我有3桌菜,葡萄酒,建议。 然后想法是用餐桌建议表把盘子和葡萄酒制作其中一个相互暗示。 我正在使用LINQ,但当一个产品没有建议时,他不会添加到json。
var query = (from m in db.Dish
join t in db.TypeDish on m.idTypeDish equals t.idTypeDish
join i in db.ImageDish on m.idDish equals i.idDish into g
join s in db.Suggestion on m.idDish equals s.idDish
join si in db.ImageWine on s.idWine equals si.idWine into f
where m.idTypeDish == dish
select new DishModel()
{
Name = m.name,
CalorificValue = m.calorificValue,
Price = m.price,
ShortName = m.shortName,
Time = m.manufactureTime,
Description = m.description,
UrlImageList = g.Select(i => _url + i.Image.urlImage).ToList(),
BeveragesList = new List<BeverageModel>()
{
new BeverageModel()
{
Name = s.Wine.name,
ShortName = s.Wine.shortName,
Price = s.Wine.price,
Description = s.Wine.description,
AlcoholContent = s.Wine.alcoholContent,
WineEnum = WineEnum.WhiteWine,
Region = s.Wine.Region.name,
WineCaste = s.Wine.wineCaste,
UrlImageList = f.Select(i => _url+i.Image.urlImage).ToList(),
}
}
}).ToList();
return query;
现在我在DB上有两个项目,他只发送一个,因为另一个没有建议。 可能是连接错误,但我是Linq的新手。
感谢。