Asp.net 并非所有路径在模型类中都返回值错误

我的模型类如下所示：

using System;
using System.Collections.Generic;
using System.Configuration;
using System.Data;
using System.Data.SqlClient;
using System.Drawing;
using System.Drawing.Imaging;
using System.Drawing.Drawing2D;
using System.Linq;
using System.Web;
using System.Web.Configuration;
using System.Web.Mvc;
using System.Web.UI;
using System.Web.UI.WebControls;

namespace OEWebPortalMVCVer5.Models
{
 class DenialModel
{
    public Dictionary< string, DataSet> DataSets { get; set; }

    public static DataSet ReadRows(DataSet dataset)
    {
        try
        {
            SqlConnection conn = new SqlConnection(WebConfigurationManager.ConnectionStrings["demo10ConnectionString"].ConnectionString);

            string startDate = "1/1/1900";
            string endDate = "12/31/2020";
            conn.Open();
            DataSet dtDS = new DataSet();
            SqlCommand dtCom = new SqlCommand("dbo.cusGenDenialReport", conn);
            dtCom.Parameters.AddWithValue("@StartDate", SqlDbType.DateTime).Value = Convert.ToDateTime(startDate);
            dtCom.Parameters.AddWithValue("@EndDate", SqlDbType.DateTime).Value = Convert.ToDateTime(endDate);
            dtCom.CommandType = CommandType.StoredProcedure;
            SqlDataAdapter dtDa = new SqlDataAdapter(dtCom);

            dtDa.Fill(dtDS);
            conn.Close();
            dtCom.Dispose();
        }
        catch (Exception ex)
        {
            string message = "It didn't work because: " + ex.Message.ToString();
        }
    }
}
}

我的第一个想法是，我没有处理参数startDate和endDate的空值，所以我硬编码了它们，稍后将解决这个问题。事实并非如此，因此我添加了一个try-catch块来获取错误。那当然没用（我一定很累了，因为我想试一试能神奇地解决这个问题）。我用谷歌搜索了这个错误，搜索了这个网站，没有找到这个问题的任何相关答案

我不知道如何让代码在此时正确编译。谢谢你的帮助

---

谢谢

错误准确地说明了问题所在：函数中没有

return

语句

public static DataSet ReadRows(DataSet dataset)
    {
        DataSet dtDS = null;
        try
        {
            SqlConnection conn = new SqlConnection(WebConfigurationManager.ConnectionStrings["demo10ConnectionString"].ConnectionString);

            string startDate = "1/1/1900";
            string endDate = "12/31/2020";
            conn.Open();
            dtDS = new DataSet();
            SqlCommand dtCom = new SqlCommand("dbo.cusGenDenialReport", conn);
            dtCom.Parameters.AddWithValue("@StartDate", SqlDbType.DateTime).Value = Convert.ToDateTime(startDate);
            dtCom.Parameters.AddWithValue("@EndDate", SqlDbType.DateTime).Value = Convert.ToDateTime(endDate);
            dtCom.CommandType = CommandType.StoredProcedure;
            SqlDataAdapter dtDa = new SqlDataAdapter(dtCom);

            dtDa.Fill(dtDS);
            conn.Close();
            dtCom.Dispose();
        }
        catch (Exception ex)
        {
            //You probably should let it break...
            string message = "It didn't work because: " + ex.Message.ToString();
        }

        return dtDS;
    }

---

错误正好说明了问题所在：函数中没有

return

语句。@Johnny5答案是这样的，您应该这样发布它。即使在使用视图时也是这样吗？请原谅我的问题，如果它是愚蠢的，但我只是了解到，错误在任何情况下都与MVC或任何东西无关。在C#中，函数必须返回值，除非其返回类型设置为void，否则在这种情况下它永远不会返回值。此外，不能在不同的情况下返回不同的类型，就像在javascript或（我认为）php等非强类型语言中一样。

public static DataSet ReadRows(DataSet dataset)
    {
        DataSet dtDS = null;
        try
        {
            SqlConnection conn = new SqlConnection(WebConfigurationManager.ConnectionStrings["demo10ConnectionString"].ConnectionString);

            string startDate = "1/1/1900";
            string endDate = "12/31/2020";
            conn.Open();
            dtDS = new DataSet();
            SqlCommand dtCom = new SqlCommand("dbo.cusGenDenialReport", conn);
            dtCom.Parameters.AddWithValue("@StartDate", SqlDbType.DateTime).Value = Convert.ToDateTime(startDate);
            dtCom.Parameters.AddWithValue("@EndDate", SqlDbType.DateTime).Value = Convert.ToDateTime(endDate);
            dtCom.CommandType = CommandType.StoredProcedure;
            SqlDataAdapter dtDa = new SqlDataAdapter(dtCom);

            dtDa.Fill(dtDS);
            conn.Close();
            dtCom.Dispose();
        }
        catch (Exception ex)
        {
            //You probably should let it break...
            string message = "It didn't work because: " + ex.Message.ToString();
        }

        return dtDS;
    }