通过xsl（msxsl）进行xml转换

Question

I need to convert a xml-file to a csv-file. Therefore I'm using msxsl. The xml-file looks like this:

(...)
<Test>
  <Field1>Blabla</Field1>
  <Values>
    <Year>2012</Year> 
    <Value>12</Value>
  </Values>
  <Values>
    <Year>2013</Year> 
    <Value>1234</Value>
  </Values>
  <Field2>abc</Field2>
</Test>

<Test>
  <Field1>Blubblub</Field1>
  <Field2>def</Field2>
</Test>
(...)

Which means: There are data sets "Test" with several nodes "Values", and others without any.

I tried to solve this with for-each:

(...)
<xsl:for-each select="Test">
(...)
<xsl:for-each select="Values">
(...)
</xsl:for-each>
</xsl:for-each>
(...)

This works (in a way), but the number of colums does not match between the data sets. I need to have in every data set the maximum number of colums for every node in the whole file.

In the end one should be able to open the csv-file with excel. Therefore each column need to contain the same data... is there any way to do this?

Thanks!

Edit: In this case the expected csv-output would look like this:

Blabla;2012;12;2013;1234;abc
Blubblub;;;;;def

... and NOT:

Blabla;2012;12;2013;1234;abc
Blubblub;def

Answer 1

Given your expected CSV output (actually SSV, semicolon-separated-values, it seems), I don't think you want an inner for-each at all. I would do it more along the lines of:

<xsl:for-each select="Test">
    <xsl:value-of select="Field1"/>
    <xsl:text>;</xsl:text>
    <xsl:value-of select="Values[1]/Year"/>
    <xsl:text>;</xsl:text>
    <xsl:value-of select="Values[1]/Value"/>
    <xsl:text>;</xsl:text>
    <xsl:value-of select="Values[2]/Year"/>
    <xsl:text>;</xsl:text>
    <xsl:value-of select="Values[2]/Value"/>
    <xsl:text>;</xsl:text>

    <!-- etc, up to the final... -->

    <xsl:value-of select="Field2"/>
</xsl:for-each>

This gives you a fixed number of fields, with empty values where no node exists.