Building Parent-Child Table Tree information

  • jnaras

    SSC-Addicted

    Points: 494

    Comments posted to this topic are about the item Building Parent-Child Table Tree information

  • Cheetah

    Old Hand

    Points: 332

    Nice code that may come in handy, esp for writing code to do cascading deletes. Expected a hierarchial tree of dependencies to be display in Management studio, something MS may consider for the future to make the life of DBA's easier.

  • D.Oc

    SSCrazy

    Points: 2855

    Ain't saying nothing until Joe Celko says it's OK 😀

    j/k, good article 🙂

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  • Atif-ullah Sheikh

    SSChampion

    Points: 12495

    Nice Code.

    I did the same thing with this query...

    select ccu.table_schema + '.' + ccu.table_name as MTablename, ccu.column_name as Mcolname,

    ccu1.table_schema + '.' + ccu1.table_name as Tablename, ccu1.column_name as colname

    ,ccu3.table_schema + '.' + ccu3.table_name as C2Tablename, ccu3.column_name as C2colname

    ,ccu5.table_schema + '.' + ccu5.table_name as C3Tablename, ccu5.column_name as C3colname

    ,ccu7.table_schema + '.' + ccu7.table_name as C4Tablename, ccu7.column_name as C4colname

    ,ccu9.table_schema + '.' + ccu9.table_name as C5Tablename, ccu9.column_name as C5colname

    ,ccu11.table_schema + '.' + ccu11.table_name as C6Tablename, ccu11.column_name as C6colname

    from information_schema.constraint_column_usage CCU

    inner join information_schema.referential_constraints RC on CCU.constraint_name=RC.unique_constraint_name

    inner join information_schema.constraint_column_usage CCU1 on RC.constraint_name=ccu1.constraint_name

    Left Outer join information_schema.constraint_column_usage CCU2 on CCU2.Table_Schema + '.' + CCU2.table_name = CCU1.Table_Schema + '.' + CCU1.table_name

    Left Outer Join information_schema.referential_constraints RC2 on CCU2.constraint_name=RC2.unique_constraint_name

    Left Outer Join information_schema.constraint_column_usage CCU3 on RC2.constraint_name=ccu3.constraint_name

    Left Outer join information_schema.constraint_column_usage CCU4 on CCU4.Table_Schema + '.' + CCU4.table_name = CCU3.Table_Schema + '.' + CCU3.table_name

    Left Outer Join information_schema.referential_constraints RC3 on CCU4.constraint_name=RC3.unique_constraint_name

    Left Outer Join information_schema.constraint_column_usage CCU5 on RC3.constraint_name=ccu5.constraint_name

    Left Outer join information_schema.constraint_column_usage CCU6 on CCU6.Table_Schema + '.' + CCU6.table_name = CCU5.Table_Schema + '.' + CCU5.table_name

    Left Outer Join information_schema.referential_constraints RC4 on CCU6.constraint_name=RC4.unique_constraint_name

    Left Outer Join information_schema.constraint_column_usage CCU7 on RC4.constraint_name=ccu7.constraint_name

    Left Outer join information_schema.constraint_column_usage CCU8 on CCU8.Table_Schema + '.' + CCU8.table_name = CCU7.Table_Schema + '.' + CCU7.table_name

    Left Outer Join information_schema.referential_constraints RC5 on CCU8.constraint_name=RC5.unique_constraint_name

    Left Outer Join information_schema.constraint_column_usage CCU9 on RC5.constraint_name=ccu9.constraint_name

    Left Outer join information_schema.constraint_column_usage CCU10 on CCU10.Table_Schema + '.' + CCU10.table_name = CCU9.Table_Schema + '.' + CCU9.table_name

    Left Outer Join information_schema.referential_constraints RC6 on CCU10.constraint_name=RC6.unique_constraint_name

    Left Outer Join information_schema.constraint_column_usage CCU11 on RC6.constraint_name=ccu11.constraint_name

    where ccu.constraint_name not in (select constraint_name from information_schema.referential_constraints)

    and ccu.table_schema + '.' + ccu.table_name in ('dbo.Portal_users')

    The thing which makes the difference is that I have to add the joins manually if I have to add the level to which i need to go to find the childs. Your code is generic in this case.

    I was just thinking to go into string processing to resolve this issue in my code, but thanks to you, now I will be using your code for my future developments.

    Thanks once again and NICE CODE...

    Atif Sheikh

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  • Jay Taylor-604520

    Old Hand

    Points: 350

    Great stuff. Thanks for doing this.

  • Ron Kunce

    SSCrazy

    Points: 2128

    Terrific code! I really like it! However it is dependent on a starting table name in which the tree only goes down. What about the parents of the starting table? Could it be written to go both ways to pick up a Person table as being the parent of the SalesPerson and possibly however many parents of Person there might be?

    Ron K.

    "Any fool can write code that a computer can understand. Good programmers write code that humans can understand." -- Martin Fowler

  • yzhang

    SSC Enthusiast

    Points: 154

    I was working on views or codes to get the same information. Thank you for the codes - saved me a lot of time. Nice code, too!

  • GSquared

    SSC Guru

    Points: 260824

    You can still end up with an infinite loop with this (well, it will loop till it hits the recursion limit).

    Table1 has an FK that references Table3

    Table2 has an FK that references Table1

    Table3 has an FK that references Table2

    So long as at least one of these keys doesn't have a Not Null constraint on the column, this data structure is possible. It's most likely to happen in many-to-many-to-many relations.

