Reading resource files

I'm trying to read a properties file in a web application. I'm using WSAD development environment. The file is located in webApplication/res folder. I'm able to get to this file from the servlet using the getServeltContext(). But I can not access this file from the other helper classes in the application. Only way I could circumvent this problem is to pass the servletContext().getRealPath(initParam) to all the classes from the servlet every time I instantiate them and this is getting to be a pain. Is there any other way to do it?
Thanks,
Ram

Why not read the property file into a ResourceBundle and store this as a public static variable of the servlet. Then all the helper classes can access it.

Similar Messages

Unable to read resource files without fully qualified file names

Hi,
I have installed 9iAS and trying to read a few resources in the OC4J container. A few points to
note :
1. The <library> element in server.xml (global and
instance specific) as well as in orion-application.xml
can seem to only scan for jars/zips.
2. Also we are using the Thread specific classloaders as in doc :
http://otn.oracle.com/tech/java/oc4j/pdf/ClassLoadingInOC4J_WP.pdf
The issue is that so long I pass the fully qualified file name for the resource, I can read the resource. However, I am not able to find a way to include the "resource directory" in the classpath of 9iAS (global) or OC4J instance and read the resource file with just the file name. So, the following code works only when the filename has the fully qualified name for the resource file (Like F:\install\some_dir\file.txt).
ClassLoader cl = Thread.currentThread().getContextClassLoader();
inStream = cl.getSystemResourceAsStream(fileName);
Can someone provide me with some pointers to how I can include resource directories in the classpath ? Any help will be appreciated.
Thanks,
Krishnendu

Hi,
Thanks for your response. Presently your solution is what we already have in place. Most of our properties are stored in a .properties file. However for all other application servers (sun one apps server, websphere, weblogic, tomcat), the classloader can read these properties without the fully qualified path but for oc4j; the code breaks.
So, even if I change my installer to hardcode the specific property to have the fully qualified file name, I would have written some code specific to OC4J that is generic to all other containers.
Krishnendu

How to read resource files?

Hello,
I need to read the settings for our LDAP server from a .properties file, but I'm not sure how to reference it. All I need to continue is to be able to get a "File" object which refers to a properties file. Here is more info:
The file is:
hello.world.MyInfo.properties
where hello.world is a package and MyInfo.properties is the file name that I need the "File" object for. Can this be done? "hello/world/MyInfo.properties" does not work. I guess that's because it's packed into the .jar file.
Please note that this is for J2EE and I need it to run on an Application Server. "hello/world/MyInfo.properties" works smoothly in J2SE as a "Java Application".
Thank you,
Pavel

ClassLoader loader =
ClassLoader.getSystemClassLoader();
InputStream in =
loader.getResourceAsStream("/hello/world/MyInfo.proper
ties");
loader.get...........() returns null for some reason.
The path is valid and the file is there. Strange.Weird, I could have sworn I've used the absolute "/" resource path before. Regardless, with my latest java 1.5, I have managed to test it successful by excluding the first slash.
java.io.InputStream in
= ClassLoader.getSystemResourceAsStream("evaluate/testprop.properties");
p.load(in);
System.out.println(p.getProperty("smack.dog"));

How to read resource file under the WEB-INF through ClassPathResource

I have an xml file under the directory WEB-INF but WEB-INF/classes, I want to read it through ClassPathResource class from Spring, and where I read this xml file is not in a Servlet class, how can I do? Thank you.

Webmethods should be able to handle it. if not you should convert the tline content into one xstring and pass it to webmethods.
othe option
if you are on WAS6.20 and above you can use BSP to achieve what you are looking for.
Regards
Raja

Problem in reading a resource file form JAR

Problem in finding a resource file from the jar.....
I have my code in the following directory structure
rootDir -
     libDir -
          XMLEntities.res
     modelDir -
          File1
          File2
          File3
Step1 - I create a jar file with root as modelDir and with all its files.
Step2 - I add XMLEntities.res also to the root of the jar
If I run my application using
java -classpath ;C:\aip_build\build_dm\rmi_server_files\dm.jar; model.data_model.ActivateDataManager
My application which uses Xalan.jar (lyong under ext directory in jdk) does not find XMLEntiries.res.
If I use my application as under without creating jar my application does find XMLEntites.res
Can anyone tell me why the application can not find the XMLEntities.res inside jar?

