencedReceipts(Receipt No, Rec_Seq_No)
其中第一个作业序列号从每个预订ID的1..N开始，每个收据号的收据序列号从1..M开始。
现在我们可以写：
SELECT B.*, J.*, R.*
FROM Booking AS B
LEFT JOIN (SELECT J1.*, J2.Job_Seq_No
FROM SequencedJobTitles AS J2
JOIN JobTitle AS J1 ON J2.BookingID = J1.BookingID
) AS J
ON B.BookingID = J.BookingID
LEFT JOIN (SELECT R1.*, R2.Rec_Seq_No
FROM SequencedReceipts AS R2
JOIN First_Graphics_Debtors AS R1 ON R1.Receipt_No = R2.Receipt_No
) AS R
WHERE J.Job_Seq_No = R.Rec_Seq_No
AND B.Receipt_No = 753263343
ORDER BY B.BookingID, COALESCE(J2.Job_Seq_No, R2.Rec_Seq_No);
是的，使用*是懒惰的;我对其进行了优化，以列出生产查询中我想要的列的确切列表。对于调试，它可以确保您查看所有数据，这可能会有所帮助。
如何创建两个中间结果集？
如果没有OLAP功能，您可以使用以下命令生成SequencedJobTitles：
SELECT J1.BookingID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID <= J2.JobTitleID
GROUP BY J1.BookingID;
与SequencedReceipts类似：
SELECT R1.Receipt_No, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID <= R2.ID
GROUP BY R1.Receipt_No;
这些表达式使用非等值连接的自连接，这是非常昂贵的。如果表格很大，您可能希望仅将相关的预订ID或收据号码限制在子查询中。请注意，您可以单独测试这些子查询;这对于TDQD（测试驱动查询设计）来说是一个好主意。
但是，我们可以简单地将这些子查询插入到之前开发的主查询中：
SELECT B.*, J.*, R.*
FROM Booking AS B
LEFT JOIN (SELECT J1.*, J2.Job_Seq_No
FROM (SELECT J1.BookingID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID <= J2.JobTitleID
GROUP BY J1.BookingID
) AS J2
JOIN JobTitle AS J1 ON J2.BookingID = J1.BookingID
) AS J
ON B.BookingID = J.BookingID
LEFT JOIN (SELECT R1.*, R2.Rec_Seq_No
FROM (SELECT R1.Receipt_No, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID <= R2.ID
GROUP BY R1.Receipt_No
) AS R2
JOIN First_Graphics_Debtors AS R1 ON R1.Receipt_No = R2.Receipt_No
) AS R
ON J.Job_Seq_No = R.Rec_Seq_No
WHERE B.Receipt_No = 753263343
ORDER BY B.BookingID, COALESCE(J2.Job_Seq_No, R2.Rec_Seq_No);
警告：未经测试的SQL！
更新：这个警告没有放在那里。无论我怎么努力，我都不相信自己第一次准确地编写复杂的查询。而且我并没有尝试制作我之前展示的轮廓结果（我总是使用两个单独的查询并将数据组合在演示工具方法中），所以我认为错误是不可避免的。 / p>
Real TDQD显示序列查询很接近，但不正确。虽然我将<=条件更改为>=条件，但这更具有外观性而非实质性。正确的查询如下：
SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID;
SELECT R1.Receipt_No, R1.ID, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID >= R2.ID
GROUP BY R1.Receipt_No, R1.ID;
这两个结果可以用相应的表连接起来，但是连接条件与我最初写的不同：
SELECT J4.*, J3.Job_Seq_No
FROM (SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID
) AS J3
JOIN JobTitle AS J4
ON J3.BookingID = J4.BookingID
AND J3.JobTitleID = J4.JobTitleID
ORDER BY J4.BookingID, J3.Job_Seq_No;
SELECT R4.*, R3.Rec_Seq_No
