1

我似乎在 SUP 中遇到了关于它处理包含对象列表的 SOAP 响应的能力的限制，我想知道是否可以编写自定义 XSLT 来处理这个问题。我正在尝试通过 SOAP 通过 getProjectsNoSchemes 方法调用 Jira。此方法返回一个 RemoteProject 对象数组。最终，我希望能够将每个节点视为表中的一行，但不幸的是，我对 XSLT 的了解不够，无法知道这是否可能。我也不知道这是否是 SUP 中的可行解决方案。

SOAP 响应的示例如下：

<?xml version="1.0" encoding="utf-8"?>
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<soapenv:Body>
    <ns1:GetProjectsNoSchemesResponse
        soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
        xmlns:ns1="http://www.webserviceX.NET">
        <GetProjectsNoSchemesReturn
            soapenc:arrayType="ns2:RemoteProject[2]" xsi:type="soapenc:Array"
            xmlns:ns2="http://beans.soap.rpc.jira.atlassian.com" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/">
            <GetProjectsNoSchemesReturn href="#id0" />
            <GetProjectsNoSchemesReturn href="#id1" />
        </GetProjectsNoSchemesReturn>
    </ns1:GetProjectsNoSchemesResponse>
    <multiRef id="id0" soapenc:root="0"
        soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
        xsi:type="ns3:RemoteProject" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
        xmlns:ns3="http://beans.soap.rpc.jira.atlassian.com">
        <description xsi:type="xsd:string">Mobile Web Project POC
        </description>
        <id xsi:type="xsd:string">10034</id>
        <issueSecurityScheme xsi:type="ns3:RemoteScheme"
            xsi:nil="true" />
        <key xsi:type="xsd:string">XLIPOC</key>
        <lead xsi:type="xsd:string">benm</lead>
        <name xsi:type="xsd:string">Redacted Project</name>
        <notificationScheme xsi:type="ns3:RemoteScheme"
            xsi:nil="true" />
        <permissionScheme xsi:type="ns3:RemotePermissionScheme"
            xsi:nil="true" />
        <projectUrl xsi:type="xsd:string"></projectUrl>
        <url xsi:type="xsd:string">https://redacted.com/browse/REDACTED</url>
    </multiRef>
    <multiRef id="id1" soapenc:root="0"
        soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
        xsi:type="ns4:RemoteProject" xmlns:ns4="http://beans.soap.rpc.jira.atlassian.com"
        xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/">
        <description xsi:type="xsd:string"></description>
        <id xsi:type="xsd:string">10017</id>
        <issueSecurityScheme xsi:type="ns4:RemoteScheme"
            xsi:nil="true" />
        <key xsi:type="xsd:string">GIC</key>
        <lead xsi:type="xsd:string">gregm</lead>
        <name xsi:type="xsd:string">REDACTED</name>
        <notificationScheme xsi:type="ns4:RemoteScheme"
            xsi:nil="true" />
        <permissionScheme xsi:type="ns4:RemotePermissionScheme"
            xsi:nil="true" />
        <projectUrl xsi:type="xsd:string"></projectUrl>
        <url xsi:type="xsd:string">https://redacted.com/browse/REDACTED</url>
    </multiRef>
</soapenv:Body>

