again: &amp; in XSL parameters - XML / XSL Discussion Boards

skoizumi291108-Jun-04 16:03

skoizumi29110

8-Jun-04 16:03

I have some xml document describing our database.
I have an element <value>
in side of that document for empty element, but
when I parse it using DOM in VB

For Each rowElement In xmlDoc.documentElement.childnodes
str_content = rowElement.childnodes(0).nodeValue
Next

when I have NULL in the element I have application error
said "NULL is not properly used". I tried to filter the
NULL by
If rowElement.childnodes(0).nodeValue = NULL then
....
however, vb doesn't recognize this null

do you think i should not use
for empty element in the first place?
there is another solution?

Thanks for help
shin

Loading XslTransform from data base/string

lkrzanik8-Jun-04 5:44

lkrzanik

8-Jun-04 5:44

How to load XslTransform from data base or just from string? I don't want to load it from file, I keep all the XSL documents in data base and I just want to load them from record.

Re: Loading XslTransform from data base/string

palbano8-Jun-04 16:32

palbano

8-Jun-04 16:32

Use the Document objects loadXML( string) method

doc.loadXML( string_from_database)

"No matter where you go, there your are." - Buckaroo Banzai

-pete

MS XML

Rhys__6668-Jun-04 0:23

Rhys__666

8-Jun-04 0:23

Apparently I'm having a problem with MS XML libraries.

We're posting an XML doc to an external site which parses the doc and provides an HTTP Response Number.

We also need the XML to be re-posted from the external site to another page of ours and have the XML saved as a doc. In standard ASP we're doing this as follows...
[code]
<%
strABBXMLFile = Server.MapPath("/spider/abb/abbn/" & Day(Now) & "-" & Month(Now) & "-" & Year(Now) & "_" & Hour(Now) & Minute(Now) & "_standardAbbreviation.xml")

Set doc = Server.CreateObject("Msxml2.DOMDocument.4.0")

doc.load(Request)
doc.save(strABBXMLFile)
%>
[/code]

The problem we have is with entity declaration and tags. With entity declarations or a DocType in the XML none of the MS XML libraries can save the XML document. If the <doctype> is removed, they xml is saved fine. Can anyone provide any further information on this issue, or if there's a work-around for it? Or do the .Net XML libraries cope with entity declarations and doctype tags, and can anyone provide any pointers to dot-netting the above?

All help much appreciated, as well as advice.

Cheers!

Rhys
"When one burns one's bridges, what a very nice fire it makes" -- Dylan Thomas
"As to marriage or celibacy, let a man take the course he will. He will be sure to repent" -- Socrates
"Doh!" -- Homer Simpson

&amp; in XSL parameters

Atara6-Jun-04 1:08

Atara

6-Jun-04 1:08

I send to the xsl string parameter as
strPathFiles = "C:\Documents and Settings.."
args.AddParam("strPathFiles", "", strPathFiles)
Me.myXslSrcTransform.Transform(oXPathNav, args, swContent, Nothing)
'(This is vb .Net code)

I want to use it in <IMG> tag as src attribute:
<IMG>
<xsl:attribute name="src">
<xsl:value-of select="$strPathFiles"/>myDir\myFile.gif<xsl:text/>
</xsl:attribute>
</IMG>
but the xsl parses it to -
"C&#58;&#92;Documents&#32;and&#32;Settings"

What string do I need to send to the xsl that will not be parsed,
or else, -
How can I force the xsl to not parse the string?

Thanks.

Atara

Re: &amp; in XSL parameters

henningbenk7-Jun-04 1:43

henningbenk

7-Jun-04 1:43

Maybe the disbale-output-escaping Attribute in xsl:value-of will help.

Re: &amp; in XSL parameters

Atara7-Jun-04 2:38

Atara

7-Jun-04 2:38

Thanks. I will try it

Atara

again: &amp; in XSL parameters

Atara7-Jun-04 19:33

Atara

7-Jun-04 19:33

I tried
<xsl:param name="strPathFiles" /> 
<xsl:variable name="strPathU" ><xsl:value-of select="$strPathFiles" disable-output-escaping="yes" /></xsl:variable>

<IMG border='0' width='19' height='18' align='middle' >
<xsl:attribute name="src">
<xsl:value-of select="$strPathU" disable-output-escaping="yes" />links_files\ref2_books.gif<xsl:text/>
</xsl:attribute>
</IMG><xsl:text/>
<xsl:value-of select="$strPathU" disable-output-escaping="yes" />myDir\myFile.gif<xsl:text/>

In the , the string is exactly as I want it, but in the <IMG> I still get -
"C&#58;&#92;Documents&#32;and&#32;Settings"

Maybe disable-output-escaping has no effect in <xsl:attribute> because it might cause illegal-xml .

