从存储过程中的两个动态创建的表构造结果集 [英] Constructing a result set from two dynamically created tables in a stored procedure

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问题描述

大家好,

我正在存储过程中动态填写两个表,并希望从每个表中提取出我想要的字段,这些字段将作为结果集成为一个表.

从结构上讲,存储过程类似于以下伪代码:

Hallo everybody,

I am dynamically filling out two tables in my stored procedure and wished to have extract out the fields those I want from each table which will become a table as a result set.

Structurally the stored procedure looks like the following pseudocode:

Declare Table_1 Table (F1, ...)
Insert InTo Table_1 Select * From X.Table

Declare Table_2 Table (F1, ...)
Insert InTo Table_2 Select * From Y.Table

Select * From Table_1, Table_2



现在,这里的最大问题是,即使其中有一个有效值,select语句也无法获取第二个表中找到的字段,这仅在第一个表中没有值时才会发生.换句话说,如果第一个表具有值,那么select语句将返回结果集;否则,如果第一个表没有值,但如果第二个表具有值,则select语句将不返回结果集.

顺便说一句,我正在使用SQL2005服务器.

如果我不清楚,请告诉我.



Now, the big problem here is that the select statement cannot fetch out fields those are found in the second table even if there is a valid value in it this happens only when there is no value in the first table. In other words, if the first table has value then the select statement returns a result set otherwise if the first table doesn''t have value but if the second table has then the select statement is not returning a result set.

By the way, I am using SQL2005 server.

If I am not clear, please let me know.

Thanks

推荐答案

如果表的列结构相同(甚至相似),则可以在两个表之间进行UNION ,并且应该可以返回所有表结果在一起.

如果表结构完全不同(或者您不想执行UNION),则可以创建一个临时表,并将第一个和第二个表的值插入其中.

select 并返回此临时表中的所有值.
If the table column structures are same (or even similar), you can do a UNION between the two tables and you should be able to return all the results together.

If the table structures are drastically different (or you do not want to do a UNION), you can create a temporary table and insert the values of both the first and the second tables into it.

The select and return all the values from this temporary table.


您可以使用 ^ ],当第一个表中没有链接记录时,也可以从第二个表中获取记录.
You could use a FULL JOIN[^] to also get the records from the second table when no linked record exists with the first.


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