C#/ .Net - 解析XML / XML的快捷方式 - > JSON

时间:2013-01-15 13:24:52

标签: c# xml json xml-parsing xmlconvert

我是XML的新手,因为现在我只和Json一起工作:( 我有一个XML文件,如下所示:

<AdapterCards>
    <cards type="MCS">
        <card>
            <id>id1</id>
            <description>desc1</description>
            <mccode>code1</mccode>
        </card>
        <card>
            <id>id2</id>
            <description>desc2</description>
            <mccode>code2</mccode>
        </card>
    </cards>
    <cards type="MCM">
        <card>
            <id>id3</id>
            <description>desc3</description>
            <mccode>code3</mccode>
        </card>
        <card>
            <id>id4</id>
            <description>desc4</description>
            <mccode>code4</mccode>
        </card>
    </cards>
    <cards type="F"/>
    <cards type="B"/>
</AdapterCards>

我想将它解析为一个json字符串,它应该是这样的:

{[{'type': 'mcs', 'id': 'id1', 'description': 'desc1', 'mccode': 'code1'},
  {'type': 'mcs', 'id': 'id2', 'description': 'desc2', 'mccode': 'code2'},
  {'type': 'mcm', 'id': 'id3', 'description': 'desc3', 'mccode': 'code3'},
  {'type': 'mcm', 'id': 'id4', 'description': 'desc4', 'mccode': 'code4'}
]}

我的问题是我根本没有使用XML(是的,对我很羞耻)。 你能不能给我一些关于如何快速解析xml的一些线索(我把它放在Stream上,我把它上传到服务器上)。我已经搜索了一些XML到Json转换器,但是根据我的需要不可能很好,因为我需要一种“特殊”格式。

谢谢你的答案:)!我正在使用C#。

3 个答案:

答案 0 :(得分:2)

我最近必须为同类型的东西写一个定制的解决方案。我使用XSLT完成了它,使用XslCompiledTransform类来运行XML输入并输出JSON。

这需要一些工作,但应该帮助你掌握基础知识(它是从我的工作中复制粘贴,改为几乎满足你的需要):

<强> AdapterCards.XML

<AdapterCards>
    <cards type="MCS">
        <card>
            <id>id1</id>
            <description>desc1</description>
            <mccode>code1</mccode>
        </card>
        <card>
            <id>id2</id>
            <description>desc2</description>
            <mccode>code2</mccode>
        </card>
    </cards>
    <cards type="MCM">
        <card>
            <id>id3</id>
            <description>desc3</description>
            <mccode>code3</mccode>
        </card>
        <card>
            <id>id4</id>
            <description>desc4</description>
            <mccode>code4</mccode>
        </card>
    </cards>
    <cards type="F"/>
    <cards type="B"/>
</AdapterCards>

<强> AdapterCards.XSL

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="no" omit-xml-declaration="yes" method="text" encoding="UTF-8" media-type="text/x-json" />

<xsl:variable name="smallcase" select="'abcdefghijklmnopqrstuvwxyz'" />
<xsl:variable name="uppercase" select="'ABCDEFGHIJKLMNOPQRSTUVWXYZ'" />

    <!-- Main template for escaping strings; used by above template and for object-properties 
       Responsibilities: placed quotes around string, and chain up to next filter, escape-bs-string -->
    <xsl:template name="escape-string">
        <xsl:param name="s"/>
        <xsl:text>"</xsl:text>
        <xsl:call-template name="escape-bs-string">
            <xsl:with-param name="s" select="$s"/>
        </xsl:call-template>
        <xsl:text>"</xsl:text>
    </xsl:template>

    <!-- Escape the backslash (\) before everything else. -->
    <xsl:template name="escape-bs-string">
        <xsl:param name="s"/>
        <xsl:choose>
            <xsl:when test="contains($s,'\')">
                <xsl:call-template name="escape-quot-string">
                    <xsl:with-param name="s" select="concat(substring-before($s,'\'),'\\')"/>
                </xsl:call-template>
                <xsl:call-template name="escape-bs-string">
                    <xsl:with-param name="s" select="substring-after($s,'\')"/>
                </xsl:call-template>
            </xsl:when>
            <xsl:otherwise>
                <xsl:call-template name="escape-quot-string">
                    <xsl:with-param name="s" select="$s"/>
                </xsl:call-template>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

    <!-- Escape the double quote ("). -->
    <xsl:template name="escape-quot-string">
        <xsl:param name="s"/>
        <xsl:choose>
            <xsl:when test="contains($s,'&quot;')">
                <xsl:call-template name="encode-string">
                    <xsl:with-param name="s" select="concat(substring-before($s,'&quot;'),'\&quot;')"/>
                </xsl:call-template>
                <xsl:call-template name="escape-quot-string">
                    <xsl:with-param name="s" select="substring-after($s,'&quot;')"/>
                </xsl:call-template>
            </xsl:when>
            <xsl:otherwise>
                <xsl:call-template name="encode-string">
                    <xsl:with-param name="s" select="$s"/>
                </xsl:call-template>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

