XML / XSL

4-Aug-06 2:10

Hi
I am now using xlst to write out my xml doc as a HTML string instead of using selectnode etc. It's neat. But....I can't work out how to specify search criteria for the xml doc. I just get it all. Sample below :-

XML DOC

<Fields>
<BatchParam name="Field 1" tab="Tab 1" Seq="1" />
<BatchParam name="Field 2" tab="Tab 1" Seq="3" />
<BatchParam name="Field 3" tab="Tab 1" Seq="2" />
<BatchParam name="Field 1" tab="Tab 2" Seq="1" />
<BatchParam name="Field 2" tab="Tab 2" Seq="3" />
<BatchParam name="Field 3" tab="Tab 2" Seq="2" />
<BatchParam name="Field 1" tab="Tab 3" Seq="1" />
</Fields>

I want to display a specific tab, eg Tab 2 in a table, ordered by seq

My XSL
<?xml version="1.0" encoding="utf-8"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match='/ | @* | node()'>
<xsl:copy>
<xsl:apply-templates select="node()">
<xsl:sort select='@seq'/>
</xsl:apply-templates>
</xsl:copy>
<tr>
<td>
<xsl:value-of select='position()'/>
</td>
<td>
<xsl:value-of select='@tab'/>
</td>
<td>
<xsl:value-of select='@name'/>
</td>
<td>
<xsl:value-of select='@seq'/>
<br/>
</td>
</tr>
</xsl:template>
</xsl:stylesheet>

and finally my C# code

System.IO.StringWriter objStream = new System.IO.StringWriter();

System.Xml.Xsl.XslTransform objTransform = new System.Xml.Xsl.XslTransform();

objTransform.Load(madconfig.Config.Get("MYXSLFILE.XSL"));

System.Xml.XPath.XPathDocument xml = new System.Xml.XPath.XPathDocument("MYXMLFILE.XML");
System.Xml.XPath.XPathNavigator xmlNav = xml.CreateNavigator();

objTransform.Transform(xmlNav, null, objStream, null);
Response.Write ("<table border=1>"+objStream.ToString()+"</table>");

And this displays ALL the entries. Before, rolling my own I did something like

string tabName = GetTabName();
System.Xml.XPath.XPathExpression expr = xmlNav.Compile("/Fields/BatchParam[@name=\""+tabName+"\"]");
System.Xml.XPath.XPathNodeIterator iterator = xmlNav.Select(expr);

Any ideas...?

Regards

Angel
*********************************************
The sooner you fall behind, the longer you have to catch up.

Malcolm Smart4-Aug-06 3:20

4-Aug-06 3:20

So...you can pass parameters down. But I am still struggling. This is where I am now...

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="tabName" select="Main"></xsl:param> <-- MY PARAM
<xsl:template match="Fields">
<table border="3">
<xsl:copy>
<xsl:apply-templates select="BatchParam">
<xsl:sort select='@tab'/>
<xsl:sort select='@seq'/>
</xsl:apply-templates>
</xsl:copy>
</table>
</xsl:template>
<xsl:template match="BatchParam[@tab='Main']"> <-- WHERE I WANT TO USE IT
<tr>
<td><xsl:value-of select='position()'/></td>
<td><xsl:value-of select='@tab'/></td>
<td><xsl:value-of select='@name'/></td>
<td><xsl:value-of select='@seq'/></td>
</tr>

</xsl:template>

</xsl:stylesheet>

If I replace the hardcoded 'Main' with the param as in

<xsl:template match="BatchParam[@tab=$tabName]">

I get an exception...

System.Xml.XPath.XPathException: 'BatchParam[@tab=$tabName]' is an invalid key pattern. It either contains a variable reference or 'key()' function.

Help!!!!

Regards

<strike>Angel</strike>Malcolm
*********************************************
The sooner you fall behind, the longer you have to catch up.

Dustin Metzgar4-Aug-06 4:21

4-Aug-06 4:21

Malcolm Smart wrote:
<xsl:param name="tabName" select="Main"></xsl:param>

This line is a little odd. This is a template parameter, not a variable. The difference is that you're supposed to be passing this parameter in when you call the stylesheet. There are also parameters for template matching sections. You can put a variable instead:

<xsl:variable name="tabName"></xsl:variable>

Also, your select is not going to put the string 'Main' into tabName. It's going to try and find a node called Main and put it there. You could stick text into it if you want:

<xsl:variable name="tabName"><xsl:text>Main</xsl:text></xsl:variable>

As for your XPath statement, I'm unsure why it gave you that error. Try changing to a variable with text and see how it reacts.