4

2 回答 2

1

SUP 期望 XML 采用非常特定的格式，因此我必须使用 for-each 构造遍历响应中的多引用：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="//ns1:getProjectsNoSchemesResponse">
    <data>
        <Record>
            <Field op_label="id" op_position="1" op_datatype="STRING" op_nullable="true">id</Field>
            <Field op_label="name" op_position="2" op_datatype="STRING" op_nullable="true">name</Field>
            <Field op_label="description" op_position="3" op_datatype="STRING" op_nullable="true">description</Field>
        </Record>
        <xsl:for-each select="//ns1:getProjectsNoSchemesResponse/getProjectsNoSchemesReturn/href*">
            <Record>
                <Field>
                    <xsl:attribute name="op_label">id</xsl:attribute>
                    <xsl:attribute name="op_position">1</xsl:attribute>
                    <xsl:attribute name="op_datatype">STRING</xsl:attribute>
                    <xsl:attribute name="op_nullable">true</xsl:attribute>
                    <xsl:value-of select="getProjectNoSchemesReturn/id" />
                </Field>
                <Field>
                    <xsl:attribute name="op_label">name</xsl:attribute>
                    <xsl:attribute name="op_position">2</xsl:attribute>
                    <xsl:attribute name="op_datatype">STRING</xsl:attribute>
                    <xsl:attribute name="op_nullable">true</xsl:attribute>
                    <xsl:value-of select="getProjectNoSchemesReturn/name" />
                </Field>
                <Field>
                    <xsl:attribute name="op_label">description</xsl:attribute>
                    <xsl:attribute name="op_position">3</xsl:attribute>
                    <xsl:attribute name="op_datatype">STRING</xsl:attribute>
                    <xsl:attribute name="op_nullable">true</xsl:attribute>
                    <xsl:value-of select="getProjectNoSchemesReturn/name" />
                </Field>
            </Record>
        </xsl:for-each>
    </data>
  </xsl:template>
</xsl:stylesheet>
于 2011-07-19T18:28:33.113 回答
1

是的，这是可能的，而且相对容易做到。

<multRef>下面是一个示例 XSLT，它为每个元素生成一个带有表格行的 HTML 文档。

的每个子元素<multiRef>首先使用标题列的元素名称呈现为表头，然后将每个<multiRef>子元素呈现为具有每个子元素的列的行：

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="/">
        <html>
            <head></head>
            <body>
                <table border="1">
                    <xsl:apply-templates select="*/*/multiRef[1]" mode="header"/>
                    <xsl:apply-templates select="*/*/multiRef" />
                </table>
            </body>
        </html>
    </xsl:template>

    <xsl:template match="multiRef" mode="header">
        <thead>
            <tr>
                <xsl:apply-templates mode="header"/>
            </tr>
        </thead>
    </xsl:template>

    <xsl:template match="multiRef/*" mode="header">
        <th>
            <xsl:value-of select="local-name()"/>
        </th>
    </xsl:template>

    <xsl:template match="multiRef">
        <tr>
            <xsl:apply-templates/>
        </tr>
    </xsl:template>

    <xsl:template match="multiRef/*">
        <td>
            <xsl:apply-templates/>
        </td>
    </xsl:template>
</xsl:stylesheet>

当应用于提供的示例 XML 时，它会生成以下 HTML：

<html>
    <head>
    <META http-equiv="Content-Type" content="text/html; charset=UTF-16">
    </head>
    <body>
        <table border="1">
            <thead>
                <tr>
                    <th>description</th>
                    <th>id</th>
                    <th>issueSecurityScheme</th>
                    <th>key</th>
                    <th>lead</th>
                    <th>name</th>
                    <th>notificationScheme</th>
                    <th>permissionScheme</th>
                    <th>projectUrl</th>
                    <th>url</th>
                </tr>
            </thead>
            <tr>
                <td>Mobile Web Project POC
                </td>
                <td>10034</td>
                <td></td>
                <td>XLIPOC</td>
                <td>benm</td>
                <td>Redacted Project</td>
                <td></td>
                <td></td>
                <td></td>
                <td>https://redacted.com/browse/REDACTED</td>
            </tr>
            <tr>
                <td></td>
                <td>10017</td>
                <td></td>
                <td>GIC</td>
                <td>gregm</td>
                <td>REDACTED</td>
                <td></td>
                <td></td>
                <td></td>
                <td>https://redacted.com/browse/REDACTED</td>
            </tr>
        </table>
    </body>
</html>
于 2011-07-15T01:46:11.060 回答