xsl将xml转换为json

Input XML

<Root>
    <Result>
        <System>
            <Name>ABC</Name>
            <ID pname="PAD">
                <value>4567</value>
            </ID>
            <lastTime>2013-11-06T17:36:46.000-05:00</lastTime>
        </System>
        <line>Metals</line>
    </Result>
    <Result>
        <System>
            <Name>CAYS</Name>
            <ID pname="PAD">
                <value>MCIERT</value>
            </ID>
            <ID pname="ATPAD">
                <value>56412</value>
            </ID>
            <lastTime>2013-12-06T16:43:36.000-05:00</lastTime>
        </System>
        <System>
            <Name>CAYS</Name>
            <ID pname="CAD">
                <value>DGSG</value>
            </ID>
            <ID pname="ARCAD">
                <value>2847114</value>
            </ID>
            <lastTime>2013-12-07T20:02:38.000-05:00</lastTime>
        </System>
        <line>Minerals</line>
    </Result>
</Root>

Output Json

{
"Root": {
"Result": [
  {
    "System": {
      "Name": "ABC",
      "ID": {
        "pname": "PAD",
        "value": "4567"
      },
      "lastTime": "2013-11-06T17:36:46.000-05:00"
    },
    "line": "Metals"
  },
  {
    "System": [
      {
        "Name": "CAYS",
        "ID": [
          {
            "pname": "PAD",
            "value": "MCIERT"
          },
          {
            "pname": "ATPAD",
            "value": "56412"
          }
        ],
        "lastTime": "2013-12-06T16:43:36.000-05:00"
      },
      {
        "Name": "CAYS",
        "ID": [
          {
            "pname": "CAD",
            "value": "DGSG"
          },
          {
            "pname": "ARCAD",
            "value": "2847114"
          }
        ],
        "lastTime": "2013-12-07T20:02:38.000-05:00"
      }
    ],
    "line": "Minerals"
  }
]
}
}

How to write the generic xslt stylesheet which will convert input xml to json

如何编写将输入xml转换为json的通用xslt样式表

Input might have so many Results under root, Systems and Names under results and also ID name and the values under systems.

输入可能在根目录下有很多结果,在结果下有系统和名称,还有ID名称和系统下的值。

1 个解决方案

#1

The below XSLT should help you to convert XML to JSON. Thanks :)

下面的XSLT应该可以帮助您将XML转换为JSON。谢谢 :)

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>

<xsl:template match="/">{
    <xsl:apply-templates select="*"/>}
</xsl:template>

<!-- Object or Element Property-->
<xsl:template match="*">
    "<xsl:value-of select="name()"/>" :<xsl:call-template name="Properties">
        <xsl:with-param name="parent" select="'Yes'"> </xsl:with-param>
    </xsl:call-template>
</xsl:template>

<!-- Array Element -->
<xsl:template match="*" mode="ArrayElement">
    <xsl:call-template name="Properties"/>
</xsl:template>

<!-- Object Properties -->
<xsl:template name="Properties">
    <xsl:param name="parent"></xsl:param>
    <xsl:variable name="childName" select="name(*[1])"/>
    <xsl:choose>            
        <xsl:when test="not(*|@*)"><xsl:choose><xsl:when test="$parent='Yes'"> <xsl:text>&quot;</xsl:text><xsl:value-of select="."/><xsl:text>&quot;</xsl:text></xsl:when>
                <xsl:otherwise>"<xsl:value-of select="name()"/>":"<xsl:value-of  select="."/>"</xsl:otherwise>
            </xsl:choose>           
        </xsl:when>                
        <xsl:when test="count(*[name()=$childName]) > 1">{ "<xsl:value-of  select="$childName"/>" :[<xsl:apply-templates select="*" mode="ArrayElement"/>] }</xsl:when>
        <xsl:otherwise>{
            <xsl:apply-templates select="@*"/>
            <xsl:apply-templates select="*"/>
            }</xsl:otherwise>
    </xsl:choose>
    <xsl:if test="following-sibling::*">,</xsl:if>
</xsl:template>

<!-- Attribute Property -->
<xsl:template match="@*">"<xsl:value-of select="name()"/>" : "<xsl:value-of select="."/>",
</xsl:template>
</xsl:stylesheet>