Logifusion[^]
If not entertaining, write your Congressman.

Malcolm Smart4-Aug-06 4:39

4-Aug-06 4:39

Hi Thanks for the info...still no joy (and I am trying - have been all ****** afternoon...)

Overall reqs are - use select tab from front end. I extract just the nodes which have the tab attribute set to this tab. So a) I need to be able to pass the tab name down from the front end into the xsl and b) use it in teh xsl. I am still struggling with b) and not even looked at a) yet!!

My code now, with your changes, still gives the exception....I need a little more hand holding. This is my first venture into XSL and really struggling, both conceptually, and syntactically.

<?xml version="1.0" encoding="utf-8"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:variable name="tabName">
    <xsl:text>Main</xsl:text>
  </xsl:variable>  <xsl:template match="Fields">
    <table border="3">
      <xsl:copy>
        <xsl:apply-templates select="BatchParam">
          <xsl:sort select='@tab'/>
          <xsl:sort select='@seq'/>
        </xsl:apply-templates>
      </xsl:copy>
    </table>
  </xsl:template>
  <xsl:template match="BatchParam[@tab=$tabName]">  <tr>
      <td><xsl:value-of select='position()'/></td>
      <td><xsl:value-of select='@tab'/></td>
      <td><xsl:value-of select='@name'/></td>
      <td><xsl:value-of select='@seq'/></td>
    </tr>

  </xsl:template>

</xsl:stylesheet>

It still doesn't like the $tabName. VS2005 highlights it as a problem in the editor before i run it!

Regards

~~Angel~~Malc
*********************************************
The sooner you fall behind, the longer you have to catch up.

Dustin Metzgar4-Aug-06 5:13

4-Aug-06 5:13

Ok, there are a few changes that you have to make to your XSL:

<?xml version="1.0" encoding="utf-8"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:variable name="tabName"><xsl:text>Tab 2</xsl:text></xsl:variable>  
  <xsl:template match="Fields">
    <table border="3">
      <xsl:apply-templates select="BatchParam[@tab=$tabName]">
        <xsl:sort select="@Seq" data-type="number"/>
      </xsl:apply-templates>
    </table>
  </xsl:template>
  <xsl:template match="BatchParam">
    <tr>
      <td><xsl:value-of select='position()'/></td>
      <td><xsl:value-of select='@tab'/></td>
      <td><xsl:value-of select='@name'/></td>
      <td><xsl:value-of select='@Seq'/></td>
    </tr>
  </xsl:template>

</xsl:stylesheet>

The first change is for the value of tabName from "Main" to "Tab 2" so it actually matches something in the XML you provided. The next thing is the statement BatchParam[@tab=$tabName]. Normally you can do this, but the problem is where you placed it. You put it in the select for the template match. You simply can't put variables there. You want that variable to be placed in the call to the template match, not in the declaration of the template match. Also, your select of @seq wasn't returning anything because the XML you provided was using Seq as the attribute. So be careful of the case sensitivity. Also, be sure to specify the "number" data type when sorting otherwise you can get something like 1, 10, 11, 12, 2, 3, etc.

Logifusion[^]
If not entertaining, write your Congressman.

Malcolm Smart4-Aug-06 5:33

4-Aug-06 5:33

Hero. Many thanks. The 'Main' bit was a typo, the code I pasted wasn't the code I am working on (not allowed....).

I appreciate your help, and looking at the number of posts on here, a lot of others appreciate your help as well!!

Regards

Malc
*********************************************

Malcolm Smart4-Aug-06 5:39

4-Aug-06 5:39

And for completeness, I swapped the xsl:variable for a xsl:param and in the code used XsltArgumentsList to populate it from the client - and it works a treat.

Thanks again

Regards

Malc
*********************************************

Dustin Metzgar4-Aug-06 5:49

Need Complete reference on XPath

4-Aug-06 5:49

Awesome. Glad to hear everything's working out. Although I am starting to feel like an XSL/XPath *cough*expert*cough*. Not sure if that's a good thing. Roll eyes | :rolleyes:

Logifusion[^]
If not entertaining, write your Congressman.

