xml 转换 - 组合节点 [英] xml transformation - combining nodes

问题描述

我有以下测试用例的数据集

i have the following data set for a test case

<?xml version="1.0" encoding="UTF-8"?>
<STUDENTS_LIST dateGenerated="2014-01-01">
    <STUDENTS>
        <MALE>
            <STUDENT_ID>10000</STUDENT_ID>
            <F_NAME>REGGIE</F_NAME>
            <M_NAME/>
            <L_NAME>MILLER</L_NAME>
            <DOB>
                <YEAR>1980</YEAR>
            </DOB>
            <STUDENT_TYPE>MORNING</STUDENT_TYPE>
            <STUDENT_REF>BLUE</STUDENT_REF>
            <JOIN_DATE>04-20-2000</JOIN_DATE>
            <NOTES/>
            <FATHER_NAME>
                <NAME>MILLER A</NAME>
            </FATHER_NAME>
            <MOTHER_NAME>
                <NAME>MILLER B</NAME>
            </MOTHER_NAME>
            <REFRESH_DATE>04-14-2014</REFRESH_DATE>
            <CORE_SUBJECTS>
                <SUBJECT_A>CALCULUS A</SUBJECT_A>
                <SUBJECT_B>CALCULUS B</SUBJECT_B>
                <SUBJECT_C>PERFORMING ARTS</SUBJECT_C>
            </CORE_SUBJECTS>
            <OPT_SUBJECTS>
                <SUBJECT_A>AMERICAN HISTORY</SUBJECT_A>
                <SUBJECT_B/>
                <SUBJECT_C/>
            </OPT_SUBJECTS>
            <STUDENT_KEY>ABC10000-1</STUDENT_KEY>
            <STUDENT_KEY_CREATION_DATE>04-20-2000</STUDENT_KEY_CREATION_DATE>
        </MALE>
        <FEMALE>
            <STUDENT_ID>10001</STUDENT_ID>
            <F_NAME>REGINA</F_NAME>
            <M_NAME/>
            <L_NAME>MOON</L_NAME>
            <DOB>
                <YEAR>1980</YEAR>
            </DOB>
            <STUDENT_TYPE>MORNING</STUDENT_TYPE>
            <STUDENT_REF>BLUE</STUDENT_REF>
            <JOIN_DATE>04-20-2000</JOIN_DATE>
            <NOTES/>
            <FATHER_NAME>
                <NAME>MOON A</NAME>
            </FATHER_NAME>
            <MOTHER_NAME>
                <NAME>MOON B</NAME>
            </MOTHER_NAME>
            <REFRESH_DATE>04-14-2014</REFRESH_DATE>
            <CORE_SUBJECTS>
                <SUBJECT_A>CALCULUS A</SUBJECT_A>
                <SUBJECT_B>CALCULUS B</SUBJECT_B>
                <SUBJECT_C>PERFORMING ARTS</SUBJECT_C>
            </CORE_SUBJECTS>
            <OPT_SUBJECTS>
                <SUBJECT_A>AMERICAN HISTORY</SUBJECT_A>
                <SUBJECT_B/>
                <SUBJECT_C/>
            </OPT_SUBJECTS>
            <STUDENT_KEY>ABC10000-1</STUDENT_KEY>
            <STUDENT_KEY_CREATION_DATE>04-20-2000</STUDENT_KEY_CREATION_DATE>
        </FEMALE>
    </STUDENTS>
</STUDENTS_LIST>

需要使用xslt转换成如下xml

which needs to be converted into the following xml using xslt

<?xml version="1.0" encoding="UTF-8"?>
<classified date="2014-01-01">
    <students>
        <student>
            <student_id>10000</student_id>
            <fname>REGGIE</fname>
            <mname/>
            <lname>MILLER</lname>
            <sex>male</sex>
            <data>student_type: MORNING, student_ref: BLUE, father_name: MILLER A,
                mother_name: MILLER B, core_subjects : CALCULUS A, CALCULUS B, PERFORMING ARTS,
                opt_subjects: AMERICAN HISTORY, student_key: ABC10000-1, student_key_creation_date:
                04-20-2000, rec_refresh_date: 04-14-2014 </data>
            <dob>
                <year>1980</year>
            </dob>
            <joindate>04-20-2000</joindate>
        </student>
        <student>
            <student_id>10001</student_id>
            <fname>REGINA</fname>
            <mname/>
            <lname>MOON</lname>
            <sex>female</sex>
            <notes>student_type: MORNING, student_ref: BLUE, father_name: MOON A,
                mother_name: MOON B, core_subjects : CALCULUS A, CALCULUS B, PERFORMING ARTS,
                opt_subjects: AMERICAN HISTORY, student_key: ABC10000-1, student_key_creation_date:
                04-20-2000, rec_refresh_date: 04-14-2014 </notes>
            <dob>
                <year>1980</year>
            </dob>
            <joindate>04-20-2000</joindate>
        </student>
    </students>
</classified>

我使用@Joel M. Lamsen 提供的以下 XSLT 来回答我的其他问题之一.

