我有一个连接两个列表并创建一个新对象列表的查询,我希望通过一个新的连接来增强它,到目前为止,我尝试的所有内容似乎都没有用。
列表是:
IList<Content> result
IList<Content> resultDefault
查询是:
IEnumerable<ContentItem> joined = from x in resultDefault
join y in result on new { x.ResourceKey, x.ResourceType, x.Application } equals new { y.ResourceKey, y.ResourceType, y.Application } into g
from o in g.DefaultIfEmpty(new Content()
{
Id = 0,
CultureCode = cultureCode,
ResourceKey = x.ResourceKey,
ResourceType = x.ResourceType,
ResourceValue = String.Empty
})
where
(
resourcesJoin == ResourcesJoin.All ||
(resourcesJoin == ResourcesJoin.Translated && o.ResourceValue != String.Empty) ||
(resourcesJoin == ResourcesJoin.NotTranslated && o.ResourceValue == String.Empty)
)
&&
(
string.IsNullOrEmpty(searchString) ||
(
x.ResourceKey.ContainsIgnoreCase(searchString) ||
x.ResourceValue.ContainsIgnoreCase(searchString) ||
o.ResourceValue.ContainsIgnoreCase(searchString)
)
)
select new ContentItem(o, x);
到目前为止效果很好。
现在,我想添加第三个列表
IList<DifferentContent> resultCollection;
我已经扩展了ContentItem以获取DifferentContent的第三个参数
所以我想要这样的东西:
IEnumerable<ContentItem> joined = from x in resultDefault
join y in result on new { x.ResourceKey, x.ResourceType, x.Application } equals new { y.ResourceKey, y.ResourceType, y.Application } into g
from o in g.DefaultIfEmpty(new Content()
{
Id = 0,
CultureCode = cultureCode,
ResourceKey = x.ResourceKey,
ResourceType = x.ResourceType,
ResourceValue = String.Empty
})
join z in resultCollection on new { x.ResourceKey, x.ResourceType, x.Application, x.CultureCode } equals new { z.ResourceKey, z.ResourceType, z.Application, defaultCultureCode } into m
from s in m.DefaultIfEmpty(new DifferentContent()
{
Id = 0,
CultureCode = defaultCultureCode,
ResourceKey = x.ResourceKey,
ResourceType = x.ResourceType,
ResourceValue = string.Empty
})
where
(
resourcesJoin == ResourcesJoin.All ||
(resourcesJoin == ResourcesJoin.Translated && o.ResourceValue != String.Empty) ||
(resourcesJoin == ResourcesJoin.NotTranslated && o.ResourceValue == String.Empty)
)
&&
(
string.IsNullOrEmpty(searchString) ||
(
x.ResourceKey.ContainsIgnoreCase(searchString) ||
x.ResourceValue.ContainsIgnoreCase(searchString) ||
o.ResourceValue.ContainsIgnoreCase(searchString)
)
)
select new ContentItem(o, x, s);
但这是无效的。错误是:
The type of one of the expressions in the join clause is incorrect. Type inference failed in the call to 'GroupJoin'.
Resharper给出了“方法的类型参数......不能被推测”。不幸的是,我无法复制粘贴。
正确的连接语法是什么?