Question

我有两张桌子：

人员表：

 ---------------------
|Person_ID | Phone_ID |
|---------------------|
|1234      | 12       |
|345       | 10       |
|43        | 33       |
|55        | 27       |
-----------------------

电话表：

 --------------------------
|Phone_ID  |  Phone_Number |
|--------------------------|
|  12      |  null         |
|  10      |  9876         |
|  33      |  9654         |
|  27      |  null         |
 --------------------------

执行查询时：

select t1.person_id, t2.phone_id 
from person t1
left join
phone t2
on t1.phone_id=t2.phone_id;

我得到了结果：

 --------------------------
| Person_ID | Phone_number |
|--------------------------|
| 1234      | null         |
| 345       | 9876         |
| 345       | null         |
| 43        | 9654         |
| 43        | null         |
| 55        | null         |
 --------------------------

如何消除已经拥有电话号码的人员ID获取null值的问题？

Answer 1

your query do not return that results.

很可能您的手机和人表格与预期不符。

您应该截断这些表并重新插入数据。然后重新执行您的查询。

 truncate table  Person ; 

truncate table  Phone ; 

 insert into person
 select 1234 ,12 from dual union all
 select 345, 10  from dual union all
 select 43  ,   33   from dual union all
 select 55 ,  27  from dual ;

 commit;

  insert into phone
 select 12 ,  null from dual union all
 select 10 ,  9876 from dual union all
 select  33  ,  9654 from dual union all
 select  27  ,null from dual ;


 commit;

 select t1.person_id, t2.phone_id 
    from person t1
    left join  phone t2
    on t1.phone_id=t2.phone_id;

Answer 2

您的查询运行良好，因为我已经检查过动态创建表，试试这个，直接在查询窗口中运行：

DECLARE @persons TABLE (
person_id INT,
phone_id INT )

DECLARE @phones TABLE (
phone_id INT,
phonenumber BIGINT )


INSERT INTO @persons VALUES(1234,12),(345,10),(43,33),(55,27)
INSERT INTO @phones VALUES(12,null),(10,9876),(33,9654),(27,null)

SELECT * FROM @persons 
SELECT * FROM @phones 

select t1.person_id, t2.phone_id 
from @persons t1
left join
@phones t2
on t1.phone_id=t2.phone_id;

如果您需要消除包含条目的null，您可以在最后使用条件：

WHERE 'column_name' IS NOT NULL

或者如果要将null替换为空白或某些值，可以使用：

ISNULL(column_name,'replacement_value') AS new_col_name

Answer 3

WITH actual_phones AS
     ( SELECT phone_id, phone_number
         FROM phone
        WHERE phone_number IS NOT NULL )
SELECT person_id, phone_number 
  FROM person NATURAL JOIN actual_phones
UNION
SELECT person_id, null AS phone_number 
  FROM person
 WHERE phone_id NOT IN ( SELECT phone_id
                            FROM actual_phones );

Left Join返回具有不同属性值的重复（两个）行（一列包含连接值，其他包含null）

3 个答案: