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Malformed URL Exception: unknown protocol: c

Franklin Phan
I use Windows XP Pro. My JAVA_HOME environment variable points to c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element "%JAVA_HOME%\bin".  I've written an XSL Transform servlet
that makes use of the package javax.xml.transform.  Why do I get the following error:

javax.servlet.ServletException:
javax.xml.transform.TransformerConfigurationException:
javax.xml.transform.TransformerException:
java.net.MalformedURLException: unknown protocol: c

The four lines above actually appear altogether in one line.  And the error appears to be due to the following piece of code where I'm trying to get the path to a folder on the local drive to
access a file:

String XML_WORK_PATH = "/WEB-INF/work_xml";
TransformerFactory tFactory = TransformerFactory.newInstance();
Transformer transformer =
       tFactory.newTransformer(new javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH) + "\\" + xslParam)); //xslParam is an XSL file name


The Malformed URL Exception does not occur on another machine running Windows XP Server.


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Re: Malformed URL Exception: unknown protocol: c

markt
I assume becuase the url you pass it starts c:\ as that is the start
of the XML_WORK_PATH. You need to prefix it with file:/// (or however
many slashes you need to get this to work in windows).

Mark

Franklin Phan wrote:

> I use Windows XP Pro. My JAVA_HOME environment variable points to
> c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
> "%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
> of the package javax.xml.transform.  Why do I get the following error:
>
> javax.servlet.ServletException:
> javax.xml.transform.TransformerConfigurationException:
> javax.xml.transform.TransformerException:
> java.net.MalformedURLException: unknown protocol: c
>
> The four lines above actually appear altogether in one line.  And the
> error appears to be due to the following piece of code where I'm trying
> to get the path to a folder on the local drive to access a file:
>
> String XML_WORK_PATH = "/WEB-INF/work_xml";
> TransformerFactory tFactory = TransformerFactory.newInstance();
> Transformer transformer =
>       tFactory.newTransformer(new
> javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
> + "\\" + xslParam)); //xslParam is an XSL file name
>
>
> The Malformed URL Exception does not occur on another machine running
> Windows XP Server.
>
>
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> For additional commands, e-mail: [hidden email]
>
>
>



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RE: Malformed URL Exception: unknown protocol: c

Jay Burgess
In reply to this post by Franklin Phan
First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
and /bin?

Second, your situation has me puzzled.  Mark's answer appears correct, as
"unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
checking the documentation, StreamSource requires a URI, so a windows filesystem
path won't work.

However...my code does something very similar, and it works. The only difference
is that I'm doing it for a StreamResult, not a StreamSource:

String root = getServletContext().getRealPath("");
String xmlFileName = root + File.separator + "WEB-INF" +
    File.separator + "TestData.xml";
TransformerFactory.newInstance().newTransformer().transform(
    new DOMSource(buffer),
    new StreamResult(xmlFileName)); // DOM into XML

I just threw in a quick println(), and "xmlFileName" is equal to
"C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".

Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
something obvious.

Jay

| Jay Burgess [Vertical Technology Group]
| "Essential Technology Links"
| http://www.vtgroup.com/


-----Original Message-----
From: Mark Thomas [mailto:[hidden email]]
Sent: Thursday, August 25, 2005 4:21 PM
To: Tomcat Users List
Subject: Re: Malformed URL Exception: unknown protocol: c

I assume becuase the url you pass it starts c:\ as that is the start
of the XML_WORK_PATH. You need to prefix it with file:/// (or however
many slashes you need to get this to work in windows).

Mark

Franklin Phan wrote:

> I use Windows XP Pro. My JAVA_HOME environment variable points to
> c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
> "%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
> of the package javax.xml.transform.  Why do I get the following error:
>
> javax.servlet.ServletException:
> javax.xml.transform.TransformerConfigurationException:
> javax.xml.transform.TransformerException:
> java.net.MalformedURLException: unknown protocol: c
>
> The four lines above actually appear altogether in one line.  And the
> error appears to be due to the following piece of code where I'm trying
> to get the path to a folder on the local drive to access a file:
>
> String XML_WORK_PATH = "/WEB-INF/work_xml";
> TransformerFactory tFactory = TransformerFactory.newInstance();
> Transformer transformer =
>       tFactory.newTransformer(new
>
javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)

> + "\\" + xslParam)); //xslParam is an XSL file name
>
>
> The Malformed URL Exception does not occur on another machine running
> Windows XP Server.
>



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Re: Malformed URL Exception: unknown protocol: c

Franklin Phan
To make things a bit more puzzling, I have a different app inside the same Tomcat 4.1.18 that uses the same XSL Transform class under its own web app context.  That app works without a hitch.
  Why is that?


