使用XSLT将XML转换为JSON，并删除JSON中的特定对名称

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <person>
        <name>John Doe</name>
        <age>30</age>
        <address>
            <street>123 Main St</street>
            <city>Anytown</city>
        </address>
    </person>
</root>

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:json="http://www.ibm.com/xmlns/prod/2009/jsonx">

    <xsl:output method="text" encoding="UTF-8"/>

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

    <xsl:template match="person">
        "name": "<xsl:value-of select="name"/>",
        "age": "<xsl:value-of select="age"/>",
        "address": {
            <xsl:apply-templates select="address"/>
        }
    </xsl:template>

    <xsl:template match="address">
        "street": "<xsl:value-of select="street"/>",
        <xsl:if test="city">
            "city": "<xsl:value-of select="city"/>"
        </xsl:if>
    </xsl:template>

</xsl:stylesheet>

{
    "person": {
        "name": "John Doe",
        "age": "30",
        "address": {
            "street": "123 Main St"
        }
    }
}