    What you're better off doing, if this kind of chain-key relationship is possible in your database, is a self-referent outer-join in the recursive portion of the CTE, with an Is Null in the Where clause, including the Level column in part of the join.

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  • Sean McDaniel-487404

    Valued Member

    Points: 68

    Try this. It takes this idea a step further and creates an entire select query based on the base table you provide.

    Right now if you run it against Northwind it will create the following query for you:

    /*

    SELECT * FROM

    [Employees] WITH (NOLOCK)

    LEFT JOIN [EmployeeTerritories] WITH (NOLOCK) ON [EmployeeTerritories].EmployeeID=Employees.EmployeeID

    INNER JOIN [Orders] WITH (NOLOCK) ON [Orders].EmployeeID=Employees.EmployeeID

    LEFT join [Order Details] WITH (NOLOCK) ON [Order Details].OrderID=Orders.OrderID

    */

    /*

    Script Name: Recursive Table Layout

    Author: Sean McDaniel

    Purpose: Will start with table and work it's way back through the foreign keys pointing to it and so forth until it reaches the highest parent

    Will also determine whether a left join or right join should be used based on whether the child table allows nulls in the column or not

    */

    set nocount on

    DECLARE @TableName varchar(200), @level int

    /*

    ------------------------------------------------------------------------------

    ------------------------------------------------------------------------------

    ------------------------------------------------------------------------------

    PUT IN THE TABLE YOU'D LIKE TO START WITH

    */

    set @TableName='Employees'

    set @level=1

    declare @ParentChildTableTree table

    (ParentTable sysname null,ChildTable sysname null,[Level] int null,Indent varchar(max) null, JOIN_INFO varchar(max));

    insert into @ParentChildTableTree select null, null, @level, null, 'SELECT * FROM

    [' + @TableName + '] WITH (NOLOCK)'

    set @level=@level+1

    insert into @ParentChildTableTree

    select distinct rkeyid,fkeyid,@level as [Level],

    @TableName + '->' + convert(varchar(max),object_name(fkeyid)) as Indent

    , case when b1.isnullable=1 then 'INNER' else 'LEFT' end + ' JOIN [' + b.name + '] WITH (NOLOCK) ON [' + b.name + '].' + b1.name + '=' + c.name + '.' + c1.name

    from sysforeignkeys a

    INNER JOIN sysobjects b ON a.fkeyid=b.id AND b.xtype='U'

    INNER JOIN syscolumns b1 ON a.fkeyid=b1.id AND a.fkey=b1.colid

    INNER JOIN sysobjects c ON a.rkeyid=c.id AND c.xtype='U'

    INNER JOIN syscolumns c1 ON a.rkeyid=c1.id AND a.rkey=c1.colid

    where rkeyid = object_id(@TableName)

    and rkeyid <> fkeyid

    while @@ROWCOUNT >0

    begin

    set @level=@level+1

    insert into @ParentChildTableTree

    select distinct rkeyid,fkeyid,@level as [Level],

    Indent + convert(varchar(max),object_name(fkeyid)) as Indent

    , case when b1.isnullable=1 then 'INNER' else 'LEFT' end + ' join [' + b.name + '] WITH (NOLOCK) ON [' + b.name + '].' + b1.name + '=' + c.name + '.' + c1.name

    from sysforeignkeys fk

    INNER JOIN sysobjects b ON fk.fkeyid=b.id AND b.xtype='U'

    INNER JOIN syscolumns b1 ON fk.fkeyid=b1.id AND fk.fkey=b1.colid

    INNER JOIN sysobjects c ON fk.rkeyid=c.id AND c.xtype='U'

    INNER JOIN syscolumns c1 ON fk.rkeyid=c1.id AND fk.rkey=c1.colid

    join @ParentChildTableTree pc on fk.rkeyid=ChildTable

    where rkeyid <> fkeyid

    and pc.Level = @level -1

    and not exists (select * from @ParentChildTableTree b where fk.rkeyid=b.ParentTable and fk.fkeyid=b.ChildTable)

    end

    select --object_name(ParentTable), Object_Name(ChildTable), Level, Indent,

    space((Level-1) * 10) + JOIN_INFO from @ParentChildTableTree order by Indent

  • Dugi

    SSCoach

    Points: 17998

    Interesting ... lot of stuff here and very rich article

    thnx and no comment!

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  • TimothyAWiseman

    SSCrazy Eights

    Points: 8819

    Very nice article, and very well done.

    You can achieve similar functionality using sp_msdependencies if you like, but that requires using an undocumented stored procedure and the bit field can be slightly difficult to get right.

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  • wyfccc

    SSC Rookie

    Points: 41

    Greate!

    But, in my routine, I often have to find the storeprocedure correlated to a table or column.

    I have a script for purpose, but it just do it by key search.

    Anyone have some good idea for it?

  • Cheetah

    Old Hand

    Points: 332

    Does somebody perhaps have the code for an dependence tree for all tables in the database.

    ie Level 1 all tables with no dependencies, Level 2 all tables that depend on Level 1 and so on..

    I can probably write it myself, but I have a time constraint, need to move data from one db to another and do some key conversions in between. So my order of processing the tables must be right.

  • Anipaul

    SSC-Insane

    Points: 24681

    Nice stuff......

  • IN_Sandeep

    SSCertifiable

    Points: 6259

    nice article and very helpful too.

    Cheers!

    Sandy.

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