Any other suggestions?

Thanks,

Atara

Re: again: &amp; in XSL parameters

henningbenk7-Jun-04 22:07

henningbenk

7-Jun-04 22:07

I just tried to reproduce your problem, but I get <img src="C:\Documents and Settings..myDir\myFile.gif"></img> even without using "disable-output-escaping".

Do you use a special encoding or xsl:output?

Re: again: &amp; in XSL parameters

Atara7-Jun-04 22:50

Atara

7-Jun-04 22:50

In the xsl file, I use
<xsl:output method="html">

It all started when I discovered that I have a problem when my path contains
special unicode chars, so I tried to replace all non-Alpha-Numeric characters
with their corresponding html entities.

so in VB .Net I get the unicodePath, I replace the characters I want,
and I send it as parameters to the XSL.

If there is a simpler way, I would like to know about it...

Thanks

Atara

Re: again: &amp; in XSL parameters

Atara7-Jun-04 22:58

Atara

7-Jun-04 22:58

(sorry, I forgot to set the Do not treat <'s as HTML tags)

In the xsl file, I use
<xsl:output method="html" />

Atara

How to load XML from string into dataset or listbox

5-Jun-04 8:54

I have an ASP.NET project with a String with XML in it. The XML was created from a DataSet using ds.GetXml(). Now, I want to put the contents of the 2nd column into a listbox. How can I read the string to extract the data to put it into a dataset or directly into the listbox?

Tor Solvang

Re: How to load XML from string into dataset or listbox

Andrew Quinn AUS6-Jun-04 22:52

Andrew Quinn AUS

6-Jun-04 22:52

Hi Tor,

The best way is to load the XML string into an XML document and then use XPath to retrieve the information (nodes) you want, e.g.

XmlDocument xmlReceive = new XmlDocument();
if (xmlReceive != null)
{
   xmlReceive.LoadXml(xmlStringFromDataSet);

   // Now use XPath to get the results and add to your listbox
   XmlNodeList collResults = xmlReceive.SelectNodes("/your/xml/tag/path/here");
   if (collResults != null && collResults.Count > 0)
   {
      for(int i=0; i < collResults.Count; i++)
      {
         XmlNode nodeResult = collResults[i];
         if (nodeResult != null)
         {
            lstBox1.Items.Add(nodeResult.text);
         }
      }
   }
}

Replace the XPath above with the one you require.

Hope this helps,
Andy

Re: Converting carriage returns to tags

Philip Fitzsimons2-Jun-04 3:22

Philip Fitzsimons

2-Jun-04 3:22

you have not actually given us your REAL code - as the template would not work " " in the template should be " ".

if you show us the actual code we might be able to help.

"When the only tool you have is a hammer, a sore thumb you will have."

Re: Converting carriage returns to tags

JussiL2-Jun-04 8:06

JussiL

2-Jun-04 8:06

Oh, sorry. I cleaned it up a little bit before posting and that small mistake slipped in. However, I already found a solution for the problem. The following template does the trick for me:

<xsl:template name="convertlinebreaks">
<xsl:param name="text"/>
<xsl:choose>
<xsl:when test="contains($text,'&#xA;')">
<xsl:value-of select="substring-before($text,'&#xA;')"/>
 
<xsl:call-template name="convertlinebreaks">
<xsl:with-param name="text">
<xsl:value-of select="substring(substring-after($text,'&#xA;'), 2, string-length(substring-after($text,'&#xA;')))"/>
</xsl:with-param>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$text"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>

But I'd still like to know how this could be implemented using the translate() function.

How to test if node exists and it's not empty?

Alvaro Mendez1-Jun-04 10:00

Alvaro Mendez

1-Jun-04 10:00

In my XSL document I need to output some HTML if a node exists and it's not empty. I've tried testing it like this but it's not working:

<xsl:if test="'{Main/Whatever}' != ''"><br />
...<br />
</xsl:if>

It always goes into the if, no matter what. What's the solution?

Thanks,
Alvaro

Give a man a fish, he owes you one fish. Teach a man to fish, you give up your monopoly on fisheries.

Re: How to test if node exists and it's not empty?

Anonymous1-Jun-04 17:25

Anonymous

1-Jun-04 17:25

Is Main/Whatever supposed to be the node your checking for? If so, your comparing strings, not checking to see if the node equals an empty string. I think what your looking for is:

<xsl:if test="Main/Whatever"><br />
...<br />
</xsl:if>

Re: How to test if node exists and it's not empty?

