Can not parse the xml in orion server.
Hi, can anyone help me on this.
I have some configuration file which should be parsed when start the server. It works fine in tomcat , JRun, and some other server, now when I test it using the orion server, It throws out the
org.apache.commons.digester.xmlrules.XmlLoadException: Relative URI "digester-rules.dtd"; can not be resolved without a document URI.
I put the dtd and xml together in web-root and specify the location using
<!DOCTYPE digester-rules PUBLIC "-//Jakarta Apache //DTD digester-rules XML V1.0//EN" "digester-rules.dtd">
can anyone tell me how to resolve this.
following is a code snippet of init the digester
Digester ruleDigester = new Digester();
ruleDigester.setUseContextClassLoader(true);
System.out.println(ruleURL);
FromXmlRuleSet ruleSet = new FromXmlRuleSet(ruleURL, ruleDigester);
Digester digester = new Digester();
digester.setUseContextClassLoader(true);
ruleSet.addRuleInstances(digester);
rules = digester.getRules();
I think you'd better ask this question in apache user mailing list
http://jakarta.apache.org/site/mail2.html#Commons
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Hi,
I install an agent 10.2.0.4 HP-UX v.11.23, 64 bit, which has an Oracle database v.10.2.0.4.
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DavidHi:
Do the following in OMS host:
[oracle@gcontrol bin]$ ./emctl secure unlock
Oracle Enterprise Manager 10g Release 4 Grid Control
Copyright (c) 1996, 2007 Oracle Corporation. All rights reserved.
Checking the security status of the OMS... Done.
Stopping the HTTP Server... Done.
Updating HTTPS Virtual Host for Enterprise Manager OMS... Done.
Starting the HTTP Server... Done.
OMS Unlocked. Non Secure Agents may upload using HTTP.
[oracle@gcontrol bin]$
In Agent host:
$AGENT_HOME/bin/emctl unsecure agent
$AGENT_HOME/bin/emctl secure agent
$AGENT_HOME/bin/emctl start agent
$AGENT_HOME/bin/emctl status agent
$AGENT_HOME/bin/emctl upload agent
but the error continues:
2009-04-16 18:08:07,700 Thread-1 WARN http: snmehl_connect: connect failed to (delfos:1830): Connection refused (error = 239)
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Even start the agent without security, but the error remains. (unsecure only)
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2009-04-16 18:07:43,354 Thread-1 ERROR main: nmectla_agentctl: Error connecting to http://delfos:1830/emd/main/. Returning status code 1
Not yet implemented $OMS_HOME/bin/emctl secure lock
Any idea?
Thanks for your help.
David -
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Simply use the JAXP API to parse/build/modify XML and the Oracle XMLParser implementation will be used behind the scenes.
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-->
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+<Trace level="3" type="T">Interface Namespace = http://xxx.de/xxx</Trace>+
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+- <!-- ************************************+
-->
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+- <!-- Aufruf eines Adapters+
-->
- <SAP:Error SOAP:mustUnderstand="1" xmlns:SAP="http://sap.com/xi/XI/Message/30" xmlns:SOAP="http://schemas.xmlsoap.org/soap/envelope/">
<SAP:Category>XIServer</SAP:Category>
<SAP:Code area="XML_VALIDATION_OUT">CX_XMS_SYSERR_VALIDATION</SAP:Code>
<SAP:P1>Schema xxx.response.xsd not found in J:\usr\sap\xxx\SYS\global\xi\runtime_server\validation\schema\00000000000000000000000000000000\ u00D3u00B9u00E1 k u00DF u00C7T HL \SI_xxx_xxx_xxx_Sync_Outbound\urnsap-comdocumentsaprfc~functions\xxx.response.xsd (J:\usr\sap\xxx\SYS\global\xi\runtime_server\validation\schema)</SAP:P1>
<SAP:P2 />
<SAP:P3 />
<SAP:P4 />
<SAP:AdditionalText>not used at the moment.</SAP:AdditionalText>
<SAP:Stack>System error occurred during XML validation</SAP:Stack>
<SAP:Retry>M</SAP:Retry>
</SAP:Error>
There seems to be an error in the adapter trying to validate the response. Does anyone has a clue as to why ?Since i cannot get i running, i helped myself using a java mapping, that i packaged into a jar
along with the xsd files. See code below:
import com.sap.aii.*;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.UnsupportedEncodingException;
import java.io.InputStreamReader;
import java.io.Reader;
import javax.xml.transform.sax.SAXResult;
import javax.xml.transform.sax.SAXSource;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.Schema;
import javax.xml.validation.SchemaFactory;
import javax.xml.validation.Validator;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;
@SuppressWarnings("unused")
public class CustomValidator extends AbstractTransformation {
public void transform(TransformationInput arg0, TransformationOutput arg1)
throws StreamTransformationException {
Validator validator;
InputStreamReader insr;
String inputPayloadString;
try {
// Trace
getTrace().addInfo("com.xxx.pi.mapping.validation.CustomValidator called");
InputStream inputPayloadStream = arg0.getInputPayload()
.getInputStream();
insr = new InputStreamReader(inputPayloadStream);
InputHeader inpHeader = arg0.getInputHeader();
String intfName = inpHeader.getInterface();
getTrace().addInfo("Interface Name is :" + intfName);
String xsdName = "";
if (intfName
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xsdName = "Z_xxx.xsd";
if (intfName.equalsIgnoreCase("SI_xxx")) {
xsdName = "Z_xxx.xsd";
if (intfName
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xsdName = "Z_xxx.xsd";
