2015-06-08 59 views
0

我想将List作为Json传递。如何创建AM_GAAM属性的成员列表? 以下是我的控制器的代码。我使用的存储过程只返回一列(AM_GAAM)和多行。将ObjectResult <string>转换为列表<T>

控制器

ObjectResult<string> obj; 
obj = objConnection.ADDRESS_MASTER_Select_Distinct_Gaam(objIErrorCode); 

if (Convert.ToInt32(objIErrorCode.Value) == 0) 
{ 
    list = obj.Select(x => new Member 
    { 
     AM_GAAM = x. 
    }).ToList(); 
} 

return Json(new { Success = true, Message = list }, JsonRequestBehavior.AllowGet); 
+0

'AM_GAAM = x.AM_GAAM' –

+0

@StephenMuecke没有得到'x.AM_GAAM'因为字符串类型的'ObjectResult ' –

回答

1

如果您ADDRESS_MASTER_Select_Distinct_Gaam方法返回一个字符串列表,你可以做到这一点

list=obj.Select(s=>new Member { AM_GAM_=x }).ToList(); 

但如果你ADDRESS_MASTER_Select_Distinct_Gaam方法返回一些自定义对象的列表AM_GAM是该定制类的属性,您可以这样做

list=obj.Select(s=>new Member { AM_GAM_=x.AM_GAM }).ToList(); 
+0

非常感谢'名单= obj.Select(s => new Member {AM_GAM_ = x})。ToList();'工作 –

1

您可以使用匿名类

var list; 
    if (Convert.ToInt32(objIErrorCode.Value) == 0) 
    { 
     list = obj.Select(x => new 
       { 
        AM_GAAM=x.ToString() //x.ToString() as list content string value 
       }); 
    } 


    return Json(new { Success = true, Message = list }, JsonRequestBehavior.AllowGet); 
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