Jay Burgess wrote:

> First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
> and /bin?
>
> Second, your situation has me puzzled.  Mark's answer appears correct, as
> "unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
> checking the documentation, StreamSource requires a URI, so a windows filesystem
> path won't work.
>
> However...my code does something very similar, and it works. The only difference
> is that I'm doing it for a StreamResult, not a StreamSource:
>
> String root = getServletContext().getRealPath("");
> String xmlFileName = root + File.separator + "WEB-INF" +
>     File.separator + "TestData.xml";
> TransformerFactory.newInstance().newTransformer().transform(
>     new DOMSource(buffer),
>     new StreamResult(xmlFileName)); // DOM into XML
>
> I just threw in a quick println(), and "xmlFileName" is equal to
> "C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".
>
> Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
> something obvious.
>
> Jay
>
> | Jay Burgess [Vertical Technology Group]
> | "Essential Technology Links"
> | http://www.vtgroup.com/
>
>
> -----Original Message-----
> From: Mark Thomas [mailto:[hidden email]]
> Sent: Thursday, August 25, 2005 4:21 PM
> To: Tomcat Users List
> Subject: Re: Malformed URL Exception: unknown protocol: c
>
> I assume becuase the url you pass it starts c:\ as that is the start
> of the XML_WORK_PATH. You need to prefix it with file:/// (or however
> many slashes you need to get this to work in windows).
>
> Mark
>
> Franklin Phan wrote:
>
>>I use Windows XP Pro. My JAVA_HOME environment variable points to
>>c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
>>"%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
>>of the package javax.xml.transform.  Why do I get the following error:
>>
>>javax.servlet.ServletException:
>>javax.xml.transform.TransformerConfigurationException:
>>javax.xml.transform.TransformerException:
>>java.net.MalformedURLException: unknown protocol: c
>>
>>The four lines above actually appear altogether in one line.  And the
>>error appears to be due to the following piece of code where I'm trying
>>to get the path to a folder on the local drive to access a file:
>>
>>String XML_WORK_PATH = "/WEB-INF/work_xml";
>>TransformerFactory tFactory = TransformerFactory.newInstance();
>>Transformer transformer =
>>      tFactory.newTransformer(new
>>
>
> javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
>
>
>>+ "\\" + xslParam)); //xslParam is an XSL file name
>>
>>
>>The Malformed URL Exception does not occur on another machine running
>>Windows XP Server.
>>
>
>
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: [hidden email]
> For additional commands, e-mail: [hidden email]
>
>
>


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RE: Malformed URL Exception: unknown protocol: c

Jay Burgess
In reply to this post by Franklin Phan
Why don't you do:

  System.out.println(getServletContext().getRealPath(XML_WORK_PATH));

And see what it tells you.  I'd be curious to see what you're passing to the
StreamSource constructor, and how it differs from my string.

Jay
 

-----Original Message-----
From: Franklin Phan [mailto:[hidden email]]
Sent: Thursday, August 25, 2005 4:52 PM
To: Tomcat Users List
Subject: Re: Malformed URL Exception: unknown protocol: c

To make things a bit more puzzling, I have a different app inside the same
Tomcat 4.1.18 that uses the same XSL Transform class under its own web app
context.  That app works without a hitch.
  Why is that?


Jay Burgess wrote:

