我有2张桌子。
合法表，包含两个字段ARP和合法日期
+--------+-----------+ | ARP | legaldate | +--------+-----------+ | 71698 | 1/22/2016 | | 82501 | 4/28/2016 | | 103451 | 5/22/2016 | | 1111 | 1/2/2016 | +--------+-----------+
摄取表，包含3个字段的摄入量，ARP和摄入日期
+--------+-------+------------+ | Intake | ARP | IntakeDate | +--------+-------+------------+ | 615729 | 71698 | 12/9/2015 | | 615891 | 71698 | 12/10/2015 | | 620697 | 71698 | 1/19/2016 | | 621681 | 71698 | 1/26/2016 | | 621711 | 71698 | 1/26/2016 | | 630455 | 82501 | 4/28/2016 | | 634946 | 82501 | 3/30/2016 | | 123009 | 1111 | 9/2/2016 | +--------+-------+------------+
我加入两个表的ARP并查找legaldate和进入日期之间的日期差异（显示两个表格中的所有字段 - 使用查询1
我发现法定日期和入学日期之间的最小日期差异（仅显示'min diff'和'ARP'） - 使用查询2
我通过日期diff / mindatediff和ARP / ARP加入查询1和查询2以获得所需结果 - 查询3 （与两个表中的所有字段的最小日期差异）
最终显示我得到（在法定日期和进入日期之间具有最短日期差异的独特ARP）
+-------+-------------+-----------+----------+ | ARP | Intake Date | legaldate | DateDiff | +-------+-------------+-----------+----------+ | 71698 | 1/19/2016 | 1/22/2016 | 3 | | 82501 | 4/28/2016 | 4/28/2016 | 0 | | 1111 | 9/2/2016 | 1/2/2016 | 244 | +-------+-------------+-----------+----------+
我运行3个查询（我发现我也可以在两个查询中得到这些结果）。 我想要有效率，并在一个查询中得到欲望结果。
我的3个查询
1）使用'摄入风险'查询
查找日期差异SELECT Intake.Intake, legal.ARP, Intake.[Intake Date], legal.legaldate, Abs([legal]![legaldate]-[Intake]![Intake Date]) AS Diff
FROM Intake RIGHT JOIN legal ON Intake.ARP = legal.ARP;
+----------+-----------+-------------+-----------+------+ | Intake | ARP | Intake Date | legaldate | Diff | +----------+-----------+-------------+-----------+------+ | 621711 | 71698 | 1/26/2016 | 1/22/2016 | 4 | | 621681 | 71698 | 1/26/2016 | 1/22/2016 | 4 | | 620697 | 71698 | 1/19/2016 | 1/22/2016 | 3 | | 615891 | 71698 | 12/10/2015 | 1/22/2016 | 43 | | 615729 | 71698 | 12/9/2015 | 1/22/2016 | 44 | | 634946 | 82501 | 3/30/2016 | 4/28/2016 | 29 | | 630455 | 82501 | 4/28/2016 | 4/28/2016 | 0 | | Blank... | 103451 | 5/22/2016 | | | | 123009 | 1111 | 9/2/2016 | 1/2/2016 | 244 | +----------+-----------+-------------+-----------+------+
2）使用'Mindatediff'查询
查找最短日期差异SELECT DISTINCTROW [legal].ARP, Min(Abs([legal]!legaldate-Intake![Intake Date])) AS datediff
FROM Intake INNER JOIN legal ON Intake.ARP=[legal].ARP
GROUP BY [legal].ARP;
+-------+-----------+ | ARP | datediff | +-------+-----------+ | 1111 | 244 | | 71698 | 3 | | 82501 | 0 | +-------+-----------+
3）查询'Mindate-legalbegin'以获得最终结果。
SELECT [intake-risk].legal.ARP, [intake-risk].[Intake Date], [intake-risk].legaldate, Mindatediff.DateDiff
FROM Mindatediff INNER JOIN [intake-risk] ON (Mindatediff.ARP = [intake-risk].Intake.ARP) AND (Mindatediff.datediff = [intake-risk].diff);
+-------+-------------+-------------+-------------+ | ARP | Intake Date | legaldate | DateDiff | +-------+-------------+-------------+-------------+ | 71698 | 1/19/2016 | 1/22/2016 | 3 | | 82501 | 4/28/2016 | 4/28/2016 | 0 | | 1111 | 9/2/2016 | 1/2/2016 | 244 | +-------+-------------+-------------+-------------+
我想通过在两个表上运行一个查询来获得最终结果！ （在法定日期和进入日期之间具有最小日期差值的独特ARP）
由于
答案 0 :(得分：0)
Pd积，
你需要的是什么？
我不确定你是如何构建数据库的，所以我重新创建了它。
DROP DATABASE LegalDB;
CREATE DATABASE LegalDB;
SHOW DATABASES ;
USE LegalDB;
CREATE TABLE Legal (
legARP INT NOT NULL,
