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  • Linq to sql(四):查询句法(四)

    对应SQL

    SELECT

        (CASE

            WHEN [t2].[value2] = 1 THEN @p1

            ELSE @p2

         END) AS [value], [t2].[value] AS [数量]

    FROM (

        SELECT COUNT(*) AS [value], [t1].[value] AS [value2]

        FROM (

            SELECT

                (CASE

                    WHEN [t0].[Freight] > @p0 THEN 1

                    WHEN NOT ([t0].[Freight] > @p0) THEN 0

                    ELSE NULL

                 END) AS [value]

            FROM [dbo].[Orders] AS [t0]

            ) AS [t1]

        GROUP BY [t1].[value]

        ) AS [t2]

    -- @p0: Input Currency (Size = 0; Prec = 19; Scale = 4) [100]

    -- @p1: Input String (Size = 1; Prec = 0; Scale = 0) []

    -- @p2: Input String (Size = 1; Prec = 0; Scale = 0) []

    distinct

     

    描述:查询顾客覆盖的国家

    查询句法:

    var 过滤相同项 = (from c in ctx.Customers orderby c.Country select c.Country).Distinct();

    对应SQL

    SELECT DISTINCT [t0].[Country]

    FROM [dbo].[Customers] AS [t0]

     

    union

     

    描述:查询城市是A打头和城市包含A的顾客并按照顾客名字排序

    查询句法:

    var 连接并且过滤相同项 = (from c in ctx.Customers where c.City.Contains("A") select c).Union

                (from c in ctx.Customers where c.ContactName.StartsWith("A") select c).OrderBy(c => c.ContactName);

    对应SQL

    SELECT [t3].[CustomerID], [t3].[CompanyName], [t3].[ContactName], [t3].[ContactTitle], [t3].[Address], [t3].[City], [t3].[Region], [t3].[PostalCode], [t3].[Country], [t3].[Phone], [t3].[Fax]

    FROM (

        SELECT [t2].[CustomerID], [t2].[CompanyName], [t2].[ContactName], [t2].[ContactTitle], [t2].[Address], [t2].[City], [t2].[Region], [t2].[PostalCode], [t2].[Country], [t2].[Phone], [t2].[Fax]

        FROM (

            SELECT [t0].[CustomerID], [t0].[CompanyName], [t0].[ContactName], [t0].[ContactTitle], [t0].[Address], [t0].[City], [t0].[Region], [t0].[PostalCode], [t0].[Country], [t0].[Phone], [t0].[Fax]

            FROM [dbo].[Customers] AS [t0]

            WHERE [t0].[City] LIKE @p0

            UNION

            SELECT [t1].[CustomerID], [t1].[CompanyName], [t1].[ContactName], [t1].[ContactTitle], [t1].[Address], [t1].[City], [t1].[Region], [t1].[PostalCode], [t1].[Country], [t1].[Phone], [t1].[Fax]

            FROM [dbo].[Customers] AS [t1]

            WHERE [t1].[ContactName] LIKE @p1

            ) AS [t2]

        ) AS [t3]

    ORDER BY [t3].[ContactName]

    -- @p0: Input String (Size = 3; Prec = 0; Scale = 0) [%A%]

    -- @p1: Input String (Size = 2; Prec = 0; Scale = 0) [A%]

     

    concat

     

    描述:查询城市是A打头和城市包含A的顾客并按照顾客名字排序,相同的顾客信息不会过滤

    查询句法:

    var 连接并且不过滤相同项 = (from c in ctx.Customers where c.City.Contains("A") select c).Concat

                (from c in ctx.Customers where c.ContactName.StartsWith("A") select c).OrderBy(c => c.ContactName);

    对应SQL

    SELECT [t3].[CustomerID], [t3].[CompanyName], [t3].[ContactName], [t3].[ContactTitle], [t3].[Address], [t3].[City], [t3].[Region], [t3].[PostalCode], [t3].[Country], [t3].[Phone], [t3].[Fax]

    FROM (

        SELECT [t2].[CustomerID], [t2].[CompanyName], [t2].[ContactName], [t2].[ContactTitle], [t2].[Address], [t2].[City], [t2].[Region], [t2].[PostalCode], [t2].[Country], [t2].[Phone], [t2].[Fax]

        FROM (

            SELECT [t0].[CustomerID], [t0].[CompanyName], [t0].[ContactName], [t0].[ContactTitle], [t0].[Address], [t0].[City], [t0].[Region], [t0].[PostalCode], [t0].[Country], [t0].[Phone], [t0].[Fax]

            FROM [dbo].[Customers] AS [t0]

            WHERE [t0].[City] LIKE @p0

            UNION ALL

            SELECT [t1].[CustomerID], [t1].[CompanyName], [t1].[ContactName], [t1].[ContactTitle], [t1].[Address], [t1].[City], [t1].[Region], [t1].[PostalCode], [t1].[Country], [t1].[Phone], [t1].[Fax]

            FROM [dbo].[Customers] AS [t1]

            WHERE [t1].[ContactName] LIKE @p1

            ) AS [t2]

        ) AS [t3]

    ORDER BY [t3].[ContactName]

    -- @p0: Input String (Size = 3; Prec = 0; Scale = 0) [%A%]

    -- @p1: Input String (Size = 2; Prec = 0; Scale = 0) [A%]

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  • 原文地址:https://www.cnblogs.com/kevin2013/p/1749089.html
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