Alvaro Mendez3-Jun-04 12:49

Alvaro Mendez

3-Jun-04 12:49

Thanks, but this only tests if the node exists. I want it to also test if it's not empty.

Regards,
Alvaro

Give a man a fish, he owes you one fish. Teach a man to fish, you give up your monopoly on fisheries.

Re: How to test if node exists and it's not empty?

Philip Fitzsimons2-Jun-04 3:25

Philip Fitzsimons

2-Jun-04 3:25

both test can be done in one go:

<xsl:if test="string-length(Main/Whatever) > 0"><br />
...<br />
</xsl:if>

"When the only tool you have is a hammer, a sore thumb you will have."

Problem with mailto link

Alvaro Mendez1-Jun-04 9:10

Alvaro Mendez

1-Jun-04 9:10

I have the following line in my XSL file:

<a href='mailto:YourFriend?Subject=I love XSL%26Body=How about you?'>Email to a Friend</a>

When I see the tranformed XML file in the IE6 browser, it shows the link just fine, and on the status bar I see it as:

mailto:YourFriend?Subject=I love XSL&Body=How about you?

The problem is that when I click on the link, I get this inside my Outlook popup:

To: YourFriend<br />
Cc: <br />
Subject: I love XSL&Body=How about you?

The message portion ends up as part of the subject. What's going on and how can I fix it?

By the way, I have to use %26 for the ampersand otherwise I get a "A semi colon character was expected." error, but I have a feeling that's part of the problem.

Thanks in advance,
Alvaro

Give a man a fish, he owes you one fish. Teach a man to fish, you give up your monopoly on fisheries.

XMLNode datagrid

dotnet_cpp31-May-04 20:20

dotnet_cpp

31-May-04 20:20

I am consuming one of my own webservices and i want to saw the result in datagrid.

XmlNode node = result.EndListEvents_Active(ar);
string info;
//XmlDocument doc = new XmlDocument();
//info =node.InnerXml;
//DataSet myDataSet = new DataSet();
//myDataSet.ReadXml(node.InnerText);
//myDataSet.GetXml();
// DataGrid1.DataSource = myDataSet;
// DataGrid1.DataBind();
TextBox1.Text = info;

I can see my result in "info" and in TextBox1 but not in datagrid .. how do i see my result in datagrid ... I have been trying since friday ... I will really appreciate your help ..

Thanks

XML Schema(s): Help Needed

mlauzon2728-May-04 10:18

mlauzon27

28-May-04 10:18

I have the following schema, which I believe needs to be fleshed out more to make it more powerful, can someone please help and also point me in the right direction...hopefully once my project gets approved on SourceForge I will be adding this as well as something else I have to the project. Anyeverhow, here is the schema as it stands right now:

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

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">

<xs:element name="student">
<xs:complexType>
<xs:sequence>

<xs:element name="stunum" type="xs:integer" />
<xs:element name="lastname" type="xs:string" />
<xs:element name="firstname" type="xs:string" />
<xs:element name="age" type="xs:integer" />
<xs:element name="dateborn" type="xs:date" />

</xs:sequence>
</xs:complexType>
</xs:element>

<xs:element name="address">
<xs:complexType>
<xs:sequence>

<xs:element name="pnum" type="xs:integer" />
<xs:element name="addnum" type="xs:integer" />
<xs:element name="street" type="xs:string" />
<xs:element name="city" type="xs:string" />
<xs:element name="country" type="xs:string" />
<xs:element name="poczip" type="xs:string" />

</xs:sequence>
</xs:complexType>
</xs:element>

</xs:schema>

If you can make any changes to it, please do...any and all help would be very much appreciated.

Someone said the following to me as well, although I am having problems understanding what he is implying, probably because he didn't include an example:

"In your case, it depends on where you're going with this, but I'd consider creating a ComplexType for a "person" and then define your "Student" as an extension of the Person type, adding perhaps the student ID. That way, if you added Instructor, you'd be able to re-use the generalized part of the definition of a person (since Instructors and Students both have some things in common)."

If I can get this up and running, I will eventually need to be able to easily change some things in it to get it working with a second project that I have on SourceForge.

Michael Lauzon, Founder
The Quill Society
http://www.quillsociety.org/
mlauzon@quillsociety.org

Re: XML Schema(s): Help Needed

JussiL2-Jun-04 8:34

JussiL

2-Jun-04 8:34

I haven't even tested if this is well formed so there might be some typos etc. but this modifed version from you schema shows the basic idea behind type extensions (or inheritance).

