Sql server Join没有返回正确的结果

我的SQL连接未返回所需的输出

表格：[访问数据]

[PAT_ID]
[VISITDATE]
[TBLE_NAME]
[CODE_ID]

[env_ID]
[NAME]

表格：[env_lkup]

[PAT_ID]
[VISITDATE]
[TBLE_NAME]
[CODE_ID]

[env_ID]
[NAME]

所需输出：

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4       dusty air
AAA000  1995-05-02  3       secondary cigarette smoke
AAA001  1999-07-07  1       air conditioning

  SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM [visit_data] v
  LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup'

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4   dusty air
AAA000  1995-05-02  3   secondary cigarette smoke
AAA001  1999-07-07  1   air conditioning

SQL:

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4       dusty air
AAA000  1995-05-02  3       secondary cigarette smoke
AAA001  1999-07-07  1       air conditioning

  SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM [visit_data] v
  LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup'

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4   dusty air
AAA000  1995-05-02  3   secondary cigarette smoke
AAA001  1999-07-07  1   air conditioning

结果输出：

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4       dusty air
AAA000  1995-05-02  3       secondary cigarette smoke
AAA001  1999-07-07  1       air conditioning

  SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM [visit_data] v
  LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup'

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4   dusty air
AAA000  1995-05-02  3   secondary cigarette smoke
AAA001  1999-07-07  1   air conditioning

---

有人能帮忙吗？

查看加入列的数据

LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID

如果ID列都是varchar（），并且其中一个ID列在值中有一个前导空格，则我可以复制结果的唯一方法

当其中一个id列为int时，SQLServer进行隐式转换并返回数据

表达式中的类型转换（CONVERT_IMPLICIT（int[e].[env_ID]，0）） 可能会影响查询计划选择、类型转换中的“CardinalityEstimate” 在表达式中（CONVERT_IMPLICIT（int[e].[env_ID]，0）=[v].[CODE_ID]）可以 影响查询计划选择中的“SeekPlan”

我是如何重现不期望的结果的：

Declare @visit_data TABLE 
    ([PAT_ID] varchar(6), [VISITDATE] date, [TBLE_NAME] varchar(11), [CODE_ID] varchar(1));
    
INSERT INTO @visit_data
    ([PAT_ID], [VISITDATE], [TBLE_NAME], [CODE_ID])
VALUES
    ('AAA000', '1995-05-02', 'COMSYS_lkup', '1'),
    ('AAA000', '1995-05-02', 'env_lkup', '4'),
    ('AAA000', '1995-05-02', 'env_lkup', '3'),
    ('AAA001', '1999-07-07', 'COMSYS_lkup', '5'),
    ('AAA001', '1999-07-07', 'env_lkup', '1');

Declare @env_lkup TABLE
    ([env_ID] varchar(2), [NAME] varchar(25));
    
INSERT INTO @env_lkup
    ([env_ID], [NAME])
VALUES
    (' 1', 'air conditioning'),
    (' 2', 'pollution'),
    (' 3', 'secondary cigarette smoke'),
    (' 4', 'dusty air');

SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM @visit_data v
  LEFT JOIN @env_lkup e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup';

INSERT INTO @env_lkup
    ([env_ID], [NAME])
VALUES
    ('1 ', 'air conditioning'),
    ('2 ', 'pollution'),
    ('3 ', 'secondary cigarette smoke'),
    ('4 ', 'dusty air');

结果（前导空格）：

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4       dusty air
AAA000  1995-05-02  3       secondary cigarette smoke
AAA001  1999-07-07  1       air conditioning

  SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM [visit_data] v
  LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup'

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4   dusty air
AAA000  1995-05-02  3   secondary cigarette smoke
AAA001  1999-07-07  1   air conditioning

注意：如果空格在后面，您将获得所需的结果：

Declare @visit_data TABLE 
    ([PAT_ID] varchar(6), [VISITDATE] date, [TBLE_NAME] varchar(11), [CODE_ID] varchar(1));
    
INSERT INTO @visit_data
    ([PAT_ID], [VISITDATE], [TBLE_NAME], [CODE_ID])
VALUES
    ('AAA000', '1995-05-02', 'COMSYS_lkup', '1'),
    ('AAA000', '1995-05-02', 'env_lkup', '4'),
    ('AAA000', '1995-05-02', 'env_lkup', '3'),
    ('AAA001', '1999-07-07', 'COMSYS_lkup', '5'),
    ('AAA001', '1999-07-07', 'env_lkup', '1');

Declare @env_lkup TABLE
    ([env_ID] varchar(2), [NAME] varchar(25));
    
INSERT INTO @env_lkup
    ([env_ID], [NAME])
VALUES
    (' 1', 'air conditioning'),
    (' 2', 'pollution'),
    (' 3', 'secondary cigarette smoke'),
    (' 4', 'dusty air');

SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM @visit_data v
  LEFT JOIN @env_lkup e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup';

INSERT INTO @env_lkup
    ([env_ID], [NAME])
VALUES
    ('1 ', 'air conditioning'),
    ('2 ', 'pollution'),
    ('3 ', 'secondary cigarette smoke'),
    ('4 ', 'dusty air');

结果（尾随空格）：

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4       dusty air
AAA000  1995-05-02  3       secondary cigarette smoke
AAA001  1999-07-07  1       air conditioning

  SELECT v.[PAT_ID]   
       , v.[VISITDATE]   
       , e.[env_ID]    
       , e.[NAME]   
  FROM [visit_data] v
  LEFT JOIN [env_lkup] e ON e.env_ID = v.CODE_ID
  WHERE v.TBLE_NAME = 'env_lkup'

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  NULL    NULL
AAA000  1995-05-02  NULL    NULL
AAA001  1999-07-07  NULL    NULL

PAT_ID  VISITDATE   env_ID  NAME
AAA000  1995-05-02  4   dusty air
AAA000  1995-05-02  3   secondary cigarette smoke
AAA001  1999-07-07  1   air conditioning

---

中的
d
是什么，其中d.TBLE='env_lkup'

？我只在

FROM

子句中看到表别名

v

和

e

。至于你的

ON

条款。您是否有一个值为

4

的

code\u ID

与

env\u ID

表中的

env\u ID

相对应？共享这两个表中与所需结果/输出匹配的一些示例数据会有所帮助。visit_表中的CODE_ID不等于4，因此这两个表不会像您认为的那样连接。我更正了别名，这是

v

中的

其中v.TBLE_NAME='env_lkup'

您的代码返回您的预期结果：您需要发布完整的表结构，包括数据类型，以防数据类型不匹配。将id列强制转换为int修复了强制转换时的

（e.ENVIRON_id as int）=强制转换（v.code_id as int）

@charlescaper，如果这些列总是要存储int，那么可以将列设置为int而不是强制转换为int吗？对联接进行强制转换会影响性能，这意味着表中的每一行，如果有许多行，则转换速度会很慢。使用强制转换：表达式中的类型转换（CONVERT（int[e].[env_ID]，0））可能会影响查询计划选择中的“CardinalityEstimate”，表达式中的类型转换（CONVERT（int[v].[CODE_ID]，0））可能会影响查询计划选择中的“CardinalityEstimate”，表达式中的类型转换（CONVERT（int[e].[env_ID]，0）=[Expr1005]）可能会影响查询计划选择中的“SeekPlan”