如何使用XSLT替换XML的节点名称中的字符-更改根元素

问题描述 投票:0回答:1

您好,我需要使用XSLT映射来转换XML,但我不知道该怎么做……您能帮我吗?>

我的源XML:

    <?xml version="1.0" encoding="UTF-8"?>
<order-header>
    <id type="integer">1</id>
    <created-at type="dateTime">2020-01-25T18:59:02-03:00</created-at>
    <updated-at type="dateTime">2020-04-23T15:28:13-03:00</updated-at>
    <po-number>1</po-number>
    <price-hidden type="boolean">false</price-hidden>
    <acknowledged-flag type="boolean">false</acknowledged-flag>
    <acknowledged-at nil="true"/>
    <status>issued</status>
    <transmission-status>sent_via_email</transmission-status>
    <version type="integer">1</version>
    <internal-revision type="integer">3</internal-revision>
    <exported type="boolean">false</exported>
    <last-exported-at nil="true"/>
    <payment-method>invoice</payment-method>
    <ship-to-attention>10422282000179</ship-to-attention>
    <coupa-accelerate-status nil="true"/>
    <change-type>revision</change-type>
    <transmission-method-override>supplier_default</transmission-method-override>
    <transmission-emails></transmission-emails>
</order-header>

我需要这个结果:

<?xml version="1.0" encoding="UTF_8"?>
<ns0:MT_CRIAALTSAP_RES xmlns:ns0="urn:xxxx:xxxxxx">
<order_header>
    <id type="integer">1</id>
    <created_at type="dateTime">2020-01-25T18:59:02-03:00</created_at>
    <updated_at type="dateTime">2020-04-23T15:28:13-03:00</updated_at>
    <po_number>1</po_number>
    <price_hidden type="boolean">false</price_hidden>
    <acknowledged_flag type="boolean">false</acknowledged_flag>
    <acknowledged_at nil="true"/>
    <status>issued</status>
    <transmission_status>sent_via_email</transmission_status>
    <version type="integer">1</version>
    <internal_revision type="integer">3</internal_revision>
    <exported type="boolean">false</exported>
    <last_exported_at nil="true"/>
    <payment_method>invoice</payment_method>
    <ship_to_attention>10422282000179</ship_to_attention>
    <coupa_accelerate_status nil="true"/>
    <change_type>revision</change_type>
    <transmission_method_override>supplier_default</transmission_method_override>
    <transmission_emails></transmission_emails>
</order_header>
</ns0:MT_CRIAALTSAP_RES>

我在下面使用此代码,这是我在论坛中获取的代码:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema" version="1.0">


    <xsl:output method="xml" indent="yes"/>


    <xsl:template match="*[contains(local-name(.), '-')]">
            <xsl:element name="{translate(local-name(),'-','_')}">
                <xsl:copy-of select="@*"/>
                <xsl:apply-templates/>

            </xsl:element>

    </xsl:template>


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

</xsl:stylesheet>

代码运行正常,但我需要根元素为<ns0:MT_CRIAALTSAP_RES xmlns:ns0="urn:xxxx:xxxxxx">

您能帮我解决这个问题吗?

先花点时间

您好,我需要使用XSLT映射来转换XML,但我不知道该怎么做……可以帮助我吗?我的源XML:...] >>] >>

[添加带有match="/"的模板,该模板创建想要的根元素作为文字结果元素,并且内部创建<xsl:apply-templates/>

xml xslt transformation
1个回答
0
投票

[添加带有match="/"的模板,该模板创建想要的根元素作为文字结果元素,并且内部创建<xsl:apply-templates/>

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