Re: Export xsl, Save diagram error

[david <daav14 gmail com>]

Hi,

<xsl:apply-templates select="//dia:object[ type='UML - State Term']"/>
<xsl:apply-templates select="//dia:object[ type='UML - State']"/>

I use these select statements to apply my templates for a particalr kind of object. It works and it's all what I want about my XSL export. If I follow you, what do you purpose to simplify the reference before the type of the object ?

Your exemple with Spaceships is interresting but I don't think that the UML - State template is called for UML - State Term objects. I will check that in few minutes...

Best regards

2009/6/4 Loek Bergman <loek bergman gmail com>

Hi David,


Why do you need that particular complicated select statement?
You might need that type of object in the end, I agree, but do you need that complicated reference?
Would a reference like UML_State_Term do as well? I always prefer using underscores instead of spaces. It is more reliable how they are interpreted.
How can you assure that both names do not point at the same object? If the first one is a subclass of the latter it could happen, isn't it?
If the second class performs two functions, namely being the superclass of the first one and a collegue of the first object at the same time, then a new collegue of the first object should be created.

Consider this hierarchy:
Spaceship
  Giantspaceship

In the hierarchy above the class Spaceship performs two functions, namely being the superclass and the collegue of the giantspaceship.

and compare it to this one:
Spaceship
  Giantspaceship
  Littlespaceship

This hierarchy resolves this problem. I got this idea from the double dispatch example on Wikipedia. There the existence of the Littlespaceship lacks.

With kind regards,


Loek

2009/6/4 david <daav14 gmail com>

Hi, thank for your answer.

I think you're right, I was working on it just before you answered.
After an export, I've tried to remove the new file (with windows explorer) and it's impossible....(file still used by an other program...)

I think that my XSL export has a problem to finalize the conversion but I don't know where.

This is an extract of my xslt file :

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:dia="http://www.lysator.liu.se/~alla/dia/"
  version="1.0">
  <xsl:output method="xml" indent="yes" encoding="ISO-8859-1"/>

  <xsl:template match="/">
    <xsl:element name="a">     
        <xsl:attribute name="id">Procedure</xsl:attribute>
       
        <xsl:call-template name="trouverDebut"/>
        <xsl:call-template name="trouverData"/>

        <xsl:apply-templates select="//dia:object[ type='UML - State Term']"/>
        <xsl:apply-templates select="//dia:object[ type='UML - State']"/>
      </xsl:element>
  </xsl:template>

</xsl:stylesheet>

I've remove all others templates because the error is still here whithout them...

Best regards

2009/6/4 Loek Bergman <loek bergman gmail com>

Hi David,


I have not seen your code, so it is a pure guess. What I can think of is that you still have a hook on the original file.
If you nullify the hook on the original file it might work.

Something like this:

File fl = getFile(fileName);
//work on file
//save file
File newFile = getFile(fileName);
newFile.save();

The first file might prevent your new file from updating. Setting the first hook on the file to null or only having a reference to a file using a local variable in your methods might solve this kind of problem.

Is this an idea?


With kind regards,


Loek

2009/6/4 Loek Bergman <loek bergman gmail com>

Hi David,


I have not seen your code, so it is a pure guess. What I can think of is that you still have a hook on the original file.
If you nullify the hook on the original file it might work.

Something like this:

File fl = getFile(fileName);
//work on file
//save file
File newFile = getFile(fileName);
newFile.save();

The first file might prevent your new file from updating. Setting the first hook on the file to null or only having a reference to a file using a local variable in your methods might solve this kind of problem.

Is this an idea?


With kind regards,


Loek

2009/6/4 david <daav14 gmail com>

Hey!

I'm still working on my own xsl export. It works : it's ok to create a new xml file from my Diagram.
But, I have a probleme : a messagebox appears after the export and says : Error while saving result: <output export file>. I don't understand because this file is full and contains my diagram converted.

A second problem (I think, it's linked) : when I wanted to save my diagram (.Dia) after an export, it's impossible : a messagebox appears and says : Unable to rename <file.dia> in <file.dia> Permission Denied.


Have you already seen this problem ? Have I made a mistake in my XSLT file ?


Thanks.

Best regards, David.

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dia-list mailing list
dia-list gnome org
http://mail.gnome.org/mailman/listinfo/dia-list
FAQ at http://live.gnome.org/Dia/Faq
Main page at http://live.gnome.org/Dia

_______________________________________________
dia-list mailing list
dia-list gnome org
http://mail.gnome.org/mailman/listinfo/dia-list
FAQ at http://live.gnome.org/Dia/Faq
Main page at http://live.gnome.org/Dia