问题描述
你好！
我在表中有一列app_Id.
它的值类似于
10/2000
10/2001
17/2001
32/2009
89/2011
我要做的是对app_id进行排序. "/"之前的部分是申请号，而"/"之后的部分是年份.
所以我想进行排序，我应该在数据表中为appNumber和year分别创建两列.
然后我用这个查询
SELECT [app_id], (Select LEFT(app_Id, CHARINDEX(''/'',app_Id )-1)) as Number,(Select Right(app_Id, CHARINDEX(''/'',app_Id )+1)) as Year ,[app_type], convert(varchar,[app_date],103) as app_date, [subject], [app_name], [gar_name], [source_vill], [source_place], [source_dist], [mode], [epat_letno],[status] ,Approved FROM [viv_app] where FinancialYear <>''2012-2013'' and Status=''Sent to treasury'' order by [Year] desc ,Number DESC
现在我面临的问题是，对于
10/2000
这样的app_id，我得到Number = 10和year = 2010，这是正确的，但是对于某些般的app_id值，我却得到number = 5，但year = 011，对于像
378/2011
这样的app_id，我得到的号码= 378，但是Year =/2011.
请告诉我如何解决此逻辑错误.您必须使用相同的逻辑
创建 表临时 (id varchar ( 50 )) 插入 插入 temp 值(' 10/2000') 插入 插入 temp 值(' 5/2000') 插入 插入 temp 值(' 345/2000') 选择 SUBSTRING(id， 1 ，charindex(' /'，id， 0 )-1)作为 ID， SUBSTRING(id，charindex(' /'，id， 0 )+ 1，LEN(id)-1)按年来自的温度
Hello!!
I have a column app_Id in a table.
It has values like
10/2000
10/2001
17/2001
32/2009
89/2011
What i have to do is sort app_id. Part before ''/'' is the application number and after ''/'' is year.
so i thought to sort i should make two columns in my datatable each for appNumber and year.
Then i used this query
SELECT [app_id], (Select LEFT(app_Id, CHARINDEX(''/'',app_Id )-1)) as Number,(Select Right(app_Id, CHARINDEX(''/'',app_Id )+1)) as Year ,[app_type], convert(varchar,[app_date],103) as app_date, [subject], [app_name], [gar_name], [source_vill], [source_place], [source_dist], [mode], [epat_letno],[status] ,Approved FROM [viv_app] where FinancialYear <>''2012-2013'' and Status=''Sent to treasury'' order by [Year] desc ,Number DESC
Now the problem that i am facing is that for app_id like
10/2000
i am getting Number = 10 and year =2010 which is right but for some values of app_id like
5/2011
i am getting number =5 but year =011 and for app_id like
378/2011
i am getting number =378 but Year= /2011.
please tell me how to solve this logical error
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