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

我以前使用过INSERT SELECT。对于那些想使用新东西的人来说,这里有一个类似的解决方案,但它要短得多:

UPDATE table1                                          // Table that's going to be updated.
LEFT JOIN                                              // Type of join.
    table2 AS tb2                                      // Second table and rename for easy.
ON
    tb2.filedToMatchTables = table1.fieldToMatchTables // Fields to connect both tables.
SET
    fieldFromTable1 = tb2.fieldFromTable2;             // Field to be updated on table1.

    field1FromTable1 = tb2.field1FromTable2,           // This is in the case you need to
    field1FromTable1 = tb2.field1FromTable2,           // update more than one field.
    field1FromTable1 = tb2.field1FromTable2;           // Remember to put ; at the end.

其他回答

UPDATE table AS a
INNER JOIN table2 AS b
ON a.col1 = b.col1
INNER JOIN ... AS ...
ON ... = ...
SET ...
WHERE ...

对于记录(以及其他像我一样的搜索),您可以在MySQL中这样做:

UPDATE first_table, second_table
SET first_table.color = second_table.color
WHERE first_table.id = second_table.foreign_id

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)

在SQL Server 2008(或更高版本)中,使用MERGE

MERGE INTO YourTable T
   USING other_table S 
      ON T.id = S.id
         AND S.tsql = 'cool'
WHEN MATCHED THEN
   UPDATE 
      SET col1 = S.col1, 
          col2 = S.col2;

或者:

MERGE INTO YourTable T
   USING (
          SELECT id, col1, col2 
            FROM other_table 
           WHERE tsql = 'cool'
         ) S
      ON T.id = S.id
WHEN MATCHED THEN
   UPDATE 
      SET col1 = S.col1, 
          col2 = S.col2;

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