Специфическая раскладка XML требуется

Question

У меня есть следующая таблица:

CREATE TABLE PLMXC_ERP_INT_XML_BOM 
    (
    admi_num int, 
    stlh_num varchar(20), 
    arti_oms_1 varchar(40), 
    arti_code varchar(20), 
    stlx_pos varchar(5), 
    stlx_aantal numeric(16,6) 
    ); 

со следующим содержанием:

INSERT INTO PLMXC_ERP_INT_XML_BOM 
(admi_num, stlh_num, arti_oms_1, arti_code, stlx_pos, stlx_aantal) 
VALUES 
(1, '3P-2000245', 'Cover 2 undercolum 1MF', '3P-2000242', '1', 1), 
(1, '3P-2000245', 'Cover 2 undercolum 1MF', '3P-2000243', '2', 1), 
(1, '3P-2000245', 'Cover 2 undercolum 1MF', '3P-101707', '3', 5), 
(1, '3P-101707', 'cover 1MF undercolumn', '3P-2000242', '1', 1), 
(1, '3P-101707', 'cover 1MF undercolumn', '3P-2000243', '2', 1); 

Когда я хочу создать XML из этого контента я использую

SELECT stlh.admi_num 
     ,stlh.stlh_num 
     ,stlh.arti_oms_1 as stlh_oms_1 
     ,(SELECT 
      admi_num 
      ,stlm.arti_code 
      ,ROW_NUMBER()OVER(PARTITION BY stlm.stlh_num ORDER BY stlm.stlh_num) as stlm_num 
      ,stlx_pos as stlm_pos 
      ,stlx_aantal as stlm_aantal 
      FROM PLMXC_ERP_INT_XML_BOM stlm 
      WHERE stlm.stlh_num = stlh.stlh_num 
      FOR XML AUTO, TYPE, ELEMENTS 
     ) 
     FROM [PLMXC_ERP_INT_XML_BOM] stlh 
     GROUP BY stlh.stlh_num, stlh.admi_num, stlh.arti_oms_1 
     FOR XML AUTO, ELEMENTS 

и это возвращает меня

<stlh> 
    <admi_num>1</admi_num> 
    <stlh_num>3P-101707</stlh_num> 
    <stlh_oms_1>cover 1MF undercolumn</stlh_oms_1> 
    <stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000242</arti_code> 
    <stlm_num>1</stlm_num> 
    <stlm_pos>1</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
    </stlm> 
    <stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000243</arti_code> 
    <stlm_num>2</stlm_num> 
    <stlm_pos>2</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
    </stlm> 
</stlh> 
<stlh> 
    <admi_num>1</admi_num> 
    <stlh_num>3P-2000245</stlh_num> 
    <stlh_oms_1>Cover 2 undercolum 1MF</stlh_oms_1> 
    <stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000242</arti_code> 
    <stlm_num>1</stlm_num> 
    <stlm_pos>1</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
    </stlm> 
    <stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000243</arti_code> 
    <stlm_num>2</stlm_num> 
    <stlm_pos>2</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
    </stlm> 
    <stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-101707</arti_code> 
    <stlm_num>3</stlm_num> 
    <stlm_pos>3</stlm_pos> 
    <stlm_aantal>5.000000</stlm_aantal> 
    </stlm> 
</stlh> 

Но что мне нужно, и я не в состоянии вернуться в

<stlh> 
    <admi_num>1</admi_num> 
    <stlh_num>3P-101707</stlh_num> 
    <stlh_oms_1>cover 1MF undercolumn</stlh_oms_1> 
</stlh> 
<stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000242</arti_code> 
    <stlm_num>1</stlm_num> 
    <stlm_pos>1</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
</stlm> 
<stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000243</arti_code> 
    <stlm_num>2</stlm_num> 
    <stlm_pos>2</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
</stlm> 

<!-- split into multiple row (result) here --> 

<stlh> 
    <admi_num>1</admi_num> 
    <stlh_num>3P-2000245</stlh_num> 
    <stlh_oms_1>Cover 2 undercolum 1MF</stlh_oms_1> 
</stlh> 
<stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000242</arti_code> 
    <stlm_num>1</stlm_num> 
    <stlm_pos>1</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
</stlm> 
<stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-2000243</arti_code> 
    <stlm_num>2</stlm_num> 
    <stlm_pos>2</stlm_pos> 
    <stlm_aantal>1.000000</stlm_aantal> 
</stlm> 
<stlm> 
    <admi_num>1</admi_num> 
    <arti_code>3P-101707</arti_code> 
    <stlm_num>3</stlm_num> 
    <stlm_pos>3</stlm_pos> 
    <stlm_aantal>5.000000</stlm_aantal> 
</stlm> 

Мне нужно написать это в отдельные файлы XML для каждого stlh_num нужен новый файл.

Возможно ли это?

Answer 1

select (
     select P1.admi_num, 
       P1.stlh_num, 
       P1.arti_oms_1 
     for xml path('stlh'), type 
     ), 
     (
     select P2.admi_num, 
       P2.arti_code, 
       row_number() over(order by P2.stlh_num) as stlm_num, 
       P2.stlx_pos as stlm_pos, 
       P2.stlx_aantal as stl_aantal 
     from PLMXC_ERP_INT_XML_BOM as P2 
     where P1.admi_num = P2.admi_num and 
      P1.stlh_num = P2.stlh_num and 
      P1.arti_oms_1 = P2.arti_oms_1 
     for xml path('stlm'), type 
     ) 
from PLMXC_ERP_INT_XML_BOM as P1 
group by P1.admi_num, 
     P1.stlh_num, 
     P1.arti_oms_1 
for xml path('') 

Чтобы разделить результат в разных строках, вы можете использовать его вместо этого.

select (
     select (
       select P1.admi_num, 
        P1.stlh_num, 
        P1.arti_oms_1 
       for xml path('stlh'), type 
      ), 
       (
       select P2.admi_num, 
        P2.arti_code, 
        row_number() over(order by P2.stlh_num) as stlm_num, 
        P2.stlx_pos as stlm_pos, 
        P2.stlx_aantal as stl_aantal 
       from PLMXC_ERP_INT_XML_BOM as P2 
       where P1.admi_num = P2.admi_num and 
        P1.stlh_num = P2.stlh_num and 
        P1.arti_oms_1 = P2.arti_oms_1 
       for xml path('stlm'), type 
      ) 
     for xml path(''), type 
     ) 
from PLMXC_ERP_INT_XML_BOM as P1 
group by P1.admi_num, 
     P1.stlh_num, 
     P1.arti_oms_1