XSLT从XML中删除特定属性

Question

i am new to XSLT need help in removing an attribute from XML. Nothing to be changed in the output from XSLT, except removing the attribute

xsi:schemaLocation="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03 
pain.001.001.03.xsd" 

from the 'Document'

<?xml version="1.0" encoding="utf-8"?>
<Document 
xsi:schemaLocation="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03 
pain.001.001.03.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-
instance" xmlns="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03">
<CstmrCdtTrfInitn>

Answer 1

I think you want to remove the namespace from the root element, try the below code to remove the namespace attributes

<xsl:template match="comment()">
    <xsl:copy>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="*">
    <!-- remove element prefix -->
    <xsl:element name="{local-name()}">
      <!-- process attributes -->
      <xsl:for-each select="@*">
        <!-- remove attribute prefix -->
        <xsl:attribute name="{local-name()}">
          <xsl:value-of select="."/>
        </xsl:attribute>
      </xsl:for-each>
      <xsl:apply-templates/>
    </xsl:element>
  </xsl:template>

Answer 2

You have to include in your script an empty template for the attribute to omit:

<xsl:template match="@*[name() = 'xsi:schemaLocation']"/>

To copy the source without any other changes, you need also the identity template .

For a working example see http://xsltransform.net/bEJaofE

Edit

The above template will actually remove every xsi:schemaLocation attribute, regardless of its position in the document. If you want to remove this attribute only from the root node , change the template to:

<xsl:template match="/*/@*[name() = 'xsi:schemaLocation']"/>

Note added /*/ to match only attributes of the root node, whatever its name.

Answer 3

Is better to use xsl:copy-of /> , try below

<xsl:template match="Document/node()">
        <xsl:copy>
            <xsl:apply-templates select="node()"/>
        </xsl:copy>
</xsl:template>`