> First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
> and /bin?
>
> Second, your situation has me puzzled.  Mark's answer appears correct, as
> "unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
> checking the documentation, StreamSource requires a URI, so a windows filesystem
> path won't work.
>
> However...my code does something very similar, and it works. The only difference
> is that I'm doing it for a StreamResult, not a StreamSource:
>
> String root = getServletContext().getRealPath("");
> String xmlFileName = root + File.separator + "WEB-INF" +
>     File.separator + "TestData.xml";
> TransformerFactory.newInstance().newTransformer().transform(
>     new DOMSource(buffer),
>     new StreamResult(xmlFileName)); // DOM into XML
>
> I just threw in a quick println(), and "xmlFileName" is equal to
> "C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".
>
> Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
> something obvious.
>
> Jay
>
> | Jay Burgess [Vertical Technology Group]
> | "Essential Technology Links"
> | http://www.vtgroup.com/
>
>
> -----Original Message-----
> From: Mark Thomas [mailto:[hidden email]]
> Sent: Thursday, August 25, 2005 4:21 PM
> To: Tomcat Users List
> Subject: Re: Malformed URL Exception: unknown protocol: c
>
> I assume becuase the url you pass it starts c:\ as that is the start
> of the XML_WORK_PATH. You need to prefix it with file:/// (or however
> many slashes you need to get this to work in windows).
>
> Mark
>
> Franklin Phan wrote:
>
>>I use Windows XP Pro. My JAVA_HOME environment variable points to
>>c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
>>"%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
>>of the package javax.xml.transform.  Why do I get the following error:
>>
>>javax.servlet.ServletException:
>>javax.xml.transform.TransformerConfigurationException:
>>javax.xml.transform.TransformerException:
>>java.net.MalformedURLException: unknown protocol: c
>>
>>The four lines above actually appear altogether in one line.  And the
>>error appears to be due to the following piece of code where I'm trying
>>to get the path to a folder on the local drive to access a file:
>>
>>String XML_WORK_PATH = "/WEB-INF/work_xml";
>>TransformerFactory tFactory = TransformerFactory.newInstance();
>>Transformer transformer =
>>      tFactory.newTransformer(new
>>
>
>
javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
>
>
>>+ "\\" + xslParam)); //xslParam is an XSL file name
>>
>>
>>The Malformed URL Exception does not occur on another machine running
>>Windows XP Server.
>>



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Re: Malformed URL Exception: unknown protocol: c

Franklin Phan
In reply to this post by Jay Burgess
Jay,

I just looked again at my Environment Variables under the Advanced tab of my System Properties window.  You are correct; it is the "Path" var that says "%JAVA_HOME%\bin;...<more paths>".

Franklin


Jay Burgess wrote:

> First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
> and /bin?
>
> Second, your situation has me puzzled.  Mark's answer appears correct, as
> "unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
> checking the documentation, StreamSource requires a URI, so a windows filesystem
> path won't work.
>
> However...my code does something very similar, and it works. The only difference
> is that I'm doing it for a StreamResult, not a StreamSource:
>
> String root = getServletContext().getRealPath("");
> String xmlFileName = root + File.separator + "WEB-INF" +
>     File.separator + "TestData.xml";
> TransformerFactory.newInstance().newTransformer().transform(
>     new DOMSource(buffer),
>     new StreamResult(xmlFileName)); // DOM into XML
>
> I just threw in a quick println(), and "xmlFileName" is equal to
> "C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".
>
> Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
> something obvious.
>
> Jay
>
> | Jay Burgess [Vertical Technology Group]
> | "Essential Technology Links"
> | http://www.vtgroup.com/
>
>
> -----Original Message-----
> From: Mark Thomas [mailto:[hidden email]]
> Sent: Thursday, August 25, 2005 4:21 PM
> To: Tomcat Users List
> Subject: Re: Malformed URL Exception: unknown protocol: c
>
> I assume becuase the url you pass it starts c:\ as that is the start
> of the XML_WORK_PATH. You need to prefix it with file:/// (or however
> many slashes you need to get this to work in windows).
>
> Mark
>
> Franklin Phan wrote:
>
>>I use Windows XP Pro. My JAVA_HOME environment variable points to
>>c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
>>"%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
>>of the package javax.xml.transform.  Why do I get the following error:
>>
>>javax.servlet.ServletException:
>>javax.xml.transform.TransformerConfigurationException:
>>javax.xml.transform.TransformerException:
>>java.net.MalformedURLException: unknown protocol: c
>>
>>The four lines above actually appear altogether in one line.  And the
>>error appears to be due to the following piece of code where I'm trying
>>to get the path to a folder on the local drive to access a file:
>>
>>String XML_WORK_PATH = "/WEB-INF/work_xml";
>>TransformerFactory tFactory = TransformerFactory.newInstance();
>>Transformer transformer =
>>      tFactory.newTransformer(new
>>
>
> javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
>
>
>>+ "\\" + xslParam)); //xslParam is an XSL file name
>>
>>
>>The Malformed URL Exception does not occur on another machine running
>>Windows XP Server.
>>
>
>
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: [hidden email]
> For additional commands, e-mail: [hidden email]
>
>
>


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Re: Malformed URL Exception: unknown protocol: c

Franklin Phan
In reply to this post by Jay Burgess
Jay,

I did that just last night.  I got:

