Question

以下是xml

<?xml version="1.0" encoding="UTF-8"?>
<hotels>
<hotel>
    <rooms>
        <room>
            <rates>
                <rate id="1" adults="1" child="0"></rate>
                <rate id="2" adults="2" child="0"></rate>
                <rate id="3" adults="1" child="0"></rate>
            </rates>
        </room>
        <room>
            <rates>
                <rate id="4" adults="1" child="0"></rate>
                <rate id="5" adults="2" child="0"></rate>
                <rate id="6" adults="2" child="0"></rate>
            </rates>
        </room>
    </rooms>
</hotel>
</hotels>

我在xslt下面尝试。 XSLT工作正常。我根据需要获得结果

<?xml version="1.0" encoding="ISO-8859-1"?>
        <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

            <xsl:key name="by-occupancy" match="rooms/room/rates/rate" use="concat(generate-id(parent::rooms),\'|\',@adults)"/>

            <xsl:template match="hotels/hotel">
                <hotel>
                <xsl:apply-templates select="rooms/room/rates/rate[generate-id() = generate-id(key(\'by-occupancy\', concat(generate-id(parent::rooms),\'|\',@adults))[1])]" mode="fun_options"/>
                </hotel>
            </xsl:template>

            <xsl:template match="rate" mode="fun_options">
            <rates>
                <xsl:for-each select="key(\'by-occupancy\', concat(generate-id(parent::rooms),\'|\',@adults))">
                <rate><xsl:value-of select="@id"/>-<xsl:value-of select="@adults"/></rate>
                </xsl:for-each>
            </rates>
            </xsl:template>

        </xsl:stylesheet>

但是当我在xml中应用xmlns：xsi时。

结果变空

<?xml version="1.0" encoding="UTF-8"?>
<hotels xmlns="http://www.test.com/schemas/messages" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" schemaLocation="http://www.test.com/schemas/messages">
<hotel >
    <rooms>
        <room>
            <rates>
                <rate id="1" adults="1" child="0"></rate>
                <rate id="2" adults="2" child="0"></rate>
                <rate id="3" adults="1" child="0"></rate>
            </rates>
        </room>
        <room>
            <rates>
                <rate id="4" adults="1" child="0"></rate>
                <rate id="5" adults="2" child="0"></rate>
                <rate id="6" adults="2" child="0"></rate>
            </rates>
        </room>
    </rooms>
</hotel>
</hotels>

低于xslt我试试。这里我applay exclude-result-prefixes =＆＃34; ms xsi＆＃34;

    <?xml version="1.0" encoding="ISO-8859-1"?>
        <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:ms="http://www.test.com/schemas/messages" exclude-result-prefixes="ms xsi">
  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

            <xsl:key name="by-occupancy" match="ms:rooms/ms:room/ms:rates/ms:rate" use="concat(generate-id(parent::ms:rooms),\'|\',@adults)"/>

            <xsl:template match="ms:hotels/ms:hotel">
                <hotel>
                <xsl:apply-templates select="ms:rooms/ms:room/ms:rates/ms:rate[generate-id() = generate-id(key(\'by-occupancy\', concat(generate-id(parent::ms:rooms),\'|\',@adults))[1])]" mode="fun_options"/>
                </hotel>
            </xsl:template>

            <xsl:template match="ms:rate" mode="fun_options">
            <rates>
                <xsl:for-each select="key(\'by-occupancy\', concat(generate-id(parent::ms:rooms),\'|\',@adults))">
                <rate><xsl:value-of select="@id"/>-<xsl:value-of select="@adults"/></rate>
                </xsl:for-each>
            </rates>
            </xsl:template>

        </xsl:stylesheet>

我需要结果如下

<hotel>
  <rates>
    <rate>1-1</rate>
    <rate>3-1</rate>
    <rate>4-1</rate>
  </rates>
  <rates>
    <rate>2-2</rate>
    <rate>5-2</rate>
    <rate>6-2</rate>
  </rates>
</hotel>

Answer 1

我看不出那些转义字符串中的任何字符串是否会编译。你试图在密钥中使用的父项不存在，所以我认为你想要一个祖先：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:ms="http://www.test.com/schemas/messages" exclude-result-prefixes="ms xsi">
  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

            <xsl:key name="by-occupancy" match="ms:rooms/ms:room/ms:rates/ms:rate" use="concat(generate-id(ancestor::ms:rooms),'|',@adults)"/>

            <xsl:template match="ms:hotels/ms:hotel">
                <hotel>
                <xsl:apply-templates select="ms:rooms/ms:room/ms:rates/ms:rate[generate-id() = generate-id(key('by-occupancy', concat(generate-id(ancestor::ms:rooms),'|',@adults))[1])]" mode="fun_options"/>
                </hotel>
            </xsl:template>

            <xsl:template match="ms:rate" mode="fun_options">
            <rates>
                <xsl:for-each select="key('by-occupancy', concat(generate-id(ancestor::ms:rooms),'|',@adults))">
                <rate><xsl:value-of select="@id"/>-<xsl:value-of select="@adults"/></rate>
                </xsl:for-each>
            </rates>
            </xsl:template>

        </xsl:stylesheet>

如何使用xml

1 个答案: