Question

我是XSLT的新手，有问题为下面的示例xml编写转换。

示例XML

<exports>
    <job>Job1</job>
    <date>Wed Sep 27 13:19:29 CDT 2017</date>
    <targets>
        <target server-name="">
            <templates>
                <template />
                <template />
            </templates>
            <components>
                <component>test2server</component>
                <component>test4server</component>
            </components>
        </target>
    </targets>
    <auditrun template="RHEL">
        <objects-audited />
        <Changed-targets>
            <target>test2server</target>
            <target>test4server</target>
        </Changed-targets>
        <unchanged-targets />
        <changed-audit-items>
            <node1>"file"</node1>
            <node2>"filename1"</node2>
            <node3>"filename1"</node3>
            <target>"test2server"</target>
            <node5>"lastnight"</node5>
        </changed-audit-items>
        <changed-audit-items>
            <node1>"registryval"</node1>
            <node2>"filename2"</node2>
            <node3>"filename2"</node3>
            <target>"test4server"</target>
            <node5>"morning"</node5>
        </changed-audit-items>
    </auditrun>
</exports>

所需输出

Changed Audit Items 1
node 1 - file
node 2 - filename1
node 3 - filename1
target - test2server
node 5 - lastnight

Changed Audit items 2
node 1 - registryval
node 2 - filename2
node 3 - filename2
target - test4server
node 5 - morning

我的转型 - 我正在使用以下for循环并添加了文字＆amp;值打印输出但未能获得所需的输出。

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output encoding="text" indent="no" />

    <xsl:template match="/exports">
        <xsl:apply-templates select="auditrun" />
    </xsl:template>

    <xsl:template match="auditrun/changed-audit-items">
        <xsl:for-each select="exports/auditrun/changed-audit-items/node3">
            <xsl:text disable-output-escaping="no"> </xsl:text>
            <xsl:text disable-output-escaping="yes"> Changed Targets: </xsl:text>
            <xsl:value-of select="//changed-audit-items/target/text()" disable-output-escaping="no" />
        </xsl:for-each>
        <xsl:variable select="1" name="counter" />
        <xsl:variable select="count(//changed-audit-items/node3)" name="CountAuditItems" />
        <Audit>
            <xsl:value-of select="count(//Changed-targets/target)" disable-output-escaping="no" />
            <xsl:text disable-output-escaping="no"> Total number of Changed Server Items: </xsl:text>
            <xsl:value-of select="count(//changed-audit-items/node3)" disable-output-escaping="no" />
        </Audit>
        <xsl:text disable-output-escaping="no"> </xsl:text>
        <xsl:text disable-output-escaping="no">node1 </xsl:text>
        <xsl:value-of select="//changed-audit-items/node1/text()" disable-output-escaping="yes" />
        <xsl:text disable-output-escaping="no"> </xsl:text>
        <xsl:text disable-output-escaping="no">node3: </xsl:text>
        <xsl:value-of select="//changed-audit-items/node3/text()" disable-output-escaping="no" />
        <xsl:text disable-output-escaping="no"> </xsl:text>
        <xsl:text disable-output-escaping="no">Node2: </xsl:text>
        <xsl:value-of select="//changed-audit-items/node2/text()" disable-output-escaping="no" />
        <xsl:text disable-output-escaping="no"> </xsl:text>
        <xsl:text disable-output-escaping="no">node4</xsl:text>
        <xsl:value-of select="//changed-audit-items/node4/text()" disable-output-escaping="no" />
        <xsl:text disable-output-escaping="no"> </xsl:text>
    </xsl:template>
</xsl:stylesheet>

我想请专家指导。

此致 [R

Answer 1

使用嵌套的<xsl:for-each>循环可以实现所需的输出。外部用于循环<changed-audit-items>节点以添加标题Changed Audit Items，内部用于循环<changed-audit-items>的所有子节点。内部循环将在输出中打印节点名称及其值。

以下是XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text" />

    <xsl:variable name="groupTitle" select="'Changed Audit Items '" />
    <xsl:variable name="newline" select="'&#10;'" />
    <xsl:variable name="hyphenSeparator" select="' - '" />

    <xsl:template match="/">
        <!-- loop for all <changed-audit-items> nodes -->
        <xsl:for-each select="//changed-audit-items">
            <xsl:value-of select="concat($groupTitle, position())" />
            <xsl:value-of select="$newline" />
            <!-- loop for all child nodes of <changed-audit-items> -->
            <xsl:for-each select="./*">
                <xsl:value-of select="local-name(.)" />
                <xsl:value-of select="$hyphenSeparator" />
                <xsl:value-of select="." disable-output-escaping="no"/>
                <xsl:value-of select="$newline" />
            </xsl:for-each>
            <xsl:value-of select="$newline" />
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

输出

Changed Audit Items 1
node1 - "file"
node2 - "filename1"
node3 - "filename1"
target - "test2server"
node5 - "lastnight"

Changed Audit Items 2
node1 - "registryval"
node2 - "filename2"
node3 - "filename2"
target - "test4server"
node5 - "morning"

具有递归节点所需输出的XSLT转换

1 个答案: