什么是将以下XML转换为JSON的XSLT

Question

<?xml version="1.0" encoding="UTF-8"?>
<!--Generated by Oracle BI Publisher -Dataengine, datamodel:_Custom_OAL_ATG_OM_Dashboard_DM_xdm -->
<DATA>
<SUCCESS>
<COUNT___>5686</COUNT___>
</SUCCESS>
<REJECT>
<COUNT___>641</COUNT___>
</REJECT>
<FAILURE>
<COUNT___>8536</COUNT___>
</FAILURE>
<ERROR>
<COUNT___>1447</COUNT___>
</ERROR>
<TERMINATED>
<COUNT___>1341</COUNT___>
</TERMINATED>
</DATA>

上面是XML。我想将以上XML转换为下面显示的JSON。

{
"appName": "PERFORMANCE",
"statsName": "Status Counts",
"DateBegin": "xxxxxx",
"DateEnd": "xxxxxxx",
"data": {
 "SUCCESS ": 1341,
"REJECT":5666,  "FAILURE":640,
 "ERROR":8515,
"TERMINATED":1447
}
}

我是XSLT样式表的新手。谁能帮助我解决上述问题？什么是将给定XML转换为JSON的XSLT样式表？

Answer 1

您可以使用以下样式表。它适用于给定的输入XML：

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"  version="1.0">
<xsl:output method="text" />
  <xsl:template match="/DATA">
      <xsl:text>{
    "appName": "PERFORMANCE",
    "statsName": "Status Counts",
    "DateBegin": "xxxxxx",
    "DateEnd": "xxxxxxx",
    "data": 
      </xsl:text>{&#xa;<xsl:apply-templates select="*" /><xsl:text>&#xa;}&#xa;}</xsl:text>
  </xsl:template>
  <xsl:template match="SUCCESS|REJECT|FAILURE|ERROR|TERMINATED">
    <xsl:value-of select="concat('&quot;',local-name(),'&quot;: ',COUNT___)" />
    <xsl:if test="position() != last()">,&#xa;</xsl:if>
  </xsl:template>
</xsl:stylesheet>

其输出是：

{
"appName": "PERFORMANCE",
"statsName": "Status Counts",
"DateBegin": "xxxxxx",
"DateEnd": "xxxxxxx",
"data": 
{
"SUCCESS": 5686,
"REJECT": 641,
"FAILURE": 8536,
"ERROR": 1447,
"TERMINATED": 1341
}
}

输出不匹配您所需的输出，因为输入XML和所需的输出XML之间存在一些不一致之处。根据您的需要更改XSLT。