2 
UPDATE Houses 
SET    lStatus = U.codesum
FROM   Houses H
JOIN   (SELECT ref, SUM(code) AS codesum
    FROM   Users
    GROUP  BY ref) AS U ON U.ref = H.ref

上面的代码获取每个房子的所有用户（Houses 表）。对所有用户的代码列（用户表）求和。最后更新房屋表的 lstatus 列中的结果。

我的问题是：我需要重写不是对代码列求和的查询。相反，我想创建案例陈述。例如：

tempvar = 0    //local variable might be required

For each user {

If code == 1 then tempvar += 5

else if code == 2 then tempvar += 10

etc

tempvar = 0;
}

一旦我们遍历了每个房子的所有用户，我们现在可以设置 lStatus = tempvar。然后应该将下一个房子的 tempvar 重置为 0。

4

3 回答 3

5

在编写 SQL 代码时，您应该尽量避免循环和其他过程结构。关系数据库不能轻易地优化这些东西，而且它们的执行速度通常比它们的声明性对应物慢几个数量级。SUM(code)在这种情况下，用您描述的CASE语句替换您似乎很简单：

UPDATE Houses 
SET    lStatus = U.codesum
FROM   Houses H
JOIN   (SELECT ref, SUM(CASE code WHEN 1 THEN 5 WHEN 2 THEN 10 ELSE 0 END) AS codesum
    FROM   Users
    GROUP  BY ref) AS U ON U.ref = H.ref

这样，SUM仍然可以处理您想象的临时变量将要执行的任务。

此外，如果您有很多案例，您可能会考虑将这些案例放在一个表格中，然后简单地加入该表格以获得您的总和。这可能会更好地维护。我在这里使用了一个表变量，但它可能如下所示：

DECLARE @codes TABLE (
    code INT NOT NULL PRIMARY KEY,
    value INT NOT NULL
)
INSERT INTO @codes SELECT 1, 5
INSERT INTO @codes SELECT 2, 10

UPDATE Houses 
SET    lStatus = U.codesum
FROM   Houses H
JOIN   (SELECT a.ref, SUM(b.value) AS codesum
    FROM   Users a
    JOIN   @codes b on a.code = b.code -- Here we get the values dynamically
    GROUP  BY a.ref) AS U ON U.ref = H.ref
于 2013-02-11T16:04:06.657 回答
0

试试这个：

UPDATE Houses 
SET    lStatus = U.codesum
FROM   Houses H
JOIN   (
             SELECT ref, SUM(
                                CASE
                                    WHEN Code = 1
                                    THEN 5
                                    WHEN Code = 2
                                    THEN 10
                                END
                            ) AS codesum
             FROM   Users
             GROUP  BY ref
       ) AS U ON U.ref = H.ref
于 2013-02-11T16:04:40.710 回答
0

试试这个 ：

UPDATE Houses 
SET    lStatus = U.code
FROM   Houses H
JOIN   (
             SELECT ref, SUM(
                                CASE
                                    WHEN Code = 1
                                    THEN 5
                                    WHEN Code = 2
                                    THEN 10
                                    ELSE 0
                                END
                            ) AS code
             FROM   Users
             GROUP  BY ref
       ) AS U ON U.ref = H.ref
于 2013-02-12T05:00:44.613 回答