在表格中添加动态列.... [英] adding a dynamic column in a table....
问题描述
有没有办法在运行时向表中添加列。这假设我有一个包含4列的表...我已经为insert，del，upd操作编写了一个存储过程。现在我想要的是不使用alter语句，当我自动编写选择查询时，我得到一个额外的列，它应该具有对应empid的价值。对于empid 100,102我得到第一个，第一个等等...
试了一下，但是进了两天......请帮我出去。提前谢谢。
先生我有一个表有4列名为empid，emp_name，emp_ADD，emp_Phno.now的表我创建了一个插入，删除，更新，搜索操作的过程。我的表格如下：
is there any way to add a column to a table during runtime. That is suppose i have a table containing 4 columns...i have written a stored procedure for insert,del,upd operations also. now what i want is without using alter statement when i write select query automatically i get a additional column and it should have values corresponding to empid..ie for empid 100,102 i get first,first and so on...
tried it but struck in it for two days ...help me out.thanks in advance.
sir i have a table that have 4 columns named empid,emp_name,emp_ADD,emp_Phno.now i have created a procedure to insert,delete,update,search operation. my table is as follows:
empid emp_name Add emp_phno
100 rachit 516-a 8696953366
101 ush 25/a 7737708131
105 unnet sfgcsy 986547896
104 bobby 76A 7899877899
103 dharam 2519 7765477654
102 kishor 77/A 8989689896
现在不使用任何改变命令..我想在使用select * from employee
时想要这个
Now without using any alter command ..i want this when use select * from employee
empid emp_name Add emp_phno grp_emp
100 rachit 516-a 8696953366 first
101 ush 25/a 7737708131 first
105 unnet sfgcsy 986547896 second
104 bobby 76A 7899877899 second
103 dharam 2519 7765477654 third
102 kishor 77/A 8989689896 third
i想要在运行时添加最后一列而不使用alter语句。我认为这对你的sir.thanks来说已经足够了。 />
[/ EDIT]
i want to add this last column on runtime without using alter statement.I think this is sufficient for u sir.thanks.
[/EDIT]
推荐答案
select empid, (case when empid between 1 and 2 then 'first' when empid=3 then 'second' else 'other' end ) as groupname from employee_test
这也可以尝试。这两个解决方案对我有用。
This can be tried also. Both solutions worked for me.
我不确定你想要实现什么...
I''m not sure what you want to achieve...
根据您希望在群组中划分员工的条件？
Based on which condition do you want to divide employees on groups?
任何条件先生。我没有得到解决这个问题的关键。你可以指导我接近这个和这类问题。谢谢
any condition sir. i am not getting key to solve this question. could you please guide me to approach this and this type of questions.thanks
任何的例子（其中任何表示： empid modulo 2 ）对员工进行分组的条件：
Example for "any" (where "any" means: empid modulo 2)condition to group employees:
SELECT empid, emp_name, [Add], emp_phno, CONVERT(INT, (empid % 2) +1 ) AS grp_emp
FROM @emp
ORDER BY CONVERT(INT, (empid % 2) +1 )
结果：
Result:
100 rachit 516-a 8696953366 1
104 bobby 76A 7899877899 1
102 kishor 77/A 8989689896 1
103 dharam 2519 7765477654 2
101 ush 25/a 7737708131 2
105 unnet sfgcsy 986547896 2
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