XML Workshop IV - FOR XML EXPLICIT

  • jacob sebastian

    SSChampion

    Points: 11812

    Comments posted here are about the content posted at http://www.sqlservercentral.com/columnists/jSebastian/3059.asp

    .

  • jacob sebastian

    SSChampion

    Points: 11812

    Hi all,

    I see that the source code listing does not wrap lines correctly in IE7. If you happened to find the same, you could try with firefox.

    thanks

    Jacob

    .

  • chazmer

    Valued Member

    Points: 68

    Great article... but maybe I read it wrong -- I took from the article that FOR XML EXPLICIT is SQL 2005 only, when you could do the same in SQL Server 2000. Nothing huge, but I thought I'd point that out.

  • jacob sebastian

    SSChampion

    Points: 11812

    You are correct. The whole XML Workshop focuses on SQL Server 2005 only. That was the reason why I did not mention it.

    The rest of the articles in the series are concentrating on XML data type, XQuery, Schemas etc, which are specific to SQL Server 2005 only.

    .

  • Mike C

    SSC-Insane

    Points: 23224

    Just a note - FOR XML EXPLICIT is deprecated in Katmai and will be removed in a future version of SQL Server.

  • John Novak-439283

    Grasshopper

    Points: 22

    I don't know if anyone else has already pointed this out, but I was able to add a root node by specifying

    FOR XML EXPLICITY, ROOT('CustomersByRegion')

    The code below generated fully formed XML:

    select

    1 as tag,

    null as parent,

    c.countryname as 'Country!1!name',

    c.currency as 'Country!1!currency',

    null as 'City!2!name',

    null as 'Customer!3!id',

    null as 'Customer!3!name',

    null as 'Customer!3!phone'

    from

    countries c

    union all

    select

    2 as tag,

    1 as parent,

    co.countryname,

    co.currency,

    ci.cityname,

    null,

    null,

    null

    from

    cities ci

    inner join

    countries co

    on

    ci.countryID = co.countryid

    union all

    select

    3 as tag,

    2 as parent,

    co.countryname as [name],

    co.currency,

    ci.cityname as [name],

    cu.customernumber as [id],

    cu.customername as [name],

    cu.phone

    from

    customers cu

    inner join

    cities ci

    on

    cu.cityid = ci.cityid

    inner join

    countries co

    on

    ci.countryid = co.countryid

    order by

    'Country!1!name',

    'City!2!name'

    for xml explicit, root('CustomersByRegion')

    John

  • yazalpizar_

    Ten Centuries

    Points: 1315

    Almost finishing these series of XML workshop, really helped a lot.

    The proposal of adding FOR XML EXPLICIT, ROOT('CustomersByRegion') works for me too, I'm using SQL 2008R2

    Again in this post the links to the code are broken.

    Here is the code with comments in spanish.

    --=================================================================================================================================

    -- XML Workshop Part IV: usando EXPLICIT

    /* EXPLICIT es mucho más complicado de usar que la sintaxis FOR XML (AUTO/RAW/PATH) pero por otra parte permite un control

    muy detallado sobre el XML que se quiere generar. Al usar EXPLICIT la query resultante debe tener una estructura

    determinada, que se configura mediante varios modificadores en la query misma (TAG, PARENT,...)

    */

    --=================================================================================================================================

    --vamos a intentar reproducir el mismo XML que usamos en el WorkShop 3

    /*

    <customersByRegion>

    <country name="USA" currency="US Dollars">

    <city name="NY">

    <customer id="MK" name="John Mark" phone="111-111-1111"/>

    <customer id="WS" name="Will Smith" phone="222-222-2222"/>

    </city>

    <city name="NJ">

    <customer id="EN" name="Elizabeth Lincoln" phone="333-333-3333"/>

    </city>

    </country>

    <country name="England" currency="Pound Sterling">

    <city name="London">

    <customer id="TH" name="Thomas Hardy" phone="444-444-4444"/>

    </city>

    </country>

    <country name="India" currency="Rupees">

    <city name="New Delhi">

    <customer id="JS" name="Jacob Sebastian" phone="555-555-5555"/>

    </city>

    </country>

    </customersByRegion>

    */

    /*

    vamos a generar en un primer paso el nodo Country del XML

    prestar atención a los modificadore TAG y PARENT.

