C# if和else if使用返回数据表c的方法获取错误#

当我使用if和elseif语句编写返回数据表的方法时，我遇到了一个错误。正确的格式是什么。提前谢谢 “并非所有代码路径都返回值”

---

如果您的两个if条件都失败，那么您的方法没有任何可返回的内容，而您的方法签名告诉编译器它将返回类型为

DataTable

的对象，在上述情况下，它将无法返回该对象

所以，应该做的是使用一个类型为

DataTable

的局部变量，在条件块中相应地设置它的值，然后从方法返回变量

重构您的方法，使其类似于：

DataTable table = null;
if (tclass ==7)
{
    table  = obj.GetData(string.Format(@"select Effect_Class_Key , From_Value , To_Value , Effect_Class_Value, Scientific_Degree, case When Scientific_Degree  is null then 'Not Related'
            when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else latin_desc  end as Spec from hr_effect_classes Left Outer Join general_cod 
            on hr_effect_classes.Scientific_Degree = general_cod .sub_cod and  main_cod = 17001  where type_class in (7) and status_class = 1  and effect_key = {0}
            Order By  Scientific_Degree  , From_Value ", efkey));
}
else if (tclass == 8)
{
    table  = obj.GetData(string.Format(@"select Effect_Class_Key ,  From_Value , To_Value , Effect_Class_Value, Scientific_Degree case When Scientific_Degree is null then 'Not Related'
            when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else L_Desc end as Spec from hr_effect_classes Left Outer Join HR_Specialty   
            on hr_effect_classes.Scientific_Degree = HR_Specialty.Spec_Key where type_class in (8) and status_class = 1 and effect_key = {0}
            Order By  Scientific_Degree  , From_Value ",efkey));
}

 return table;

---

现在，我们返回一个类型为

DataTable

的引用，尽管如果两个条件都失败，它将为null，但它将使编译器认为有效的代码没有任何构建时错误。

无论发生什么情况，您都需要始终返回某些内容，因此如果两个条件都不满足，请在if语句外部添加return null

[HttpGet]
public DataTable getcareerdata (int tclass, int efkey)
{
    if (tclass ==7)
    {
        return obj.GetData(string.Format(@"select Effect_Class_Key , From_Value , To_Value , Effect_Class_Value, Scientific_Degree, case When Scientific_Degree  is null then 'Not Related'
        when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else latin_desc  end as Spec from hr_effect_classes Left Outer Join general_cod 
        on hr_effect_classes.Scientific_Degree = general_cod .sub_cod and  main_cod = 17001  where type_class in (7) and status_class = 1  and effect_key = {0}
        Order By  Scientific_Degree  , From_Value ", efkey));
    }
    else if (tclass == 8)
    {
        return obj.GetData(string.Format(@"select Effect_Class_Key ,  From_Value , To_Value , Effect_Class_Value, Scientific_Degree case When Scientific_Degree is null then 'Not Related'
        when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else L_Desc end as Spec from hr_effect_classes Left Outer Join HR_Specialty   
        on hr_effect_classes.Scientific_Degree = HR_Specialty.Spec_Key where type_class in (8) and status_class = 1 and effect_key = {0}
        Order By  Scientific_Degree  , From_Value ",efkey));
    }
return null;
}

---

在tclass不=7或8的情况下，没有返回数据表。如何处理取决于您需要什么。是否需要一个catch all

else

？您可以在末尾添加一个

return null

，以避免编译错误。在类的顶部声明一个数据表（在if之外），然后在if语句中填充该数据表，并在底部的if之外返回。我认为您可以找到此错误的数千个重复项。在下次发布之前，请尝试搜索

[HttpGet]
public DataTable getcareerdata (int tclass, int efkey)
{
    if (tclass ==7)
    {
        return obj.GetData(string.Format(@"select Effect_Class_Key , From_Value , To_Value , Effect_Class_Value, Scientific_Degree, case When Scientific_Degree  is null then 'Not Related'
        when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else latin_desc  end as Spec from hr_effect_classes Left Outer Join general_cod 
        on hr_effect_classes.Scientific_Degree = general_cod .sub_cod and  main_cod = 17001  where type_class in (7) and status_class = 1  and effect_key = {0}
        Order By  Scientific_Degree  , From_Value ", efkey));
    }
    else if (tclass == 8)
    {
        return obj.GetData(string.Format(@"select Effect_Class_Key ,  From_Value , To_Value , Effect_Class_Value, Scientific_Degree case When Scientific_Degree is null then 'Not Related'
        when Scientific_Degree  = 0 then 'Not Related' when Scientific_Degree  = -1 then 'Not Related' else L_Desc end as Spec from hr_effect_classes Left Outer Join HR_Specialty   
        on hr_effect_classes.Scientific_Degree = HR_Specialty.Spec_Key where type_class in (8) and status_class = 1 and effect_key = {0}
        Order By  Scientific_Degree  , From_Value ",efkey));
    }
return null;
}