C:\Program Files\Apache Group\Tomcat 4.1\webapps\htmaint\WEB-INF\work_xml

Franklin Phan
Cygna Energy Services
www.cygna.net

Jay Burgess wrote:

> Why don't you do:
>
>   System.out.println(getServletContext().getRealPath(XML_WORK_PATH));
>
> And see what it tells you.  I'd be curious to see what you're passing to the
> StreamSource constructor, and how it differs from my string.
>
> Jay
>  
>
> -----Original Message-----
> From: Franklin Phan [mailto:[hidden email]]
> Sent: Thursday, August 25, 2005 4:52 PM
> To: Tomcat Users List
> Subject: Re: Malformed URL Exception: unknown protocol: c
>
> To make things a bit more puzzling, I have a different app inside the same
> Tomcat 4.1.18 that uses the same XSL Transform class under its own web app
> context.  That app works without a hitch.
>   Why is that?
>
>
> Jay Burgess wrote:
>
>>First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
>>and /bin?
>>
>>Second, your situation has me puzzled.  Mark's answer appears correct, as
>>"unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
>>checking the documentation, StreamSource requires a URI, so a windows filesystem
>>path won't work.
>>
>>However...my code does something very similar, and it works. The only difference
>>is that I'm doing it for a StreamResult, not a StreamSource:
>>
>>String root = getServletContext().getRealPath("");
>>String xmlFileName = root + File.separator + "WEB-INF" +
>>    File.separator + "TestData.xml";
>>TransformerFactory.newInstance().newTransformer().transform(
>>    new DOMSource(buffer),
>>    new StreamResult(xmlFileName)); // DOM into XML
>>
>>I just threw in a quick println(), and "xmlFileName" is equal to
>>"C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".
>>
>>Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
>>something obvious.
>>
>>Jay
>>
>>| Jay Burgess [Vertical Technology Group]
>>| "Essential Technology Links"
>>| http://www.vtgroup.com/
>>
>>
>>-----Original Message-----
>>From: Mark Thomas [mailto:[hidden email]]
>>Sent: Thursday, August 25, 2005 4:21 PM
>>To: Tomcat Users List
>>Subject: Re: Malformed URL Exception: unknown protocol: c
>>
>>I assume becuase the url you pass it starts c:\ as that is the start
>>of the XML_WORK_PATH. You need to prefix it with file:/// (or however
>>many slashes you need to get this to work in windows).
>>
>>Mark
>>
>>Franklin Phan wrote:
>>
>>
>>>I use Windows XP Pro. My JAVA_HOME environment variable points to
>>>c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
>>>"%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
>>>of the package javax.xml.transform.  Why do I get the following error:
>>>
>>>javax.servlet.ServletException:
>>>javax.xml.transform.TransformerConfigurationException:
>>>javax.xml.transform.TransformerException:
>>>java.net.MalformedURLException: unknown protocol: c
>>>
>>>The four lines above actually appear altogether in one line.  And the
>>>error appears to be due to the following piece of code where I'm trying
>>>to get the path to a folder on the local drive to access a file:
>>>
>>>String XML_WORK_PATH = "/WEB-INF/work_xml";
>>>TransformerFactory tFactory = TransformerFactory.newInstance();
>>>Transformer transformer =
>>>     tFactory.newTransformer(new
>>>
>>
>>
> javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
>
>>
>>>+ "\\" + xslParam)); //xslParam is an XSL file name
>>>
>>>
>>>The Malformed URL Exception does not occur on another machine running
>>>Windows XP Server.
>>>
>
>
>
>
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> For additional commands, e-mail: [hidden email]
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OT: soap interface

MW Janssen
In reply to this post by Franklin Phan
Have someone of you a servlet implementing a SOAP interface? For a job I
must implement a SOAP interface, thus calling a SOAP webservice and deling
with the response.

Regards,

Maarten
--
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.338 / Virus Database: 267.10.13/78 - Release Date: 19-8-2005


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RE: Malformed URL Exception: unknown protocol: c

Jay Burgess
In reply to this post by Franklin Phan
Then I'm at loss as to what the issue is.  The fact that it works in some cases,
and not others, has me puzzled.  I'd suggest digging into the javax.xml source
and see if you can figure out the condition that throws the exception, etc.

And I'm still not fully clear on whether our "C:\..." strings are considered
valid URI's according to the spec, though they must be, as everything seems to work.