if (intfName
.equalsIgnoreCase("SI_xxxx_Sync_Outbound")) {
xsdName = "Z_xxx.xsd";
getTrace().addInfo("Using schema: " + xsdName);
if (xsdName.equals(""))
throw new StreamTransformationException("Schema für "
+ intfName + " nicht gefunden.");
BufferedReader reader = new BufferedReader(new InputStreamReader(
getClass().getResourceAsStream(
"/com/xxx/pi/mapping/xsd/" + xsdName)));
// prepare document validator:
String schemaLang = "http://www.w3.org/2001/XMLSchema";
SchemaFactory jaxp = SchemaFactory.newInstance(schemaLang);
Schema schema = jaxp.newSchema(new StreamSource(reader));
validator = schema.newValidator();
} catch (SAXException saxe) {
throw new StreamTransformationException(saxe.getMessage());
// validate
try {
SAXSource source = new SAXSource(new InputSource(insr));
validator.validate(source);
} catch (Exception e) {
getTrace().addInfo(e.getMessage());
String trace = e.getMessage();
throw new StreamTransformationException(trace, e.getCause());
// Write the response
try {
inputPayloadString = convertStreamToString(arg0.getInputPayload()
.getInputStream());
arg1.getOutputPayload().getOutputStream()
.write(inputPayloadString.getBytes("UTF-8"));
getTrace().addInfo("Schema Validierung erfolgreich !");
} catch (Exception e) {
throw new StreamTransformationException(e.getMessage());
// helper to convert to String
private static String convertStreamToString(InputStream in) {
StringBuffer sb = new StringBuffer();
try {
InputStreamReader isr = new InputStreamReader(in);
BufferedReader reader = new BufferedReader(isr);
String line;
while ((line = reader.readLine()) != null) {
sb.append(line);
catch (Exception exception) {
return sb.toString(); -
Why i can not parse xml in client jar successfully?
If the length of xml is short in 300bytes(maybe another amount),i can read the string of xml from the file in java web start client jar and can parse it exactly. But if length is more than the amount, the xml string will be cut off so that i can not parse it.
is this a bug of java web start? or my fault?
display my function below:
className :FileUtils
private static byte[] getData(String fileName) throws IOException
byte[] data = null;
File file = null;
try
file = new File(fileName);
}catch(Exception ex)
ex.printStackTrace();
InputStream in = null;
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in = new FileInputStream(file);
data = new byte[in.available()];
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in = FileUtil.class.getClassLoader() .
getResourceAsStream(fileName);
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in.read(data);
in.close();
return data;
please help me,thank you.
best regards!in = new FileInputStream(file);
data = new byte[in.available()];available tells how many bytes are available at the moment in the buffer. For smaller files fitting into the buffer it might be okay but for longer ones it is definitively not.
Use a ByteArrayOutputStream to store the data in and return its toByteArray().
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I have :
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Thanks,To solve this problem you can:
Windows 7
1) Click on Start
2) Type cmd and press enter
you will see a new window
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4) restart the machine
should work.
If still does not work you must disable the firewall and antivirus
Para solucionar este problema debes hacer lo siguiente:
en Windows 7
1) click en Inicio
2) escribe CMD y pulsa enter
te va a aparecer una nueva ventana
3) escribe "netsh winsock reset"
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Slow-Start Up and "can not find the server" dialog box
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2 year old Ibook G4. When I try to open either Safari or Netscape the "can not find the server" box pops up.. I have a solid connection wi-fi for the wireless signal is on full strength. I have cleared my cache. Also I am experiencing start-ups that are taking 2-3 minutes. The color wheel just keeps spinning.Also, I have done repair and I renewed the DHCD. Not sure what else to do?as a guess, spinning wheel could be caused by failing hard drive.
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Safari is not able to open this page because it can not find the server.
I too need help with this problem with my Ipod touch. Currently, I am in the Ukraine, go to a wifi cafe, access the wifi and get a clear powerful signal. Then I try to load the Google page, any page from my bookmark or write the web address in and I receive the same message; "Safari can not open this page because it can not find the server." Something is wrong with my settings and I tried to reset the options but it does not work. Can anyone help me with this problem? Thanks!!
Hello and Welcome to Apple Discussions.
The 169.254.xxx.xxx address indicates that there is a problem with DHCP server in your BT Digital Hub (Home hub?) because the iPod is taking a default IP rather than been given a suitable one. At the moment the iPod is not even in the same subnet as the router so it's no wonder that Safari can't find the Internet.
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mrtotes -
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Have you seen this article?
iTunes for Windows: iTunes cannot contact the iPhone, iPad, or iPod software update server
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