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

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。

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

从select语句更新的另一种方法:

UPDATE A
SET A.col = A.col,B.col1 = B.col1
FROM  first_Table AS A
INNER JOIN second_Table AS B  ON A.id = B.id WHERE A.col2 = 'cool'

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

以下解决方案适用于MySQL数据库:

UPDATE table1 a , table2 b 
SET a.columname = 'some value' 
WHERE b.columnname IS NULL ;