<?xml version="1.0" encoding="utf-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">

<xs:element name="address" type="addressType"/>
<xs:element name="student" type="studentType"/>
<xs:element name="instructor" type="instructorType"/>

<xs:complexType name="personType">
<xs:sequence>
<xs:element name="stunum" type="xs:integer" />
<xs:element name="lastname" type="xs:string" />
<xs:element name="firstname" type="xs:string" />
<xs:element name="age" type="xs:integer" />
<xs:element name="dateborn" type="xs:date" />
</xs:sequence>
</xs:complexType>

<xs:complexType name="studentType">
<xs:complexContent>
<xs:extension base="personType">
<xs:sequence>
<xs:element name="stunum" type="xs:integer" />
</xs:sequence>
</xs:extension>
</xs:complexContent>
</xs:complexType>

<xs:complexType name="instructorType">
<xs:complexContent>
<xs:extension base="personType">
<xs:sequence>
<xs:element name="someinfo" type="xs:string" />
</xs:sequence>
</xs:extension>
</xs:complexContent>
</xs:complexType>

<xs:complexType name="addressType">
<xs:sequence>
<xs:element name="pnum" type="xs:integer" />
<xs:element name="addnum" type="xs:integer" />
<xs:element name="street" type="xs:string" />
<xs:element name="city" type="xs:string" />
<xs:element name="country" type="xs:string" />
<xs:element name="poczip" type="xs:string" />
</xs:sequence>
</xs:complexType>

</xs:schema>

Re: XML Schema(s): Help Needed

mlauzon272-Jun-04 8:47

mlauzon27

2-Jun-04 8:47

Can you -- or someone -- please give me an example XML document based on the schema you posted here...so I can reference it to something.

Thanks,

Michael Lauzon, Founder
The Quill Society
http://www.quillsociety.org/
mlauzon@quillsociety.org

Re: XML Schema(s): Help Needed

JussiL2-Jun-04 23:55

JussiL

2-Jun-04 23:55

The schema itself was buggy as it had no root element and so on, but here's a refined version from it and below it you'll fin a sample xml-file. You still have to modify the schema so that it suites you purposes..

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">

<xs:element name="record">
<xs:complexType>
<xs:sequence>
<xs:element name="student" type="studentType"></xs:element>
<xs:element name="instructor" type="instructorType"></xs:element>
</xs:sequence>
</xs:complexType>
</xs:element>

<xs:complexType name="personType">
<xs:sequence>
<xs:element name="lastname" type="xs:string"></xs:element>
<xs:element name="firstname" type="xs:string"></xs:element>
<xs:element name="age" type="xs:integer"></xs:element>
<xs:element name="dateborn" type="xs:date"></xs:element>
<xs:element name="address" type="addressType"></xs:element>
</xs:sequence>
</xs:complexType>

<xs:complexType name="studentType">
<xs:complexContent>
<xs:extension base="personType">
<xs:sequence>
<xs:element name="stunum" type="xs:integer"></xs:element>
</xs:sequence>
</xs:extension>
</xs:complexContent>
</xs:complexType>

<xs:complexType name="instructorType">
<xs:complexContent>
<xs:extension base="personType">
<xs:sequence>
<xs:element name="someinfo" type="xs:string"></xs:element>
</xs:sequence>
</xs:extension>
</xs:complexContent>
</xs:complexType>

<xs:complexType name="addressType">
<xs:sequence>
<xs:element name="pnum" type="xs:integer"></xs:element>
<xs:element name="addnum" type="xs:integer"></xs:element>
<xs:element name="street" type="xs:string"></xs:element>
<xs:element name="city" type="xs:string"></xs:element>
<xs:element name="country" type="xs:string"></xs:element>
<xs:element name="poczip" type="xs:string"></xs:element>
</xs:sequence>
</xs:complexType>

</xs:schema>

and here's a piece of xml which can be validated with the schema:

<?xml version="1.0" encoding="UTF-8"?>
<record
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:noNamespaceSchemaLocation='schema.xsd'>
<student>
<lastname>Will</lastname>
<firstname>Smith</firstname>
<age>20</age>
<dateborn>1970-06-02</dateborn>
<address>
<pnum>235</pnum>
<addnum>234</addnum>
<street>Street</street>
<city>City</city>
<country>Country</country>
<poczip>123</poczip>
</address>
<stunum>123</stunum>
</student>
<instructor>
<lastname>John</lastname>
<firstname>Smith</firstname>
<age>20</age>
<dateborn>1960-06-02</dateborn>
<address>
<pnum>432</pnum>
<addnum>432</addnum>
<street>Street2</street>
<city>City2</city>
<country>Country2</country>
<poczip>321</poczip>
</address>
<someinfo>Some information</someinfo>
</instructor>
</record>

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