FROM (SELECT R1.Receipt_No, R1.ID, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID >= R2.ID
GROUP BY R1.Receipt_No, R1.ID
) AS R3
JOIN First_Graphics_Debtors AS R4
ON R3.Receipt_No = R4.Receipt_No
AND R3.ID = R4.ID
ORDER BY R4.Receipt_No, R3.Rec_Seq_No;
然后你可以运行这个查询：
SELECT B.*, J.*, R.*
FROM Booking AS B
LEFT JOIN (SELECT J4.*, J3.Job_Seq_No
FROM (SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID
) AS J3
JOIN JobTitle AS J4
ON J3.BookingID = J4.BookingID
AND J3.JobTitleID = J4.JobTitleID
) AS J
ON B.BookingID = J.BookingID
LEFT JOIN (SELECT R4.*, R3.Rec_Seq_No
FROM (SELECT R1.Receipt_No, R1.ID, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID >= R2.ID
GROUP BY R1.Receipt_No, R1.ID
) AS R3
JOIN First_Graphics_Debtors AS R4
ON R3.Receipt_No = R4.Receipt_No
AND R3.ID = R4.ID
) AS R
ON B.Receipt_No = R.Receipt_No
WHERE B.Receipt_No = 753263343
AND J.Job_Seq_No = R.Rec_Seq_No
ORDER BY B.BookingID, NVL(J.Job_Seq_No, R.Rec_Seq_No);
不幸的是，它只在样本数据上产生三行;它省略了4个收据号码中的最后一个。这就是我做TDQD的原因;当我采取捷径时，我发现存在问题。
最后一个查询的结构需要使用FULL OUTER JOIN（通常）连接J和R子查询，因此它们需要共同的收据号或预订ID。我选择将收据编号添加到JobTitle信息中;将预订ID添加到First Graphics Debtors信息可以正常工作;对称性表明两者兼顾，但这也可能被视为过度杀伤。
SELECT B.Receipt_No, J4.*, J3.Job_Seq_No
FROM (SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID
) AS J3
JOIN JobTitle AS J4
ON J3.BookingID = J4.BookingID
AND J3.JobTitleID = J4.JobTitleID
JOIN Booking AS B
ON B.BookingID = J4.BookingID
ORDER BY J4.BookingID, J3.Job_Seq_No;
然后，将两个排序列表与FULL OUTER JOIN组合：
SELECT *
FROM (SELECT B.Receipt_No, J4.*, J3.Job_Seq_No
FROM (SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID
) AS J3
JOIN JobTitle AS J4
ON J3.BookingID = J4.BookingID
AND J3.JobTitleID = J4.JobTitleID
JOIN Booking AS B
ON B.BookingID = J4.BookingID
) AS J
FULL OUTER JOIN
(SELECT R4.*, R3.Rec_Seq_No
FROM (SELECT R1.Receipt_No, R1.ID, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID >= R2.ID
GROUP BY R1.Receipt_No, R1.ID
) AS R3
JOIN First_Graphics_Debtors AS R4
ON R3.Receipt_No = R4.Receipt_No
AND R3.ID = R4.ID
) AS R
ON J.Receipt_No = R.Receipt_No AND J.Job_Seq_No = R.Rec_Seq_No
ORDER BY NVL(J.Receipt_No, R.Receipt_No), NVL(J.Job_Seq_No, R.Rec_Seq_No);
最后将其嵌入到主查询中会给出（一个非常详细的版本）所需的结果：
SELECT B.*, JR.*
FROM Booking AS B
JOIN (SELECT NVL(J.Receipt_No, R.Receipt_No) AS Rec_No,
NVL(J.Job_Seq_No, R.Rec_Seq_No) AS Seq_No,
J.*, R.*
FROM (SELECT B.Receipt_No, J4.*, J3.Job_Seq_No
FROM (SELECT J1.BookingID, J1.JobTitleID, COUNT(*) AS Job_Seq_No