legLegalDate DATE NOT NULL,
PRIMARY KEY (legARP)
);
CREATE TABLE Intake(
intIntake INT NOT NULL,
intARP INT NOT NULL,
intIntakeDate DATE NOT NULL,
PRIMARY KEY (intIntake)
);
INSERT INTO Legal
(legARP, legLegalDate)
VALUES
(71698, '2016-01-22'),
(82501, '2016-04-28'),
(103451, '2016-05-22'),
(1111, '2016-01-02');
INSERT INTO Intake
(intIntake, intARP, intIntakeDate)
VALUES
(615729,71698 ,'2015-12-09'),
(615891, 71698, '2015-12-10'),
(620697, 71698 , '2016-01-19' ),
(621681, 71698 , '2016-01-26'),
(621711, 71698 , '2016-01-26'),
(630455, 82501, '2016-04-28'),
(634946, 82501 , '2016-03-30'),
(123009, 1111, '2016-09-02');
-- --------------------------------------
-- this query SHOWS +/- sign
-- --------------------------------------
SELECT * FROM (SELECT legARP,
(SELECT datediff(intIntakeDate, legLegalDate)) as dateDifference
FROM Legal
RIGHT JOIN Intake
ON intARP = legARP
ORDER BY legARP, ABS(dateDifference)) as selection
GROUP BY legARP;
This will produce the following:
+--------+----------------+
| legARP | dateDifference |
+--------+----------------+
| 1111 | 244 |
| 71698 | -3 |
| 82501 | 0 |
+--------+----------------+
-- --------------------------------------
-- this query DOES NOT show +/- sign
-- --------------------------------------
SELECT * FROM (SELECT legARP, (SELECT ABS(datediff(intIntakeDate, legLegalDate))) as dateDifference
FROM Legal
RIGHT JOIN Intake
ON intARP = legARP
ORDER BY legARP, dateDifference) AS SELECTION
GROUP BY legARP;
This will produce the following:
+--------+----------------+
| legARP | dateDifference |
+--------+----------------+
| 1111 | 244 |
| 71698 | 3 |
| 82501 | 0 |
+--------+----------------+
添加显示的日期进行编辑。
SELECT * FROM (SELECT legARP, intIntakeDate, legLegalDate,
(SELECT ABS(datediff(intIntakeDate, legLegalDate)))
as dateDifference
FROM Legal
RIGHT JOIN Intake
ON intARP = legARP
ORDER BY legARP, dateDifference) AS SELECTION
GROUP BY legARP;
这将产生所要求的确切输出：
+--------+---------------+--------------+----------------+
| legARP | intIntakeDate | legLegalDate | dateDifference |
+--------+---------------+--------------+----------------+
| 1111 | 2016-09-02 | 2016-01-02 | 244 |
| 71698 | 2016-01-19 | 2016-01-22 | 3 |
| 82501 | 2016-04-28 | 2016-04-28 | 0 |
+--------+---------------+--------------+----------------+
请注意，问题中显示的日期不是标准。 日期应存储为YYYY-MM-DD，如Mysql手册https://dev.mysql.com/doc/refman/5.7/en/datetime.html
中所述。答案 1 :(得分：0)
MySQL缺乏ROW_NUMBER等分析功能。这使得MySQL中的这些任务变得困难，而且查询速度相当慢。
我看到MySQL在一个查询中有两种方法可以做到这一点。其中哪一个更快取决于表格大小和其他一些东西。
1）最小日期差异是不存在较小日期差异的日期：
select
l.arp,
i.intakedate,
l.legaldate,
abs(datediff(l.legaldate, i.intakedate)) as datediff
from legal l
join intake i on i.arp = l.arp
where not exists
(
select *
from intake i2
where i2.arp = i.arp
and abs(datediff(l.legaldate, i2.intakedate)) < abs(datediff(l.legaldate, i.intakedate))
);
2）选择最小日期差异并使用它来再次访问该表：
select
agg.arp,
i2.intakedate,
agg.legaldate,
agg.datediff
from
(
select
l.arp,
l.legaldate,
min(abs(datediff(l.legaldate, i1.intakedate))) as datediff
from legal l
join intake i1 on i1.arp = l.arp
group by l.arp, l.legaldate
) agg
join intake i2 on i2.arp = agg.arp
and abs(datediff(agg.legaldate, i2.intakedate)) = agg.datediff;