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使用LINQ到实体生成信息

[英]Using LINQ-To-Entities to Generate Information

I am working on a website where a user can add tags to their posted books, much like is currently done for questions on Stack Overflow. 我正在一个网站上工作,在该网站上,用户可以将标签添加到已发布的书本上,就像目前针对Stack Overflow的问题所做的一样。

Classes : 课程

Books
{
bookId,
Title
}

Tags
{
Id
Tag
}

BooksTags
 {
 Id
 BookId
 TagId
 }

Here are few sample records. 以下是一些示例记录。

Books
BookId Title
113421  A
113422  B

Tags
Id Tag
1  ASP 
2  C#
3  CSS
4  VB
5  VB.NET
6  PHP
7  java
8  pascal

 BooksTags   
 Id  BookId  TagId
 1  113421    1
 2  113421    2
 3  113421    3
 4  113421    4
 5  113422    1
 6  113422    4
 7  113422    8

Questions 问题

  1. I need to write something in LINQ to entity queries which gives me data according to the tags: 我需要在LINQ中为实体查询写一些东西,它根据标签为我提供数据:

    Query : bookIds where tagid = 1 查询bookIds where tagid = 1
    Returns : bookid: 113421, 113422 返回bookid: 113421, 113422

    Query 2 : tags 1 and 2 查询2tags 1 and 2
    Returns : 113421 回报113421

  2. I need tags and their count to to show in related tags, so in first case my related tags class should have following result. 我需要标签及其计数才能显示在相关标签中,因此在第一种情况下,我的相关标签类应具有以下结果。

    RelatedTags Tag Count 2 1 3 1 4 2 8 1 相关标签标签计数2 1 3 1 4 2 8 1

Second Case : 第二种情况

RelatedTags
Tag Count
3   1
4   1

How do I do this in LINQ? 如何在LINQ中做到这一点?

Just do Foreign keys mapping the tables in 1:N or 1:1 relations, and let the designer create navigation properties for you. 只需执行以1:N或1:1关系映射表的外键,然后让设计者为您创建导航属性。 (Books:BooksTags maps 1:N from Books.BookID to BooksTags.BookID, and BooksTags.TagID maps 1:1 to Tags.TagID). (Books:BooksTags将1:N从Books.BookID映射到BooksTags.BookID,而BooksTags.TagID则将1:1映射到Tag.TagID)。 This is actually a disguised N:M relation. 这实际上是伪装的N:M关系。 I don't know if the designer picks this up directly, but with some fiddling you can get the navigation properties right. 我不知道设计者是否直接选择了它,但是通过一些摆弄可以使导航属性正确。

Now for the questions: 现在开始提问:

  1. model.Tags.Where(t => t.ID == 1).Books.Select(b => b.ID)

  2. Get all the tags present for a book, and join that table on the BooksTags, by this you can simply use Count() to get the count. 获取一本书的所有标签,然后将该表连接到BooksTags上,这样您就可以简单地使用Count()来获取计数。

On the first part, the interesting restriction is that the book has to match every tag entered, so a where clause of "where tagid == someId" wouldn't really work. 在第一部分,有趣的限制是,该书必须匹配输入的每个标签,因此“ where tagid == someId”的where子句实际上不会起作用。 I envision something like this (LINQ-to-objects example) 我设想这样的事情(LINQ到对象的示例)

List<int> selectedTagIds = new List<int>() { 1, 2 };
var query = from book in books
            join booktag in booktags 
            on book.Id equals booktag.BookId 
            join selectedId in selectedTagIds 
            on booktag.TagId equals selectedId 
            group book by book into bookgroup 
            where bookgroup.Count() == selectedTagIds.Count
            select bookgroup.Key;

Which basically performs a join from books to booktags and also to the list of selected tag ids and restricts the selection to where the count of book->tag matches equals the count of selected tag ids. 它基本上执行从书籍到书签的连接,也执行到选定标签ID列表的连接,并将选择范围限制为book-> tag匹配的数目等于选定标签ID的数目。