FROM JobTitle AS J1
JOIN JobTitle AS J2
ON J1.BookingID = J2.BookingID
WHERE J1.JobTitleID >= J2.JobTitleID
GROUP BY J1.BookingID, J1.JobTitleID
) AS J3
JOIN JobTitle AS J4
ON J3.BookingID = J4.BookingID
AND J3.JobTitleID = J4.JobTitleID
JOIN Booking AS B
ON B.BookingID = J4.BookingID
) AS J
FULL OUTER JOIN
(SELECT R4.*, R3.Rec_Seq_No
FROM (SELECT R1.Receipt_No, R1.ID, COUNT(*) AS Rec_Seq_No
FROM First_Graphics_Debtors AS R1
JOIN First_Graphics_Debtors AS R2
ON R1.Receipt_No = R2.Receipt_No
WHERE R1.ID >= R2.ID
GROUP BY R1.Receipt_No, R1.ID
) AS R3
JOIN First_Graphics_Debtors AS R4
ON R3.Receipt_No = R4.Receipt_No
AND R3.ID = R4.ID
) AS R
ON J.Receipt_No = R.Receipt_No AND J.Job_Seq_No = R.Rec_Seq_No
) AS JR
ON B.Receipt_No = JR.Rec_No
WHERE B.Receipt_No = 753263343
ORDER BY B.BookingID, Seq_No;
有一个非常有选择性的列列表：
SELECT B.BookingID AS Booking,
B.Receipt_No AS Receipt,
B.Client AS Client,
JR.JobTitleID AS Title
JR.JobTitle AS JobTitle,
JR.Old_Balance AS Old_Balance,
JR.New_Balance AS New_Balance
FROM Booking AS B
...
我从样本数据中获得的结果数据是：
booking receipt client title jobtitle old_balance new_balance
208 753263343 Chaka 460 A3 Full Colour 152161 0
208 753263343 Chaka 461 Flex Banner 152161 137637
208 753263343 Chaka 462 A2 Single Colour 152161 150161
208 753263343 Chaka 152161 151161
唯一剩下的问题是MySQL是否支持FULL OUTER JOIN表示法。如果没有，您可以对此样本数据使用RIGHT OUTER JOIN（作业标题信息中的行数少于FGD信息中的行数。）
更新：这是基于现在已知的错误查询。必须在此处进行类似的更改，以使其按要求运行。
使用OLAP函数，我认为您能够像这样创建排序结果集：
SELECT BookingID,
ROW_NUMBER() OVER (PARTITION BY BookingID ORDER BY JobTitleID) AS Job_Seq_No
FROM JobTitle;
SELECT Receipt_No,
ROW_NUMBER() OVER (PARTITION BY Receipt_No ORDER BY ID) AS Rec_Seq_No
FROM First_Graphics_Debtors;
但是，我对SQL的这一部分信心不足。但是，生成的大查询更简单：
SELECT B.*, J.*, R.*
FROM Booking AS B
LEFT JOIN (SELECT J1.*, J2.Job_Seq_No
FROM (SELECT BookingID,
ROW_NUMBER()
OVER (PARTITION BY BookingID ORDER BY JobTitleID)
AS Job_Seq_No
FROM JobTitle
) AS J2
JOIN JobTitle AS J1 ON J2.BookingID = J1.BookingID
) AS J
ON B.BookingID = J.BookingID
LEFT JOIN (SELECT R1.*, R2.Rec_Seq_No
FROM (SELECT Receipt_No,
ROW_NUMBER() OVER (PARTITION BY Receipt_No ORDER BY ID)
AS Rec_Seq_No
FROM First_Graphics_Debtors
) AS R2
JOIN First_Graphics_Debtors AS R1 ON R1.Receipt_No = R2.Receipt_No
) AS R
WHERE J.Job_Seq_No = R.Rec_Seq_No
AND B.Receipt_No = 753263343
ORDER BY B.BookingID, COALESCE(J2.Job_Seq_No, R2.Rec_Seq_No);
警告：更多未经测试的SQL！