Good luck.

| Jay Burgess [Vertical Technology Group]
| "Essential Technology Links"
| http://www.vtgroup.com/
 

-----Original Message-----
From: Franklin Phan [mailto:[hidden email]]
Sent: Thursday, August 25, 2005 5:09 PM
To: Tomcat Users List
Subject: Re: Malformed URL Exception: unknown protocol: c

Jay,

I did that just last night.  I got:

C:\Program Files\Apache Group\Tomcat 4.1\webapps\htmaint\WEB-INF\work_xml

Franklin Phan
Cygna Energy Services
www.cygna.net

Jay Burgess wrote:

> Why don't you do:
>
>   System.out.println(getServletContext().getRealPath(XML_WORK_PATH));
>
> And see what it tells you.  I'd be curious to see what you're passing to the
> StreamSource constructor, and how it differs from my string.
>
> Jay
>  
>
> -----Original Message-----
> From: Franklin Phan [mailto:[hidden email]]
> Sent: Thursday, August 25, 2005 4:52 PM
> To: Tomcat Users List
> Subject: Re: Malformed URL Exception: unknown protocol: c
>
> To make things a bit more puzzling, I have a different app inside the same
> Tomcat 4.1.18 that uses the same XSL Transform class under its own web app
> context.  That app works without a hitch.
>   Why is that?
>
>
> Jay Burgess wrote:
>
>>First, I assume you mean CLASSPATH and /lib, or PATH and /bin, but not CLASSPATH
>>and /bin?
>>
>>Second, your situation has me puzzled.  Mark's answer appears correct, as
>>"unknown protocol: c" is typically the "C:\" of a Windows filesystem path.  And
>>checking the documentation, StreamSource requires a URI, so a windows filesystem
>>path won't work.
>>
>>However...my code does something very similar, and it works. The only difference
>>is that I'm doing it for a StreamResult, not a StreamSource:
>>
>>String root = getServletContext().getRealPath("");
>>String xmlFileName = root + File.separator + "WEB-INF" +
>>    File.separator + "TestData.xml";
>>TransformerFactory.newInstance().newTransformer().transform(
>>    new DOMSource(buffer),
>>    new StreamResult(xmlFileName)); // DOM into XML
>>
>>I just threw in a quick println(), and "xmlFileName" is equal to
>>"C:\tomcat-5.0.19\webapps\TestApp\WEB-INF\TestData.xml".
>>
>>Can anyone explain why mine works and Franklin's fails?  Maybe I'm missing
>>something obvious.
>>
>>Jay
>>
>>| Jay Burgess [Vertical Technology Group]
>>| "Essential Technology Links"
>>| http://www.vtgroup.com/
>>
>>
>>-----Original Message-----
>>From: Mark Thomas [mailto:[hidden email]]
>>Sent: Thursday, August 25, 2005 4:21 PM
>>To: Tomcat Users List
>>Subject: Re: Malformed URL Exception: unknown protocol: c
>>
>>I assume becuase the url you pass it starts c:\ as that is the start
>>of the XML_WORK_PATH. You need to prefix it with file:/// (or however
>>many slashes you need to get this to work in windows).
>>
>>Mark
>>
>>Franklin Phan wrote:
>>
>>
>>>I use Windows XP Pro. My JAVA_HOME environment variable points to
>>>c:\j2sdk1.4.2_05.  The CLASSPATH is set to have as the first element
>>>"%JAVA_HOME%\bin".  I've written an XSL Transform servlet that makes use
>>>of the package javax.xml.transform.  Why do I get the following error:
>>>
>>>javax.servlet.ServletException:
>>>javax.xml.transform.TransformerConfigurationException:
>>>javax.xml.transform.TransformerException:
>>>java.net.MalformedURLException: unknown protocol: c
>>>
>>>The four lines above actually appear altogether in one line.  And the
>>>error appears to be due to the following piece of code where I'm trying
>>>to get the path to a folder on the local drive to access a file:
>>>
>>>String XML_WORK_PATH = "/WEB-INF/work_xml";
>>>TransformerFactory tFactory = TransformerFactory.newInstance();
>>>Transformer transformer =
>>>     tFactory.newTransformer(new
>>>
>>
>>
>
javax.xml.transform.stream.StreamSource(getServletContext().getRealPath(XML_WORK_PATH)
>
>>
>>>+ "\\" + xslParam)); //xslParam is an XSL file name
>>>
>>>
>>>The Malformed URL Exception does not occur on another machine running
>>>Windows XP Server.
>>>



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