我有一个放射学报告数据库，我已经为肺结节事件开采了。每位患者都有一个病历号，每个程序都有一个唯一的登记号。因此，MRN可以具有多个用于差异过程的登录号。入藏号码是递增的，因此如果患者有多个入藏号，则最大的入藏号是最新的过程。我需要：
我相信使用相关子查询可以解决这个问题。但是，我还不熟悉SQL来解决这个问题。我尝试自己加入表并找到每个子查询的最大加入。下面的一些示例代码用于制作数据集：
CREATE TABLE Stack_Example (Rank, Accession1, MRN1, Textbox2, Textbox47,Textbox43,Textbox45,ReadBy,SignedBy,Addendum1,ReadDate,SignedDate,Textbox49,Result,Impression,max_size_nodule, max_nodule_loc, max_nodule_type)
INSERT INTO Stack_Example
VALUES ("10", "33399", "001734", "5/21/1965", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "COMB, YAHAIR", "YES", "12/19/2014 11:48", "12/19/2014 17:50", "TEXT", "Results of Nodules!","Impressions of Nodules","3.0", "right middle lobe","None Found")
INSERT INTO Stack_Example
VALUES ("9", "33104", "001734", "5/21/1965", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "12/21/2013 06:52", "01/21/2014 06:52", "TEXT", "Results of Nodules!","Impressions of Nodules","3.7", "right upper lobe","None Found")
INSERT INTO Stack_Example
VALUES ("9", "33374", "001734", "5/21/1965", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "01/21/2014 08:19", "01/21/2014 06:52", "TEXT", "Results of Nodules!","Impressions of Nodules","2.1", "right lower lobe","None Found")
INSERT INTO Stack_Example
VALUES ("1", "34453", "001734", "5/21/1965", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "03/14/2014 09:14", "03/14/2014 09:14", "TEXT", "Results of Nodules!","Impressions of Nodules","1.4", "left upper lobe","None Found")
INSERT INTO Stack_Example
VALUES ("1", "27122", "80592", "1/14/1984", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "06/26/2013 10:20", "06/26/2013 10:20", "TEXT", "Results of Nodules!","Impressions of Nodules","2.5", "left upper lobe","None Found")
INSERT INTO Stack_Example
VALUES ("1", "27248", "80592", "1/14/1984", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "08/01/2013 06:23", "08/01/2013 06:23", "TEXT", "Results of Nodules!","Impressions of Nodules","4.0", "left lower lobe","None Found")
INSERT INTO Stack_Example
VALUES ("1", "28153", "35681", "03/01/1990", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "09/14/2012 05:00", "09/14/2012 05:00", "TEXT", "Results of Nodules!","Impressions of Nodules","4.0", "left lower lobe","None Found")
INSERT INTO Stack_Example
VALUES ("1", "29007", "35681", "03/01/1990", "CTS", "3341", "ROUTINE", "TUCK, YOURPANTSIN", "PICK, YASELFUP", "YES", "11/16/2012 08:23", "11/16/2012 08:23", "TEXT", "Results of Nodules!","Impressions of Nodules","3.5", "right lower lobe","None Found")
显然这是假数据。我一直在尝试做的是用相关的子查询加入表格。像这样：
SELECT DISTINCT a.Accession1, a.MRN1, a.ReadDate, p.Accession1, p.ReadDate
FROM Stack_Example as a
INNER JOIN Stack_Example as p on a.MRN1 = p.MRN1
WHERE a.Accession1 =
(SELECT max(Accession1)
FROM Stack_Example as b
WHERE a.MRN1 = b.MRN1 AND
a.Accession1 != p. Accession1)
ORDER BY a.MRN1
理想情况下，我想要的是一个主表，每个患者在行和每个MRN作为列的加入时有一个MRN（除了加入的日期等）。像这样：
| MRN | Accession (First Follow-up) | Date First Followup |Accession (Second Follow-up)..| Date Second Follow up | etc.
|:-----------|----------------------------:|:-------------------:|
| 001734 | 33374 | ......
| 80592 | 27248 | ......
