Question

我想提取＆＃34; KEY＆＃34;的值。和＆＃34; VALUE＆＃34;并使用带有EXTRACT的XPath连接这些结果。

示例XML如下。

<PivotSet>
  <item>
    <column name = "KEY">RET_1</column>
    <column name = "VALUE">A</column>
  </item>
  <item>
    <column name = "KEY">RET_2</column>
    <column name = "VALUE">R</column>
  </item>
  <item>
    <column name = "KEY">RET_3</column>
    <column name = "VALUE">P</column>
  </item>
</PivotSet>

我需要在SQL中返回类似

的内容

RET_1@A;RET_2@R;RET_3@P

分隔符＆＃39; @＆＃39;分隔键/值和＆＃39;：＆＃39;分开项目。我想用EXTRACT和XPath来做。

Answer 1

您可以将ListAgg与XmlTable一起使用，请参阅下文：

select LISTAGG(keyItem || '@' || keyValue, ';') WITHIN GROUP (ORDER BY keyItem) AS Concated
from XmlTable( '/PivotSet/item' PASSING XmlType('<PivotSet>
  <item>
    <column name = "KEY">RET_1</column>
    <column name = "VALUE">A</column>
  </item>
  <item>
    <column name = "KEY">RET_2</column>
    <column name = "VALUE">R</column>
  </item>
  <item>
    <column name = "KEY">RET_3</column>
    <column name = "VALUE">P</column>
  </item>
</PivotSet>') COLUMNS keyItem PATH '*:column[@name = "KEY"]',
                      keyValue PATH '*:column[@name = "VALUE"]');

您可以使用ExtractValue，但是您需要知道XML中有多少//item个元素，然后在提取调用的路径部分中指出，但这有点混乱IMO：

SELECT ExtractValue(xmlOut, '//item[1]/column[@name="KEY"]') || '@' || ExtractValue(xmlOut, '//item[1]/column[@name="VALUE"]') || ';' ||
       ExtractValue(xmlOut, '//item[2]/column[@name="KEY"]') || '@' || ExtractValue(xmlOut, '//item[2]/column[@name="VALUE"]') || ';' ||
       ExtractValue(xmlOut, '//item[3]/column[@name="KEY"]') || '@' || ExtractValue(xmlOut, '//item[3]/column[@name="VALUE"]')  As Concatted
FROM
(
    SELECT XmlType('<PivotSet>
  <item>
    <column name = "KEY">RET_1</column>
    <column name = "VALUE">A</column>
  </item>
  <item>
    <column name = "KEY">RET_2</column>
    <column name = "VALUE">R</column>
  </item>
  <item>
    <column name = "KEY">RET_3</column>
    <column name = "VALUE">P</column>
  </item>
</PivotSet>') as xmlOut FROM dual
);

XPath concat与EXTRACT函数Oracle

1 个答案: