使用xslt从xml中删除特定的xmlns

Question

<Orders xmlns="hi/hi" xmlns:shipping="hello/hi" xmlns:message="hello/hi" message:timestamp="2018-04-23T14:28:39Z">
    <orderLineStatusUpdate orderNumber="CCORD002" line="0">
        <orderLineStatus>inPicking</orderLineStatus>
        <shipmentNumber>1</shipmentNumber>
        <leadTime unit="days">
            <shipping:min>1</shipping:min>
            <shipping:max>1</shipping:max>
        </leadTime>
    </orderLineStatusUpdate>
</Orders>

预期产出：

<Orders message:timestamp="2018-04-23T14:28:39Z">
    <orderLineStatusUpdate orderNumber="CCORD002" line="0">
        <orderLineStatus>inPicking</orderLineStatus>
        <shipmentNumber>1</shipmentNumber>
        <leadTime unit="days">
            <shipping:min>1</shipping:min>
            <shipping:max>1</shipping:max>
        </leadTime>
    </orderLineStatusUpdate>
</Orders>

这是我正在使用的XSLT。它会从代码中删除shipping:min和shipping:max。我只想删除xmlns。

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

    <xsl:template match="*">
        <xsl:element name="{local-name(.)}">
            <xsl:apply-templates select="@* | node()" />
        </xsl:element>
    </xsl:template>
    <xsl:template match="@*">
        <xsl:attribute name="{local-name(.)}">
        <xsl:value-of select="." />
       </xsl:attribute>
    </xsl:template>
    <xsl:template match="@*[local-name(.)='message']" />
</xsl:stylesheet>

但我的XSLT代码正在删除shipping:min和shipping:max