我需要使用XSLT fron字段在另一個XML文件中提取xml。XSLT - 從另一個XML中獲取xml
<?xml version="1.0" encoding="UTF-8" ?>
<ns:JDBC_SELECT xmlns:ns="namepace/nt">
<row>
<KEY>19</KEY>
<QUEUE_TYPE>2</QUEUE_TYPE>
<EVENT_STATUS>0</EVENT_STATUS>
<EVENT_TYPE>New Work Package Request</EVENT_TYPE>
<CONTENT><?xml version="1.0" encoding="utf-8" standalone="yes"?><event xmlns:ns="namespace/nt" reference="0000000000" source="NBA" target="SAP" timestamp="2013-02-04T14:32:01.836+00:00" type="New Work Package Request" version="1.0" id="19"><new-work-package-request-event><work-order-id>11</work-order-id><personnel-id>ra_a</personnel-id><request-timestamp>2013-02-04T14:32:01.836+00:00</request-timestamp><new-notification-request><type>Z2</type><requirement-code>RI0D</requirement-code><main-work-centre>0102</main-work-centre><planner-group>L00</planner-group><functional-location-id>1024</functional-location-id><short-text>SET UP/REMOVE ISOLATION RUN</short-text><long-text>SET UP/REMOVE ISOLATION RUN, 1024, NORTON FOXLEY (PRV) NORTON FOXLEY Nightline Increase: (l/h)</long-text><personnel-resp-id>12157</personnel-resp-id><cost-centre>3600</cost-centre></new-notification-request><new-work-order-request><type>ZPM2</type></new-work-order-request></new-work-package-request-event></event></CONTENT>
</row>
</ns:JDBC_SELECT>
<xsl:transform xmlns:xsl="namespace/nt" version="1.0">
<xslutput method="xml"/>
<xsl:template match="/">
<xsl:copy-of select="//CONTENT"/>
</xsl:template>
</xsl:transform>
但不是運氣。 有什麼想法?
如何生成此XML文件? –