如何将Xpath语法转换为XSL / XSLT语法？

Question

I understand how XPath syntax works, and can write Xpath commands to extract certain information from a XML file.
I would like to convert my XPath commands to XSLT scripts so that someone else can just run the script over the XML file to get the same output.

eg

I have a XML file that, let's say, looks like follows:

<?xml version="1.0" encoding="UTF-8"?>
  <library>
        <section id="109196796">
            <master_information>
                <shelf_identifier>
                    <identifier type="CodeX" type_id="2">LB1500605917</identifier>
                    <identifier type="Common Code" type_id="15">150060591</identifier>
                </shelf_identifier>
                <shelf_master>
                    <section_type>1</section_type>
                    <book_type>3</book_type>
                </shelf_master>
            </master_information>
        </section>
        <section id="109196798">
            <master_information>
                <shelf_identifier>
                    <identifier type="CodeX" type_id="2">LB0777775917</identifier>
                    <identifier type="Common Code" type_id="15">077777591</identifier>
                </shelf_identifier>
                <shelf_master>
                    <section_type>1</section_type>
                    <book_type>3</book_type>
                </shelf_master>
            </master_information>
        </section>
        <section id="109196800">
            <master_information>
                <shelf_identifier>
                    <identifier type="CodeX" type_id="2">LB2589165917</identifier>
                    <identifier type="Common Code" type_id="15">258916591</identifier>
                </shelf_identifier>
                <shelf_master>
                    <section_type>1</section_type>
                    <book_type>3</book_type>
                </shelf_master>
            </master_information>
        </section>
  </library>

If I run the XPath command below,

//identifier[@type='CodeX']

I get the output:

LB1500605917
LB0777775917
LB2589165917

..which is expected. Now, I tried converting the XPath command to an XSL syntax as below:

<?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="text"/>
    <xsl:variable name="return">
    <xsl:text>
</xsl:text> <!-- defined a line break -->
    </xsl:variable>
    <xsl:template match="//library"></xsl:template>
    <xsl:template match="//section/master_information/shelf_identifier/identifier"> 
        <xsl:value-of select="@type='CodeX'"/>
        <xsl:value-of select="$return"/> <!-- this basically puts a line break -->
    </xsl:template>
</xsl:stylesheet>

The XSL seems to be correct. However, no output is generated.

I am new to XSL/XSLT. What am I doing wrong?

Answer 1

What you could do is add a template matching the same nodes as your xpath, then outputting the value along with a newline...

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="text"/>
  <!--The strip-space isn't completely necessary. I just always include it in my
  default stylesheets. It strips whitespace. You can preserve whitespace with
  xsl:preserve-space. See https://www.w3.org/TR/xslt#strip for more details.-->
  <xsl:strip-space elements="*"/>

  <!--Suppress output of text nodes by built-in templates.-->
  <xsl:template match="text()"/>

  <!--Match "indentifier" elements that contain a "type" attribute
  with the value of "CodeX".-->
  <xsl:template match="identifier[@type='CodeX']">
    <!--Output the value of the current context ("identifier") concatenated with
    a newline. ("&#xA;" is a hex entity reference. You could also use a decimal
    reference ("&#10;")). You could use either of these references as the value
    of a variable too (or even declare it as an entity).
    I use normalize-space() instead of . to clean up any additional spaces.
    See https://www.w3.org/TR/xpath/#function-normalize-space for details.-->
    <xsl:value-of select="concat(normalize-space(),'&#xA;')"/>
  </xsl:template>

</xsl:stylesheet>

Notice the empty template that matches text() . This is added to suppress the output of text() nodes by XSLT's built-in template rules .

Also note that I didn't use // in my match. This is also because of the built-in rules; they allow recursive processing by default.

Answer 2

Why don't you do simply:

XSLT 1.0

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

<xsl:template match="/">
    <xsl:for-each select="//identifier[@type='CodeX']">
        <xsl:value-of select="."/>
        <xsl:text>&#10;</xsl:text>
    </xsl:for-each>
</xsl:template>

</xsl:stylesheet>