    <!-- Replace tab, line feed and/or carriage return by its matching escape code. Can't escape backslash
       or double quote here, because they don't replace characters (&#x0; becomes \t), but they prefix 
       characters (\ becomes \\). Besides, backslash should be seperate anyway, because it should be 
       processed first. This function can't do that. -->
    <xsl:template name="encode-string">
        <xsl:param name="s"/>
        <xsl:choose>
            <!-- tab -->
            <xsl:when test="contains($s,'&#x9;')">
                <xsl:call-template name="encode-string">
                    <xsl:with-param name="s" select="concat(substring-before($s,'&#x9;'),'\t',substring-after($s,'&#x9;'))"/>
                </xsl:call-template>
            </xsl:when>
            <!-- line feed -->
            <xsl:when test="contains($s,'&#xA;')">
                <xsl:call-template name="encode-string">
                    <xsl:with-param name="s" select="concat(substring-before($s,'&#xA;'),'\n',substring-after($s,'&#xA;'))"/>
                </xsl:call-template>
            </xsl:when>
            <!-- carriage return -->
            <xsl:when test="contains($s,'&#xD;')">
                <xsl:call-template name="encode-string">
                    <xsl:with-param name="s" select="concat(substring-before($s,'&#xD;'),'\r',substring-after($s,'&#xD;'))"/>
                </xsl:call-template>
            </xsl:when>
            <xsl:otherwise>
                <xsl:value-of select="$s"/>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>


    <xsl:template match="card">
        <xsl:text>{</xsl:text>

        <xsl:text>"type":</xsl:text>
        <xsl:call-template name="escape-string">
            <xsl:with-param name="s" select="translate(../@type, $uppercase, $smallcase)"/>
        </xsl:call-template>

        <xsl:text>,"id":</xsl:text>
        <xsl:call-template name="escape-string">
            <xsl:with-param name="s" select="id"/>
        </xsl:call-template>

        <xsl:text>,"description":</xsl:text>
        <xsl:call-template name="escape-string">
            <xsl:with-param name="s" select="description"/>
        </xsl:call-template>

        <xsl:text>,"mccode":</xsl:text>
        <xsl:call-template name="escape-string">
            <xsl:with-param name="s" select="mccode"/>
        </xsl:call-template>

        <xsl:if test="following::card">},</xsl:if>
        <xsl:if test="not(following::card)">}</xsl:if>
    </xsl:template>


    <xsl:template match="/AdapterCards">
        <xsl:text>{[</xsl:text>

        <xsl:apply-templates select="cards/card" />

        <xsl:text>]}</xsl:text>
  </xsl:template>
</xsl:stylesheet>

<强> C#

// Load XML document
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("AdapterCards.XML");

// Transform the XML into JSON
XslCompiledTransform transformer = new XslCompiledTransform();
using (var xslStylesheetFile = File.Open("AdapterCards.XSL", FileMode.Open))
{
    using (var xmlReader = new XmlTextReader(xslStylesheetFile))
    {
        transformer.Load(xmlReader);
    }
}
var sourceNavigator = xmlDoc.CreateNavigator();
using (MemoryStream ms = new MemoryStream())
{
    transformer.Transform(sourceNavigator, null, ms);
    ms.Position = 0;
    using (var sr = new StreamReader(ms))
    {
        return sr.ReadToEnd(); // <-- this is your JSON
    }
}

我在Notepad ++中执行了上面的XSL并获得了以下内容:

{[{"type":"mcs","id":"id1","description":"desc1","mccode":"code1"},
{"type":"mcs","id":"id2","description":"desc2","mccode":"code2"},
{"type":"mcm","id":"id3","description":"desc3","mccode":"code3"},
{"type":"mcm","id":"id4","description":"desc4","mccode":"code4"}]}

正如你所看到的,在他们需要的地方缺少逗号。但它几乎就在那里!!

更新了XSL和输出,它现在可以在正确的位置使用逗号。接下来要修复的是“类型”的情况。我想你可能需要使用XSL 2.0才能访问xpath function fn:lower-case()

EDIT3 :已完成 - 案例现已转换为较低with the help of this answer

参考文献:

答案 1 :(得分:0)

使用包含帮助方法的JsonConvert类来实现这个目的:

// To convert an XML node contained in string xml into a JSON string   
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
string jsonText = JsonConvert.SerializeXmlNode(doc);

// To convert JSON text contained in string json into an XML node
XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(json);

此处的文档:Converting between JSON and XML with Json.NET

答案 2 :(得分:-3)

XDocument xDoc = XDocument.Load(queryURL);                               
var x1 = from el in xDoc.Root.Descendants("results").Descendants("div").Descendants("div").Descendants("span").Attributes("class")
where el.Value == "pr"
  select el.Parent.Value;                
   stockFeed.Price = Decimal.Parse(x1.First().ToString());

LINQ是提取数据的最佳方式。