内部JOIN返回重复的行

serverNewbie 发表于 Dev

服务器新手

我具有以下表结构，并且在每个表中都包含了主键和外键：

CREATE TABLE `Table1` (
`Table1_ID` int(6) ,
`Table2_FK` int(6) ,
 **Other Fields***
) 

CREATE TABLE `Table2` (
`Table2_ID` int(6) ,
`Table3_FK` int(11),
 **Other Fields***
) 
CREATE TABLE `Table3` (
`Table3_ID` int(6) ,
`Table2_FK` int(11),
 **Other Fields***
) 

CREATE TABLE `Table4` (
`Table4_ID` int(6) ,
`Table3_FK` int(11),
 **Other Fields***
) 

CREATE TABLE `Table5` (
`Table5_ID` int(6) ,
`Table4_FK` int(6),
 **Other Fields***
)

我已经设置了以下外键：

 ALTER TABLE `Table5`
 ADD CONSTRAINT `table5_ibfk_4` FOREIGN KEY (`Table4_FK `) REFERENCES   `Table4` (`Table4_ID`);

 ALTER TABLE `Table4`
 ADD CONSTRAINT `table4_ibfk_3` FOREIGN KEY (`Table3_FK `) REFERENCES `Table3` (`Table3_ID`);

 ALTER TABLE `Table1`
ADD CONSTRAINT `table1_ibfk_2` FOREIGN KEY (`Table2_FK `) REFERENCES     `Table2` (`Table2_ID `);

 ALTER TABLE `Table2`
 ADD CONSTRAINT `table2_ibfk_1` FOREIGN KEY (`Table3_FK`) REFERENCES `Table3` (`Table3_ID `);

我的问题是当我运行以下INNER JOIN查询时：

 SELECT *
 FROM `Table1 `
 INNER JOIN `Table2` ON `Table1`.`Table2_FK` =`Table2`.`Table2_ID` 
 INNER JOIN `Table3` ON `Table2`.`Table3_FK` = `Table3`.`Table3_ID` 
 INNER JOIN `Table4` ON `Table3`.`Table3_ID` = `Table4`.`Table3_FK ` 
 INNER JOIN `Table5` ON `Table4`.`Table4_ID` = `Table5`.`Table4_FK ` 
 WHERE (`Table1`.`Table1_ID ` ='43');

我希望将返回两行，因为只有两个记录的ID为43，如“ WHERE”子句中所述。相反，它返回ID为43的8条记录，我认为INNER Join仅会返回真实的结果，而不是所有结果。

更新

当前数据如下：

INSERT INTO `Table1` (`Table1_ID `, `OtherData`, `Table2_FK `, `OtherData2`, `Date`) VALUES
(42, 1, 1, 'New', '2015-03-10 17:41:50'),
(43, 1, 1, 'New', '2015-03-10 17:44:35'),
(44, 1, 1, 'New', '2015-03-10 17:50:34'),
(45, 1, 1, 'New', '2015-03-10 17:55:20'),
(46, 1, 1, 'New', '2015-03-10 18:10:47');

INSERT INTO `Table2` (`Table2_ID `, `OtherData3`, `OtherData4 `,     `OtherData5`, `OtherData6`) VALUES
(1, 'blahtype', NULL, 1, '2015-03-13 00:00:00');

INSERT INTO `Table3` (`Table3_ID `, `Table2_FK `, `OtherData6`) VALUES
(1, 1, 'blahname');

INSERT INTO `Table4` (`Table4_ID`, `Table3_FK `, `OtherData6`, `OtherData7`,     `OtherData7`) VALUES
(2, 1, 'blahfieldname', 'blahcont', 'blahtype'),
(3, 1, 'blahfieldname2', 'blahcont', 'blahtype');

INSERT INTO `Table5` (`Table5_ID `, `OtherData`, `Table4_FK`, `OtherData`) VALUES
(1, 'test2', 2, 42),
(2, 'test3', 3, 42),
(3, 'Test4', 2, 43),
(4, 'test5', 3, 43),
(5, 'test6', 2, 44),
(6, 'test7', 3, 44),
(9, 'test8', 2, 78),
(10, 'test9',3, 78);

电流输出为：

 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |2|1| blahfieldname | blahcont | blahtype |1|test2|2|42
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |2|1| blahfieldname | blahcont | blahtype |3|test3|2|43
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |2|1| blahfieldname | blahcont | blahtype |5|test4|2|44
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |2|1| blahfieldname | blahcont | blahtype |9|test5|2|78
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |3|1| blahfieldname2| blahcont | blahtype |2|test6|3|42
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |3|1| blahfieldname2| blahcont | blahtype |4|test7|3|43
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |3|1| blahfieldname2| blahcont | blahtype |6|test8|3|44
 |43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |3|1| blahfieldname2| blahcont | blahtype |10|test9|3|78

预期输出为：

|43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |2|1| blahfieldname | blahcont | blahtype |3|test3|2|43
|43|1|1|New|2015-03-10 17:44:35|1| blahtype |NULL|1|2015-03-13 00:00:00|1|1| blahname |3|1| blahfieldname2| blahcont | blahtype |4|test7|3|43

麦坎

您说在表1中有两条ID为43的记录。但这是在table2，table3 ... table5中引用的。

最后，在表1中显示与ID为43的这两行数据的每个关系。

table1
ID    name
1     T1-Firstrow
2     T1-Secondrow

table2
ID    FK    name
1     1     T2-Firstrow
2     1     T2-Secondrow
3     2     T2-Thirdrow

如果从表1中选择ID = 1，那么如果您加入表2，仍然会得到两行结果。

编辑：

使用您问题中的数据更新，选择ID 43：

table1 has 1 row matching
table2 has 1 row matching
table3 has 1 row matching
table4 has 2 rows matching
table5 has 8 rows matching

在table5中有两列称为“ otherdata”的列，但其中之一似乎是table1的FK。如果是这样，请使用以下命令：

SELECT *
 FROM `Table1 `
 INNER JOIN `Table2` ON `Table1`.`Table2_FK` =`Table2`.`Table2_ID` 
 INNER JOIN `Table3` ON `Table2`.`Table3_FK` = `Table3`.`Table3_ID` 
 INNER JOIN `Table4` ON `Table3`.`Table3_ID` = `Table4`.`Table3_FK` 
 INNER JOIN `Table5` ON `Table4`.`Table4_ID` = `Table5`.`Table4_FK` AND 
                        `Table5`.`OtherDataFK` = `Table1`.`Table1_ID`
 WHERE (`Table1`.`Table1_ID ` ='43');

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