我相信我的子查询需要一系列左连接;但是，有更好的方法吗？一些患者有7次以上的随访。感谢任何帮助，并为长时间的解释感到抱歉。希望格式化没问题。
答案 0 :(得分：1)
你走在正确的轨道上。你可以使用自联接和子查询来完成它。该表应在MRN1上与其自身连接，并且后一记录的Accession1等于该MRN1的最小Accession1，其大于第一记录的MRN1（下一个MRN1）。左连接允许查询报告所有记录，甚至是最后一个（没有后继记录）。
此查询生成所有相邻研究对：
Select a.ReadDate ARead, b.ReadDate BRead,
b.ReadDate-A.ReadDate elapsed,
a.*, b.*,
From table a
left Join table b
on b.MRN1 = a.MRN1
and b.Accession1 =
(Select min(Accession1) From table
where MRN1 = a.MRN1
and Accession1 > a.Accession1)
此查询生成前三个研究：
Select a.ReadDate ARead, b.ReadDate BRead, c.ReadDate CRead,
b.ReadDate-A.ReadDate elapsedAB,
c.ReadDate-b.ReadDate elapsedBCB
From table a
left Join table b
on b.MRN1 = a.MRN1
and b.Accession1 =
(Select min(Accession1) From table
where MRN1 = a.MRN1
and Accession1 > a.Accession1)
left Join table c
on c.MRN1 = a.MRN1
and c.Accession1 =
(Select min(Accession1) From table
where MRN1 = a.MRN1
and Accession1 > b.Accession1)
Where A.ReadDate =
(Select Min(readDate) from table
where MRN1 = a.MRN1)
答案 1 :(得分：0)
不确定您是想要所有范围还是前两个范围。查尔斯查询我相信提供所有。
这只是前两个。
SELECT *
FROM YourTable as O -- oldest
LEFT JOIN YourTable as neO -- next oldest
ON O.MRN1 = neO.MRN1
WHERE
O.Accession1 = (SELECT MIN(Accession1)
FROM YourTable A
WHERE A.MRN1 = O.MRN1)
AND neO.Accession1 = (SELECT MIN(Accession1)
FROM YourTable A
WHERE A.MRN1 = O.MRN1
AND A.Accession1 <> O.Accession1)
答案 2 :(得分：0)
您可能希望发布minimal working example，您的示例包含许多不需要的列，并使事情变得复杂。
以下架构在SQL Fiddle上，请参阅下文。我将Accession 34453的年份更改为2015，加入顺序和日期错误。
CREATE TABLE Stack_Example (
Accession VARCHAR(32),
MRN VARCHAR(32),
ReadDate DATETIME
);
INSERT INTO Stack_Example
VALUES ("33399", "001734", STR_TO_DATE("12/19/2014 11:48", "%m/%d/%Y %h:%i" )),
("33104", "001734", STR_TO_DATE("12/21/2013 06:52", "%m/%d/%Y %h:%i" )),
("33374", "001734", STR_TO_DATE("01/21/2014 08:19", "%m/%d/%Y %h:%i" )),
("34453", "001734", STR_TO_DATE("03/14/2015 09:14", "%m/%d/%Y %h:%i" )),
("27122", "80592", STR_TO_DATE("06/26/2013 10:20", "%m/%d/%Y %h:%i" )),
("27248", "80592", STR_TO_DATE("08/01/2013 06:23", "%m/%d/%Y %h:%i" )),
("28153", "35681", STR_TO_DATE("09/14/2012 05:00", "%m/%d/%Y %h:%i" )),
("29007", "35681", STR_TO_DATE("11/16/2012 08:23", "%m/%d/%Y %h:%i" ));
群组Concat所有以前的种质
您似乎希望拥有可变数量的列，或者动态创建这些列。据我所知，这不起作用。正如在其他答案中提出的那样，您必须为每列添加LEFT JOIN。但是，在MySQL中，您可以使用GROUP_CONCAT来连接组的值。您的值将不再位于单个列中，但结果可能接近您的预期。
接下来为了生成两个后续日期之间的差异，在PostgreSQL中你有窗口函数来实现这一点。在MySQL中，您可以使用nested sets or adjacent lists。
SELECT S.MRN,
GROUP_CONCAT( 'Acc: ', S.Accession,
' Date: ', S.ReadDate,
' Days to prev.: ', IFNULL(Diff, 0)
ORDER BY Accession SEPARATOR ' :: ' )
FROM (
SELECT S0.MRN,
S0.Accession,
S0.ReadDate,
TIMESTAMPDIFF(DAY, S1.ReadDate, S0.ReadDate) AS Diff
