Question

我正在一个将JSON消息转换为XML的项目中，并且我有以下代码：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet 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:exslt="http://exslt.org/common" extension-element-prefixes="exsl"
                xmlns:exsl="http://exslt.org/common" xmlns:fcn="http://www.re.com/2018/local-functions"
                exclude-result-prefixes="xs fn fcn exsl" version="3.0">
<xsl:param as="xs:string" name="json">[{
    "results": [
        {
            "id": "750584f1-ce77-4e55-92b4-aae0c0ac37ae",
            "profileId": "4aebb5de-5461-425c-b75e-959cef17048b",
            "_embedded": {
                "notes": {
                    "results": [
                        {
                            "content": "<div>General notes<div>"
                        },
                        {
                            "content": "General notes"
                        }
                    ],
                    "paging": {
                        "pageNumber": 1,
                        "pageSize": 20,
                        "totalCount": 2
                    }
                }
            }
        },
         {
            "id": "750584f1-ce77-4e55-92b4-aae0c0ac37ae",
            "profileId": "4aebb5de-5461-425c-b75e-959cef17048b",
            "_embedded": {
                "notes": {
                    "results": [
                        {
                            "content": "General notes"
                        },
                        {
                            "content": "General notes"
                        }
                    ],
                    "paging": {
                        "pageNumber": 1,
                        "pageSize": 20,
                        "totalCount": 2
                    }
                }
            }

    ]
}]
</xsl:param>
<xsl:variable name="input-as-xml" select="json-to-xml($json)"/>
<xsl:output indent="yes" method="text"/>
 <xsl:template match="/" name="xsl:initial-template">
        <xsl:variable name="transformed-xml">
            <array xmlns="http://www.w3.org/2005/xpath-functions">
                <xsl:for-each select="$input-as-xml/fn:array/fn:map/fn:array[@key = 'results']/*">
                    <map xmlns="http://www.w3.org/2005/xpath-functions">
                      <string key="id">
                        <xsl:value-of select="*[@key='id']" />
                      </string>
                    </map>
                </xsl:for-each>
            </array>
        </xsl:variable>
        <xsl:value-of select="xml-to-json($transformed-xml, map {'indent': true()})"/>
    </xsl:template>
</xsl:stylesheet>

当我尝试运行样式表时，我收到此错误消息：

Error executing XSLT at line 7 : A sequence of more than one item is not allowed as the value of variable $json ("[{ "results": [ {...", "General notes")

我猜我在匹配模板时犯了一些错误，但是我无法弄清楚我到底在做什么错。从我的测试看来，<div>标签似乎也加剧了这个问题。一些帮助，将不胜感激。

Answer 1

在这种情况下（嵌入式标记）是必要的，并且通常更安全的做法是将您拥有的JSON数据（如果您将其内联）放入CDATA部分：

LoadModule php7

不允许超过一个项目的序列进行XSLT转换

1 个答案: