xsl 将 xml 转换为 json [英] xsl to convert xml to json

问题描述

输入 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>

输出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"
  }
]
}
}

如何编写通用的 xslt 样式表，将输入的 xml 转换为 json

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

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

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

推荐答案

我从 here 应该可以帮助您将 XML 转换为 JSON.谢谢:)

The below XSLT which I copied and pasted from here should help you to convert XML to JSON. Thanks :)

<?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>

这篇关于xsl 将 xml 转换为 json的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！