To pull the related tags, maybe something like this 要拉相关标签,也许像这样

var relatedTags = from book in query // use original query as base
                    join booktag in booktags
                    on book.Id equals booktag.BookId
                    join tag in tags
                    on booktag.TagId equals tag.Id
                    where !selectedTagIds.Contains(tag.Id) // exclude selected tags from related tags
                    group tag by tag into taggroup
                    select new
                    {
                        Tag = taggroup.Key,
                        Count = taggroup.Count()
                    };

Full code for the quick example. 快速示例的完整代码。 Not fully OOP, but you get the idea. 并不是完全面向对象的,但是您知道了。

using System;
using System.Collections.Generic;
using System.Linq;

namespace StackOverflow
{
    class Program
    {
        static void Main()
        {
            List<Book> books = new List<Book>() 
            {
                new Book() { Id = 113421, Title = "A" },
                new Book() { Id = 113422, Title = "B" }
            };

            List<Tag> tags = new List<Tag>()
            {
                new Tag() { Id = 1, Name = "ASP" },
                new Tag() { Id = 2, Name = "C#" },
                new Tag() { Id = 3, Name = "CSS" },
                new Tag() { Id = 4, Name = "VB" },
                new Tag() { Id = 5, Name = "VB.NET" },
                new Tag() { Id = 6, Name = "PHP" },
                new Tag() { Id = 7, Name = "Java" },
                new Tag() { Id = 8, Name = "Pascal" }
            };

            List<BookTag> booktags = new List<BookTag>()
            {
                new BookTag() { Id = 1, BookId = 113421, TagId = 1 },
                new BookTag() { Id = 2, BookId = 113421, TagId = 2 },
                new BookTag() { Id = 3, BookId = 113421, TagId = 3 },
                new BookTag() { Id = 4, BookId = 113421, TagId = 4 },
                new BookTag() { Id = 5, BookId = 113422, TagId = 1 },
                new BookTag() { Id = 6, BookId = 113422, TagId = 4 },
                new BookTag() { Id = 7, BookId = 113422, TagId = 8 }
            };


            List<int> selectedTagIds = new List<int>() { 1,2 };

            // get applicable books based on selected tags

            var query = from book in books
                        join booktag in booktags
                        on book.Id equals booktag.BookId
                        join selectedId in selectedTagIds
                        on booktag.TagId equals selectedId
                        group book by book into bookgroup
                        where bookgroup.Count() == selectedTagIds.Count
                        select bookgroup.Key;

            foreach (Book book in query)
            {
                Console.WriteLine("{0}\t{1}",
                    book.Id,
                    book.Title);
            }

            // get related tags for selected tags

            var relatedTags = from book in query // use original query as base
                              join booktag in booktags
                              on book.Id equals booktag.BookId
                              join tag in tags
                              on booktag.TagId equals tag.Id
                              where !selectedTagIds.Contains(tag.Id) // exclude selected tags from related tags
                              group tag by tag into taggroup
                              select new
                              {
                                  Tag = taggroup.Key,
                                  Count = taggroup.Count()
                              };

            foreach (var relatedTag in relatedTags)
            {
                Console.WriteLine("{0}\t{1}\t{2}",
                    relatedTag.Tag.Id,
                    relatedTag.Tag.Name,
                    relatedTag.Count);
            }

            Console.Read();
        }
    }

    class Book
    {
        public int Id { get; set; }
        public string Title { get; set; }
    }

    class Tag
    {
        public int Id { get; set; }
        public string Name { get; set; }
    }

    class BookTag
    {
        public int Id { get; set; }
        public int BookId { get; set; }
        public int TagId { get; set; }
    } 
}

So for selected tags 1 & 2, you'll get book A, and the related tags would be 3 (CSS) and 4 (VB). 因此,对于选定的标签1和2,您将获得书籍A,而相关的标签将是3(CSS)和4(VB)。

这与答案本身没有直接关系,但是您可能想看看linqpad,因为它可以帮助您直接从数据库中构建L2S语句。

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