根据登录用户显示项目

Show projects on the basis of login user

我有一个包含 Project Namemkey 的查询。查询如下

select  
    distinct
    comp.company_name as 'Company',EMP.emp_name as 'Employee',
     EMP.emp_card_no as 'Card_No',
     EMP.mkey as 'Mkey',
    isnull(hub.type_desc,'--NA--') as 'HUB_Name'
    ,isnull(Program_name,'--NA--') as 'Site',isnull(Project_name,'--NA--') as 'Project'
    from emp_mst EMP
    JOIN emp_mst1 PINFO ON EMP.MKEY=PINFO.MKEY 
    left join company_mst comp on EMP.comp_mkey=comp.mkey and fa_year=2008
    left join type_mst_a b1 on EMP.New_Dept_mkey=b1.master_mkey --and b1.type_code='d11'
    left join type_mst_a c1 on EMP.New_Subdept_mkey=c1.master_mkey-- and c1.type_code='d2'
    left join type_mst_a d1 on EMP.new_design_mkey=d1.master_mkey --and d1.type_code='dm1'
    left join type_mst_a e1 on EMP.ctc_hdr_mkey=e1.Add_IINFo1 and e1.type_code='gr'

    left join type_mst_a f1 on EMP.department_mkey=f1.master_mkey-- and f1.type_code='d1'
    left join type_mst_a g1 on EMP.designation_mkey=g1.master_mkey --and g1.type_code='dm'
    join emp_mst s1 on s1.mkey=EMP.reporting_to
    left join emp_mst s2 on s2.mkey=EMP.reporting_to2
    left join Emp_Address_Trl addr on addr.mkey=emp.mkey 
    left join state_mst as s on addr.state_mkey=s.mkey 
    left join city_mst as c on addr.city_mkey = c.mkey 
    left join view_p_program pg on EMP.project_mkey=pg.mkey 
    left join view_p_project pr on EMP.building_mkey=pr.mkey 
    left join type_mst_a hub on EMP.hub_mkey=hub.master_mkey and hub.type_code='HWH'
    WHERE 
    EMP.STATUS IN ('A','S')

所以我想要的是,我想在表单上显示指定用户的 PROJECT NAME

我想在页面的 page_load 上显示。

<td align="right" class="Header" colspan="1">
            Location : <asp:label ID="lblproject" runat="server" Visible="false" Text='<%#= %>'></asp:label>
        </td>

如何绑定查询以便显示相关的 project name

代码隐藏

var dtAsEnum= ds.Tables[0].AsEnumerable();
string ProjectName= (from r in dtAsEnum
              where r.Field<string>("Employee") == 'LoggedInUser'
              select r.Field<string>("Project")).First<string>();

 Label.Text = ProjectName;

这可能对您有帮助或给您一个想法!

一个参考。 link: Select Value