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  • sql树形结果,查询所有子类

    新建测试表

    CREATE TABLE [dbo].[test](
    [Id] [bigint] IDENTITY(1,1) NOT NULL,
    [parentId] [bigint] NULL,
    [name] [nvarchar](50) NULL,
    [jb] [int] NULL,
    )
    GO

    插入测试数据

    INSERT INTO test ( Id, parentId, name, jb) values (1,0,'1',1);
    INSERT INTO test ( Id, parentId, name, jb) values (2,0,'2',1);
    INSERT INTO test ( Id, parentId, name, jb) values (3,1,'11',2);
    INSERT INTO test ( Id, parentId, name, jb) values (4,1,'12',2);
    INSERT INTO test ( Id, parentId, name, jb) values (5,2,'21',2);
    INSERT INTO test ( Id, parentId, name, jb) values (6,2,'22',2);
    INSERT INTO test ( Id, parentId, name, jb) values (7,3,'111',3);
    INSERT INTO test ( Id, parentId, name, jb) values (8,3,'112',3);
    INSERT INTO test ( Id, parentId, name, jb) values (9,5,'221',3);
    INSERT INTO test ( Id, parentId, name, jb) values (10,0,'3',1);

    树形结构,查询分类下所有子类

    ;with cte as
    (
    select * from test where id=1
    union all
    select a.* from test a join cte b on a.parentId=b.Id
    )

    select * from cte

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  • 原文地址:https://www.cnblogs.com/zylong/p/4623635.html
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