在SQL Server中，可以使用insert将行插入到表中。。SELECT语句：
INSERT INTO Table (col1, col2, col3)
SELECT col1, col2, col3
FROM other_table
WHERE sql = 'cool'
是否也可以使用SELECT更新表？我有一个包含这些值的临时表，并希望使用这些值更新另一个表。也许是这样的：
UPDATE Table SET col1, col2
SELECT col1, col2
FROM other_table
WHERE sql = 'cool'
WHERE Table.id = other_table.id
当前回答
另一种方法是使用派生表：
UPDATE t
SET t.col1 = a.col1
,t.col2 = a.col2
FROM (
SELECT id, col1, col2 FROM @tbl2) a
INNER JOIN @tbl1 t ON t.id = a.id
样本数据
DECLARE @tbl1 TABLE (id INT, col1 VARCHAR(10), col2 VARCHAR(10))
DECLARE @tbl2 TABLE (id INT, col1 VARCHAR(10), col2 VARCHAR(10))
INSERT @tbl1 SELECT 1, 'a', 'b' UNION SELECT 2, 'b', 'c'
INSERT @tbl2 SELECT 1, '1', '2' UNION SELECT 2, '3', '4'
UPDATE t
SET t.col1 = a.col1
,t.col2 = a.col2
FROM (
SELECT id, col1, col2 FROM @tbl2) a
INNER JOIN @tbl1 t ON t.id = a.id
SELECT * FROM @tbl1
SELECT * FROM @tbl2
其他回答
Use:
drop table uno
drop table dos
create table uno
(
uid int,
col1 char(1),
col2 char(2)
)
create table dos
(
did int,
col1 char(1),
col2 char(2),
[sql] char(4)
)
insert into uno(uid) values (1)
insert into uno(uid) values (2)
insert into dos values (1,'a','b',null)
insert into dos values (2,'c','d','cool')
select * from uno
select * from dos
或者：
update uno set col1 = (select col1 from dos where uid = did and [sql]='cool'),
col2 = (select col2 from dos where uid = did and [sql]='cool')
OR:
update uno set col1=d.col1,col2=d.col2 from uno
inner join dos d on uid=did where [sql]='cool'
select * from uno
select * from dos
如果两个表中的ID列名相同，则只需将表名放在要更新的表之前，并为所选表使用别名，即：
update uno set col1 = (select col1 from dos d where uno.[id] = d.[id] and [sql]='cool'),
col2 = (select col2 from dos d where uno.[id] = d.[id] and [sql]='cool')
下面是另一个有用的语法：
UPDATE suppliers
SET supplier_name = (SELECT customers.name
FROM customers
WHERE customers.customer_id = suppliers.supplier_id)
WHERE EXISTS (SELECT customers.name
FROM customers
WHERE customers.customer_id = suppliers.supplier_id);
它使用“WHERE EXIST”检查它是否为空。
如果使用MySQL而不是SQL Server，语法为：
UPDATE Table1
INNER JOIN Table2
ON Table1.id = Table2.id
SET Table1.col1 = Table2.col1,
Table1.col2 = Table2.col2
像其他人一样，必须指出MySQL或MariaDB使用不同的语法。它还支持非常方便的USING语法（与T/SQL相反）。INNER JOIN也是JOIN的同义词。因此，原始问题中的查询最好在MySQL中实现：
UPDATE
Some_Table AS Table_A
JOIN
Other_Table AS Table_B USING(id)
SET
Table_A.col1 = Table_B.col1,
Table_A.col2 = Table_B.col2
WHERE
Table_A.col3 = 'cool'
我在其他答案中没有看到问题的答案，因此我的两分钱。（在PHP 7.4.0 MariaDB 10.4.10上测试）
使用别名：
UPDATE t
SET t.col1 = o.col1
FROM table1 AS t
INNER JOIN
table2 AS o
ON t.id = o.id
