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
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）
其他回答
单向
UPDATE t
SET t.col1 = o.col1,
t.col2 = o.col2
FROM
other_table o
JOIN
t ON t.id = o.id
WHERE
o.sql = 'cool'
UPDATE YourTable
SET Col1 = OtherTable.Col1,
Col2 = OtherTable.Col2
FROM (
SELECT ID, Col1, Col2
FROM other_table) AS OtherTable
WHERE
OtherTable.ID = YourTable.ID
选项1：使用内部联接：
UPDATE
A
SET
A.col1 = B.col1,
A.col2 = B.col2
FROM
Some_Table AS A
INNER JOIN Other_Table AS B
ON A.id = B.id
WHERE
A.col3 = 'cool'
选项2：共同相关子查询
UPDATE table
SET Col1 = B.Col1,
Col2 = B.Col2
FROM (
SELECT ID, Col1, Col2
FROM other_table) B
WHERE
B.ID = table.ID
通过CTE进行更新比此处的其他答案更具可读性：
;WITH cte
AS (SELECT col1,col2,id
FROM other_table
WHERE sql = 'cool')
UPDATE A
SET A.col1 = B.col1,
A.col2 = B.col2
FROM table A
INNER JOIN cte B
ON A.id = B.id
UPDATE table AS a
INNER JOIN table2 AS b
ON a.col1 = b.col1
INNER JOIN ... AS ...
ON ... = ...
SET ...
WHERE ...