I use the following XSLT provided by @Joel M. Lamsen in response to one of my other questions.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

    <xsl:strip-space elements="*"/>
    <xsl:output indent="yes"/>

    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="STUDENT_TYPE|STUDENT_REF|FATHER_NAME|MOTHER_NAME|STUDENT_KEY|STUDENT_KEY_CREATION_DATE">
<xsl:value-of select="concat(name(), ': ', .)"/><xsl:text>&#xA;</xsl:text>
    </xsl:template>

    <xsl:template match="NOTES"/>

    <xsl:template match="REFRESH_DATE">
        <xsl:value-of select="concat('REC_REFRESH_DATE', ': ', .)"/><xsl:text>&#xA;</xsl:text>
    </xsl:template>

    <xsl:template match="CORE_SUBJECTS|OPT_SUBJECTS">
        <xsl:value-of select="concat(name(), ': ')"/>
        <xsl:for-each select="*[.!='']">
            <xsl:choose>
                <xsl:when test="position() != last()">
                    <xsl:value-of select="."/>
                    <xsl:text>, </xsl:text>
                </xsl:when>
                <xsl:otherwise>
                    <xsl:value-of select="."/>
                </xsl:otherwise>
            </xsl:choose>
        </xsl:for-each>
        <xsl:text>&#xA;</xsl:text>
    </xsl:template>

    <xsl:template match="STUDENT">
        <xsl:copy>
            <xsl:apply-templates select="STUDENT_ID|F_NAME|M_NAME|L_NAME|NOTES|DOB"/>
            <NOTES>
                <xsl:text>&#xA;</xsl:text>
                <xsl:apply-templates select="STUDENT_TYPE"/>
                <xsl:apply-templates select="STUDENT_REF"/>
                <xsl:apply-templates select="FATHER_NAME"/>
                <xsl:apply-templates select="MOTHER_NAME"/>
                <xsl:apply-templates select="CORE_SUBJECTS"/>
                <xsl:apply-templates select="OPT_SUBJECTS"/>
                <xsl:apply-templates select="STUDENT_KEY"/>
                <xsl:apply-templates select="STUDENT_KEY_CREATION_DATE"/>
                <xsl:apply-templates select="REFRESH_DATE"/>
            </NOTES>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

转换后需要按照准确的映射顺序，并且需要按男性/女性进行排序.

the converted needs to be in the exact mapping order and needs to be sorted on male/female basis.

在这方面的任何帮助将不胜感激.

any help in this regard would be appreciated.

前面的例子已经解决了.

a previous example has already been solved.

问候雷吉.

推荐答案

我已经改编了@helderdarocha 下面的样式表:

I have adapted the stylesheet below from @helderdarocha:

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

    <xsl:strip-space elements="*"/>

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

    <xsl:template match="/">
        <classified date="{STUDENTS_LIST/@dateGenerated}">
            <students>
                <xsl:apply-templates select="STUDENTS_LIST/STUDENTS/MALE|STUDENTS_LIST/STUDENTS/FEMALE"/>
            </students>
        </classified>
    </xsl:template>

    <xsl:template match="*" mode="notes">
        <xsl:value-of select="translate(name(.), $uppercase, $lowercase)"/>
        <xsl:text>: </xsl:text>
        <xsl:choose>
            <xsl:when test="child::text()">
                <xsl:value-of select="child::text()"/>
                <xsl:if test="not(self::REFRESH_DATE)">
                    <xsl:text>, </xsl:text> 
                </xsl:if>
            </xsl:when>
            <xsl:otherwise>
                <xsl:for-each select="*[.!='']">
                    <xsl:value-of select="."/>
                    <xsl:text>, </xsl:text> 
                </xsl:for-each>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

    <xsl:template match="*" mode="tag">
        <xsl:element name="{translate(name(.), $uppercase, $lowercase)}">
            <xsl:choose>
                <xsl:when test="name() = 'DOB'">
                    <year><xsl:value-of select="."/></year>
                </xsl:when>
                <xsl:otherwise>
                    <xsl:value-of select="."/>
                </xsl:otherwise>
            </xsl:choose>
        </xsl:element>
    </xsl:template>

    <xsl:template match="MALE|FEMALE">
        <student>
            <xsl:apply-templates select="STUDENT_ID|F_NAME|M_NAME|L_NAME" mode="tag"/>
            <sex><xsl:value-of select="translate(name(), $uppercase, $lowercase)"/></sex>

            <xsl:if test="*[not(STUDENT_ID|F_NAME|M_NAME|L_NAME|DOB|JOIN_DATE)]">
                <notes>
                    <xsl:apply-templates select="STUDENT_TYPE|STUDENT_REF|FATHER_NAME|MOTHER_NAME|CORE_SUBJECTS|OPT_SUBJECTS|STUDENT_KEY|STUDENT_KEY_CREATION_DATE" mode="notes"/>
                    <xsl:apply-templates select="REFRESH_DATE" mode="notes"/>
                </notes>
            </xsl:if>
            <xsl:apply-templates select="DOB|JOIN_DATE" mode="tag"/>

        </student>
    </xsl:template>


</xsl:stylesheet>

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