    TAG: esta columna es obligada, informa al generador del XML el nivel del elemento en la jerarquía del XML que queremos obtener

    en el ejemplo tenemos "1" que significa que este será el nodo principal del XML

    Parent: es la 2da columna obligada. Dice al generador XML cual será el nodo padre del elemento seleccionado.

    en el ejemplo es NULL ya que Country no tiene nodo padre

    'Country!1!name' : "Country" es el nombre del elemento

    "1" especifica el nivel del nodo

    "name" es el nombre del atributo

    'Country!1!currency' : igual que con Country, tenemos que Country es el nombre del elemento, "1" su nivel y "currency" el atributo

    */

    SELECT 1 AS Tag,

    NULL AS Parent,

    c.CountryName AS 'Country!1!name',

    c.Currency AS 'Country!1!currency'

    FROM Countries c

    /*

    veamos que tenemos en el nodo City

    prestar atención a:

    TAG tiene valor 2, para hacer corresponder la jerarquía en el XML que queremos como resultado

    PARENT tiene valor 1 para hacer corresponder con el tag parent superior, que sería Country

    */

    SELECT 2 AS Tag,

    1 AS Parent,

    Country.CountryName,

    Country.Currency,

    City.CityName

    FROM Cities City

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    --el resultado lo uniremos con lo que tenemos para el nodo Country

    SELECT 1 AS Tag,

    NULL AS Parent,

    c.CountryName AS 'Country!1!name',

    c.Currency AS 'Country!1!currency',

    NULL AS 'City!2!name'

    FROM Countries c

    UNION ALL

    SELECT 2 AS Tag,

    1 AS Parent,

    Country.CountryName,

    Country.Currency,

    City.CityName

    FROM Cities City

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    ORDER BY 'Country!1!name', 'City!2!name'

    FOR XML EXPLICIT

    /*

    so far so good...añadiremos ahora el nodo Customer, siguiendo la misma filosofía de usar el UNION ALL

    al igual que antes prestar atención a los valores de TAG y PARENT, 3 y 2 respectivamente para hacer la correspondencia

    con el elemento City

    */

    SELECT

    1 AS Tag,

    NULL AS Parent,

    c.CountryName AS 'Country!1!name',

    c.Currency AS 'Country!1!currency',

    NULL AS 'City!2!name',

    NULL AS 'Customer!3!id',

    NULL AS 'Customer!3!name',

    NULL AS 'Customer!3!phone'

    FROM

    Countries c

    UNION ALL

    SELECT

    2 AS Tag,

    1 AS Parent,

    Country.CountryName,

    Country.Currency,

    City.CityName,

    NULL,

    NULL,

    NULL

    FROM Cities City

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    UNION ALL

    SELECT

    3 AS Tag,

    2 AS Parent,

    Country.CountryName AS [name],

    Country.Currency,

    City.CityName AS [name],

    Customer.CustomerNumber AS [id],

    Customer.CustomerName AS [name],

    Customer.Phone

    FROM

    Customers Customer

    INNER JOIN Cities City ON (City.CityID = Customer.CityID)

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    ORDER BY 'Country!1!name', 'City!2!name'

    FOR XML EXPLICIT

    /*

    Aún falta por añadir el elemento root. Para lograrlo añadiremos un elemento padre superior con todos los campos a NULL

    y solo el campo TAG=1, modificando todos los demás valores de TAG para tener la correspondencia correcta

    */

    SELECT 1 AS Tag,

    NULL AS Parent,

    NULL AS 'CustomersByRegion!1', -- empty root element

    NULL AS 'Country!2!name',

    NULL AS 'Country!2!currency',

    NULL AS 'City!3!name',

    NULL AS 'Customer!4!id',

    NULL AS 'Customer!4!name',

    NULL AS 'Customer!4!phone'

    UNION ALL

    SELECT

    2 AS Tag,

    1 AS Parent,

    NULL,

    c.CountryName AS 'Country!1!name',

    c.Currency AS 'Country!1!currency',

    NULL AS 'City!2!name',

    NULL AS 'Customer!3!id',

    NULL AS 'Customer!3!name',

    NULL AS 'Customer!3!phone'

    FROM

    Countries c

    UNION ALL

    SELECT

    3 AS Tag,

    2 AS Parent,

    NULL,

    Country.CountryName,

    Country.Currency,

    City.CityName,

    NULL,

    NULL,

    NULL

    FROM Cities City

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    UNION ALL

    SELECT

    4 AS Tag,

    3 AS Parent,

    NULL,

    Country.CountryName AS [name],

    Country.Currency,

    City.CityName AS [name],

    Customer.CustomerNumber AS [id],

    Customer.CustomerName AS [name],

    Customer.Phone

    FROM

    Customers Customer

    INNER JOIN Cities City ON (City.CityID = Customer.CityID)

    INNER JOIN Countries Country ON (Country.CountryID = City.CountryID)

    ORDER BY 'Country!2!name', 'City!3!name', Parent

    FOR XML EXPLICIT

  • Mary.Stephens 54133

    SSC Rookie

    Points: 47

    But is there a way to add <?xml version="1.0" encoding="ISO-8859-1" ?> to the resulting xml file

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