FROM stack_example S0
-- join on previous accession
LEFT JOIN stack_example S1
ON S1.MRN = S0.MRN
AND S1.Accession = ( SELECT MAX(S2.Accession)
FROM stack_example S2
WHERE S2.MRN = S0.MRN
AND S2.Accession < S0.Accession )
) S
GROUP BY MRN;
可能接近你要找的东西，Results在SQL小提琴上。
| MRN | Result |
|--------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| 001734 | Acc: 33104 Date: 2013-12-21 06:52:00 Days to prev.: 0 :: Acc: 33374 Date: 2014-01-21 08:19:00 Days to prev.: 31 :: Acc: 33399 Date: 2014-12-19 11:48:00 Days to prev.: 332 :: Acc: 34453 Date: 2015-03-14 09:14:00 Days to prev.: 84 |
| 35681 | Acc: 28153 Date: 2012-09-14 05:00:00 Days to prev.: 0 :: Acc: 29007 Date: 2012-11-16 08:23:00 Days to prev.: 63 |
| 80592 | Acc: 27122 Date: 2013-06-26 10:20:00 Days to prev.: 0 :: Acc: 27248 Date: 2013-08-01 06:23:00 Days to prev.: 35 |
加入固定数量的先前种质
以下查询与Charles Bretana已发布的查询相同。它加入了固定数量的种质。这个查询的缺点是，您不会获得最新的种质，但是最早的种类是七/四种。
SELECT S0.MRN,
S0.Accession, S0.ReadDate,
0,
S1.Accession, S1.ReadDate,
TIMESTAMPDIFF(DAY, S0.ReadDate, S1.ReadDate),
S2.Accession, S2.ReadDate,
TIMESTAMPDIFF(DAY, S1.ReadDate, S2.ReadDate),
S3.Accession, S3.ReadDate,
TIMESTAMPDIFF(DAY, S2.ReadDate, S3.ReadDate)
FROM stack_example S0
LEFT JOIN stack_example S1
ON S1.MRN = S0.MRN
AND S1.Accession = ( SELECT MIN(SX.Accession)
FROM stack_example SX
WHERE SX.MRN = S0.MRN
AND SX.Accession > S0.Accession )
LEFT JOIN stack_example S2
ON S2.MRN = S0.MRN
AND S2.Accession = ( SELECT MIN(SX.Accession)
FROM stack_example SX
WHERE SX.MRN = S1.MRN
AND SX.Accession > S1.Accession )
LEFT JOIN stack_example S3
ON S3.MRN = S0.MRN
AND S3.Accession = ( SELECT MIN(SX.Accession)
FROM stack_example SX
WHERE SX.MRN = S2.MRN
AND SX.Accession > S2.Accession )
WHERE S0.Accession = ( SELECT MIN(SX.Accession)
FROM stack_example SX
WHERE SX.MRN = S0.MRN )
;
结果
| MRN | Accession | ReadDate | 0 | Accession | ReadDate | TIMESTAMPDIFF | Accession | ReadDate | TIMESTAMPDIFF | Accession | ReadDate | TIMESTAMPDIFF |
|--------|-----------|-----------------------------|---|-----------|----------------------------|---------------|-----------|----------------------------|---------------|-----------|-------------------------|---------------|
| 001734 | 33104 | December, 21 2013 06:52:00 | 0 | 33374 | January, 21 2014 08:19:00 | 31 | 33399 | December, 19 2014 11:48:00 | 332 | 34453 | March, 14 2015 09:14:00 | 84 |
| 80592 | 27122 | June, 26 2013 10:20:00 | 0 | 27248 | August, 01 2013 06:23:00 | 35 | (null) | (null) | (null) | (null) | (null) | (null) |
| 35681 | 28153 | September, 14 2012 05:00:00 | 0 | 29007 | November, 16 2012 08:23:00 | 63 | (null) | (null) | (null) | (null) | (null) | (null) |