Rizwan Bashir3-Aug-06 23:36

Rizwan Bashir

3-Aug-06 23:36

Hello There
can any one mail me or post me a link for complete reference on Xpath ?

rbashir@alm-soft.com

Rizwan Bashir
ALM Soft[^]

Re: Need Complete reference on XPath

Dustin Metzgar4-Aug-06 4:24

How to match several different elements in the middle of XPath?

4-Aug-06 4:24

Try Google[^]

Logifusion[^]
If not entertaining, write your Congressman.

CherezZaboro3-Aug-06 9:46

CherezZaboro

3-Aug-06 9:46

Consider this xml:

<br />
<rrx id="someother">...</rrx><br />
<rrx id="Zones"><br />
 <f name="NM"><br />
  <p name="xoxoxo"/><br />
 </f><br />
 <s name="GB"><br />
  <p name="xexexe"/><br />
  <p name="xaxaxa"/><br />
 </s><br />
 <a name="SP"><br />
  <p name="hmhmhm"/><br />
 </a><br />
 <b name="Alien"><br />
  <p name="opsops"/><br />
 </b><br />
</rrx><br />

I need to select all "p" elements in "f","s" or "a" with a given "name" but not "b" which are in "rrx" with particular "id". So I'm trying to come up with or statement for f,s,a in the middle of XPath.
Using XSLT 2.0 and Xpath 2.0 this works fine:

//rrx[@id='Zones']/(f|s|a)[@name='GB']/p - I should get xexexe and xaxaxa and if
//rrx[@id='Zones']/(f|s|a)[@name='SP']/p - I need hmhmhm

But good old 1.0 says it's invalid. That () are not allowed there. How else would the XPath then look?

OMG | :OMG:

Re: How to match several different elements in the middle of XPath?

Erik Molenaar8-Aug-06 5:08

Erik Molenaar

8-Aug-06 5:08

The feature you try to use is new to XPath 2.0

One of the ways you could solve it in XPath 1.0 would be:
instead of:
//rrx[@id='Zones']/(f|s|a)[@name='SP']/p
write:
//p[(parent::f[(@name='SP')and(parent::rrx[@id='Zones'])]) or (parent::s[(@name='SP')and(parent::rrx[@id='Zones'])]) or (parent::a)[(@name='SP')and(parent::rrx[@id='Zones'])]]

It does not look nice, but would be an example.

Page width in xsl

Support1232-Aug-06 22:31

Support123

2-Aug-06 22:31

Hi all,
how would i set the page width in xsl to 80mm?

i want to create a template for a slip printer....

He who laughs last...

Re: Page width in xsl

Support1232-Aug-06 22:52

Support123

2-Aug-06 22:52

Okay thanks anyways, i'll just use a table and set it according.

Sorry, i thought of it a i posted the message.

He who laughs last...

Re: Page width in xsl I WAS WRONG

Support1233-Aug-06 0:09

Support123

3-Aug-06 0:09

Sorry, the table idea didn't work. is it possible to set the page width, because if i try to print now, through the slip printer, it prints it fine but i keeps spitting out paper... my friend suggested that it could be because the xsl is using an A4 and the slip is only 80mm in width, actually 79.5mm, and it is not working correctly because of this...

Please help Cry | :((

He who laughs last...

Re: Page width in xsl I WAS WRONG

Dustin Metzgar3-Aug-06 6:27

XPath syntax to find all nodes that don't have some attribute at all? [modified]

3-Aug-06 6:27

The simple-page-master node has an available page-width property.

Logifusion[^]
If not entertaining, write your Congressman.

CherezZaboro1-Aug-06 6:41

CherezZaboro

1-Aug-06 6:41

Excuse me for this quick noob question.
Let's say I have XML:

<br />
<a name="a1">text</a><br />
<a>text 2</a><br />
<a name="a3">text 3</a><br />

What's the XPath syntax to locate that secon "a" element that doen't have the "name" attribute?

-- modified at 12:42 Tuesday 1st August, 2006

Re: XPath syntax to find all nodes that don't have some attribute at all?

Dustin Metzgar1-Aug-06 7:36