Question

我正在尝试使用for-each转换JSON对象，但没有任何根元素。这是我的对象，节点号可以更多。

[       
{
  "id": "1",
  "href": "string",
  "description": "string",
  "isBundle": true,
  "isCustomerVisible": true,
  "name": "string",
  "productSerialNumber": "string",
  "productNumber": "string",
  "startDate": "2018-11-27T13:26:22.783Z",
  "endDate": "2018-11-27T13:26:22.783Z",
  "status": "created"
},
{
  "id": "2",
  "href": "string",
  "description": "string",
  "isBundle": true,
  "isCustomerVisible": true,
  "name": "string",
  "productSerialNumber": "string",
  "productNumber": "string",
  "startDate": "2018-11-27T13:26:22.783Z",
  "endDate": "2018-11-27T13:26:22.783Z",
  "status": "created"
},
{
  "id": "3",
  "href": "string",
  "description": "string",
  "isBundle": true,
  "isCustomerVisible": true,
  "name": "string",
  "productSerialNumber": "string",
  "productNumber": "string",
  "startDate": "2018-11-27T13:26:22.783Z",
  "endDate": "2018-11-27T13:26:22.783Z",
  "status": "created"
}
]

我正在尝试使用xsl转换：

<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema"
    xmlns:fn="http://www.w3.org/2005/xpath-functions" xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">
    <xsl:output method="xml" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
    <xsl:template match="/">
        <jsonObject xmlns:json="http://json.org/">
            <requestResponse>
                <xsl:choose>
                    <xsl:when test="count(//id) > 0">
                        <returnCode>100</returnCode>
                        <returnMessage>SUCCESS</returnMessage>
                    </xsl:when>
                    <xsl:otherwise>
                        <returnCode>9999</returnCode>
                        <returnMessage>BUSINESS_FAULT</returnMessage>
                    </xsl:otherwise>
                </xsl:choose>
            </requestResponse>

            <xsl:for-each select="@*|node()">           
                    <xsl:if test="id">
                        <id>
                            <xsl:value-of select="/id" />
                        </id>
                    </xsl:if>
            </xsl:for-each>                     
        </jsonObject>
    </xsl:template>
</xsl:stylesheet>

如果我有对象的根名称，则可以处理它，但是在这里我需要一些帮助。预先感谢您的任何想法！

Answer 1

在无根名称数组json中，可以在xslt下面使用<id>标签。关键点是<jsonArray><jsonElement></jsonElement></jsonArray>方法。但是您需要注意<jsonElement>，它应该放在<xsl:for-each select="//jsonArray/jsonElement">块中。为了不暴露每种症候群，您不应为其中的字段编写根元素。例如<xsl:if test="id">。

<xsl:stylesheet version="2.0"
        xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
        xmlns:xs="http://www.w3.org/2001/XMLSchema"
        xmlns:fn="http://www.w3.org/2005/xpath-functions"
        xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
        xmlns:ns1="http://www.qas.com/OnDemand-2011-03"
        xmlns:json="http://json.org">
    <xsl:output method="xml" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
    <xsl:template match="/">
        <jsonArray>
            <xsl:for-each select="//jsonArray/jsonElement">
                <jsonElement>   
                    <xsl:if test="id">
                        <id><xsl:value-of select="id" /> </id>
                    </xsl:if>
                </jsonElement>
            </xsl:for-each>
        </jsonArray>
    </xsl:template>
</xsl:stylesheet>

转换没有根名的JSON对象

1 个答案: