Question

如何简化此代码？只需将整个Null列更新为具有相应ID的名称。

update salesman 
set name = 'James Hoog'
where salesman_id = 5001

update salesman
set name = 'Nail Knite' 
where salesman_id = 5002

update salesman
set name = 'Pit Alex'
where Salesman_id = 5005

update salesman
set name = 'Mc Lyon' 
where salesman_id = 5006

Answer 1

使用CASE语句标识每个NAME

的salesman_id

试试这种方式

UPDATE salesman 
SET    NAME = CASE salesman_id 
                WHEN 5001 THEN 'James Hoog' 
                WHEN 5002 THEN 'Nail Knite' 
                WHEN 5005 THEN 'Pit Alex' 
                WHEN 5006 THEN 'Mc Lyon' 
              END 
WHERE  salesman_id IN ( 5001, 5002, 5005, 5006 )

Answer 2

如果您在另一个表中有名称和ID，则执行UPDATE / JOIN

vi

Answer 3

尝试MERGE：

MERGE salesman AS t
USING (VALUES
    (5001, 'James Hoog'),
    (5002, 'Nail Knite'),
    (5005, 'Pit Alex'),
    (5006, 'Mc Lyon'))
    AS s (salesmain_id, name)
ON (t.salesman_id = s.salesman_id)
WHEN MATCHED THEN
    UPDATE SET name = s.name;

Answer 4

UPDATE salesman SET
      NAME = CASE WHEN salesman_id = 5001 THEN 'James Hoog' 
                  WHEN salesman_id = 5002 THEN 'Nail Knite' 
                  WHEN salesman_id = 5005 THEN 'Pit Alex' 
                  WHEN salesman_id = 5006 THEN 'Mc Lyon' END
WHERE salesman_id IN (5001,5002,5005,5006)

这可以简单说明吗？

4 个答案: