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
简单的方法是：
UPDATE
table_to_update,
table_info
SET
table_to_update.col1 = table_info.col1,
table_to_update.col2 = table_info.col2
WHERE
table_to_update.ID = table_info.ID
其他回答
UPDATE TQ
SET TQ.IsProcessed = 1, TQ.TextName = 'bla bla bla'
FROM TableQueue TQ
INNER JOIN TableComment TC ON TC.ID = TQ.TCID
WHERE TQ.IsProcessed = 0
要确保正在更新所需内容，请先选择
SELECT TQ.IsProcessed, 1 AS NewValue1, TQ.TextName, 'bla bla bla' AS NewValue2
FROM TableQueue TQ
INNER JOIN TableComment TC ON TC.ID = TQ.TCID
WHERE TQ.IsProcessed = 0
UPDATE
Table_A
SET
Table_A.col1 = Table_B.col1,
Table_A.col2 = Table_B.col2
FROM
Some_Table AS Table_A
INNER JOIN Other_Table AS Table_B
ON Table_A.id = Table_B.id
WHERE
Table_A.col3 = 'cool'
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）
Oracle SQL（使用别名）：
UPDATE Table T
SET T.col1 = (SELECT OT.col1 WHERE OT.id = T.id),
T.col2 = (SELECT OT.col2 WHERE OT.id = T.id);
SQLite3对我很有用，在INNER SELECT之后用SELECT更新行。
UPDATE clients
SET col1 = '2023-02-02 18:51:30.826621'
FROM (
SELECT * FROM clients dc WHERE dc.phone NOT IN (
SELECT do.phone FROM dclient_order do WHERE do.order_date > '2023-01-01' GROUP BY do.phone
)
) NewTable
WHERE clients.phone = NewTable.phone;
