Question

表1

表2

我想显示table1的所有行，并像这样匹配table2的行

输出表

ID status

001 1 - Found
002 - 0
003 -0

1 - 发现 0 - 未发现

如何在SQL查询中实现

Answer 1

select table1.id, case when table2.id is not null 
                       then 'found' 
                  end as status
from table1
left join table2 on table1.id = table2.id

Answer 2

如果第二个表格可能有重复项，那么exists和case可能会更好：

select t1.*,
       (case when exists (select 1 from table2 t2 where t1.id = t2.id)
             then 'Found'
             else 'Not Found'
        end)
from table1 t1;

Answer 3

如何使用LEFT JOIN和CASE？

像

这样的东西

SELECT  t1.*, 
        CASE 
            WHEN t2.ID IS NOT NULL 
                THEN 'FOUND' 
            ELSE 'NOT FOUND' 
        END [Status]
FROM    Table1 t1 LEFT JOIN 
        Table2 t2 ON t1.ID = t2.ID

Answer 4

使用LEFT JOIN：

SELECT T1.Id, CASE WHEN T2.id IS NULL THEN 0 ELSE 1 END AS Status
FROM Table1 T1 LEFT JOIN
     Table2 T2 ON T1.Id=T2.Id

结果：

ID  STATUS
1   1
2   0
3   0

结果为SQL Fiddle。

如果您想显示Found和Not Found：

SELECT T1.Id, CASE WHEN T2.id IS NULL THEN 'Not Found' ELSE 'Found' END AS Status
FROM Table1 T1 LEFT JOIN
     Table2 T2 ON T1.Id=T2.Id

结果：

ID  STATUS
1   Found
2   Not Found
3   Not Found

小提琴示例here。

Answer 5

你可以在以下情况下使用案例：

SELECT
    Table1.id,
    CASE WHEN
        Table2.id IS NOT NULL THEN 1 ELSE 0
    END
FROM
    Table1
LEFT JOIN Table2 ON Table1.id = Table2.id

如何在更新时匹配行

5 个答案: