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XML Formatting

Posted on 2008-10-09
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Last Modified: 2013-11-18
Is there any way to add line breaks to an XML file after each element using XSLT or C#?

The problem is I have multiple files (several hundred thousand) that were all exported into XML as a single line. For the input program to accept them, they have to include line breaks to be more properly formatted. Any ideas on how to accomplish this?


Before
 

<document xmlns="http://www.ademero.com/XmlSchemas/ContentCentral/XmlCaptureDescriptorV1.1" xmlns:addData="http://www.abbyy.com/FlexiCapture/Schemas/Export/AdditionalFormData.xsd" xmlns:form="http://www.abbyy.com/FlexiCapture/Schemas/Export/FormData.xsd"><electronicFile>Images\AIRCO-W8371_1.pdf</electronicFile><destination><catalog documentType="default">Accounts Payable</catalog></destination><fields><field name="Invoice">W8371</field><field name="Vendor">AIRCO INC</field><field name="Date">12/18/98</field></fields></document>
 
 

AFTER
 

<document xmlns="http://www.ademero.com/XmlSchemas/ContentCentral/XmlCaptureDescriptorV1.1" xmlns:addData="http://www.abbyy.com/FlexiCapture/Schemas/Export/AdditionalFormData.xsd" xmlns:form="http://www.abbyy.com/FlexiCapture/Schemas/Export/FormData.xsd">

  <electronicFile>Images\AIRCO-W8371_1.pdf</electronicFile>

  <destination>

    <catalog documentType="default">Accounts Payable</catalog>

  </destination>

  <fields>

    <field name="Invoice">W8371</field>

    <field name="Vendor">AIRCO INC</field>

    <field name="Date">12/18/98</field>

  </fields>

</document>

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0
Comment
Question by:developernetwork
11 Comments
 
LVL 18

Expert Comment

by:zc2
ID: 22678775
Try to use an identity transformation, it should output with line breaks, especially if you use <xsl:output> with the indent="yes" attribute
<?xml version="1.0"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

	<xsl:output method="xml" indent="yes" encoding="iso-8859-1"/>
 

	<xsl:template match="@*|node()">

	  <xsl:copy>

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

	  </xsl:copy>

	</xsl:template>	

</xsl:stylesheet>

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Author Comment

by:developernetwork
ID: 22678925
I'm a little confused.

Here, let me show you the XSLT stylesheet we're already using.
I have the indent="yes" in there and it fails to do that.
<?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" media-type="text/xml" omit-xml-declaration="no" version="1.0"/>

	<xsl:template match="/" xmlns:addData="http://www.abbyy.com/FlexiCapture/Schemas/Export/AdditionalFormData.xsd"

	xmlns:form="http://www.abbyy.com/FlexiCapture/Schemas/Export/FormData.xsd">

		<document xmlns="http://www.ademero.com/XmlSchemas/ContentCentral/XmlCaptureDescriptorV1.1">

			<electronicFile>

				<xsl:value-of select="form:Documents/*/@addData:ImagePath" />

			</electronicFile>

			<destination>

				<catalog documentType="default">Accounts Payable</catalog>

			</destination>

			<fields>

				<field name="Invoice">

					<xsl:value-of select="form:Documents/*/_Document/_Invoice" />

				</field>

				<field name="Vendor">

					<xsl:value-of select="form:Documents/*/_Document/_Vendor" />

				</field>

				<field name="Date">

					<xsl:value-of select="form:Documents/*/_Document/_Date" />

				</field>

			</fields>

		</document>

	</xsl:template>

</xsl:stylesheet>

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LVL 18

Expert Comment

by:zc2
ID: 22680013
May I ask you what the xsl processor are you using?
I tried on MSXML v3 and v6, it does obey the indent attribute.
0
 

Author Comment

by:developernetwork
ID: 22681459
I'm using a simple Xsl.Transform command in Microsoft C# version 3.5
0
 

Author Comment

by:developernetwork
ID: 22681524
I'm not familiar with how to use MSXML v6, I've just written a little program in C# that does the transform.
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Author Comment

by:developernetwork
ID: 22681723
Did you make changes to the XSLT that I posted to make it do that? I'm trying again and I'm just not able to get it to format the document. I'm extremely new to working with XML, and they've put me on a very short deadline, so I haven't had the chance to read a lot of what I would have if I had been granted more time on this project.
0
 
LVL 18

Expert Comment

by:zc2
ID: 22683800
You don't need a complex xslt to just reformat the xml with the same structure it already has.
All you need is a copy xslt which I have provided you in the first comment.

I tried to use the xsl from a C# program, and you right it does ignore the indent="yes" attribute.

Try to use the copy xsl with the following command line utility, it uses MSXML v3 as far as I know.
http://www.microsoft.com/downloads/details.aspx?FamilyID=2fb55371-c94e-4373-b0e9-db4816552e41&DisplayLang=en
0
 

Author Comment

by:developernetwork
ID: 22684123
Well, this isn't the original document. The XSLT transforms it from a different document.

The reason I use the C# tool is because I have to execute this on a couple hundred thousand documents.
0
 
LVL 10

Assisted Solution

by:Hans Langer
Hans Langer earned 200 total points
ID: 22684191
Just load the XML on a XML object and then save it or call his outerxml property. Both method ident the data.

xml.xmlDocument oXML = new xml.xmldocument;
oXML.loadxml(sXML)


oXML.outterXML

or

oXML.save("c:\myfile.xml")
0
 
LVL 23

Accepted Solution

by:
Tiggerito earned 300 total points
ID: 22687638
This might help...
xmlNode = the xml data
 

StringWriter stringWriter = new StringWriter();

XmlTextWriter xmlWriter = new XmlTextWriter(stringWriter);

xmlWriter.Formatting = Formatting.Indented;

xmlWriter.Indentation = 4;

xmlWriter.IndentChar = ' ';

xmlWriter.QuoteChar = '"';
 

xmlNode.WriteTo(xmlWriter);
 

return stringWriter.ToString();

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Author Comment

by:developernetwork
ID: 22688594
Thanks, using all of this I was able to design a working solution. I appreciate it!
0

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