Sorry I missed one line............
Problem in finding a resource file from the jar.....
I have my code in the following directory structure
rootDir -
----libDir -
--------XMLEntities.res
----modelDir -
--------File1
--------File2
--------File3
Step1 - I create a jar file with root as modelDir and
with all its files.
Step2 - I add XMLEntities.res also to the root of the
jar
If I run my application using
java -classpath
;C:\aip_build\build_dm\rmi_server_files\dm.jar;
model.data_model.ActivateDataManager
My application which uses Xalan.jar (lyong under ext
directory in jdk) does not find XMLEntiries.res.
If I use my application as under without creating jar
my application does find XMLEntites.res
java -classpath
;C:\aip_build\build_dm\rmi_server_files\;C:\aip_build\b
ild_dm\rmi_server_files\lib
model.data_model.ActivateDataManager
Can anyone tell me why the application can not find
the XMLEntities.res inside jar?try loading the resources as a InputStream
something like :
InputStream is = NameOfLoadingClass.class.getClassLoader().getResourceAsStream("XMLEntities.res");
Hope this might help you in some way.

Unable to read a resources file from my page

Hi,
I am trying to read a file from the project resources and I can't have access to the physical file. I have the following code in my page:
javax.faces.context.ExternalContext ec = this.getExternalContext();
URL url = ec.getResource("/resources/displayPassengers.xsl");
InputSource inputSourceXSL = new InputSource(url.toExternalForm());
InputStream inpStreamXSL = null;
if (inputSourceXSL.getByteStream() != null) {
log("Loading xml file: " + url.toExternalForm());
inpStreamXSL = inputSourceXSL.getByteStream();
} else {
log("Unable to extract URL from Page context: WEB-INF/displayPassengers.xsl");
The url is not null.
The inputSourceXSL is not null and inputSourceXSL.getByteStream() is always null.
Please let me know how to access the file.
Rgrds.

this should work
InputSource inputSourceXSL = new InputSource(url.openStream());

How to Read all files inside resource Folder inside Jar?

I have a Jar file,,,, The program reads for resource files in the resource folder inside the Jar. I want my program to read all files inside this folder without knowing the names or the number of files in the folder.
I am using this to make an Applet easy to be updated with spicific files. I just want to add a file inside the resource folder and Jar the program and automatically the program reads what ever is in there!!
I used the File class to get all file names inside the resource folder. it works fine before Jarring the program. After I jar I recieve a URI not Herarichy Exception!!
File fold=new java.io.File(getClass().getResource(folder).toURI());
String[] files=fold.list();
I hope the question is clear!!

How to get the directory and jar file that you class resides in:
// returns path and jarfile (ex: /home/mydir/my.jar)
public String getJarfileName()
        // Get the location of the jar file and the jar file name
        java.net.URL outputURL = YourClass.class.getProtectionDomain().getCodeSource().getLocation();
        String outputString = outputURL.toString();
        String[] parseString;
        int index1 = outputString.indexOf(":");
        int index2 = outputString.lastIndexOf(":");
        if (index1!=index2) // Windows/DOS uses C: naming convention
           parseString = outputString.split("file:/");
        else
           parseString = outputString.split("file:");
        String jarFilename = parseString[1];
       return jarFilename;
}If your my.jar was in /home/mydir, it will store "/home/mydir/my.jar" in jarFilename.
note: getLocation returns "file:/C:/home/mydir/my.jar" for Windows/DOS and "file:/home/mydir/my.jar" for windows, thus why I look for the first and last index of the colon so I can correctly split the string.
If you want to grab a file in the jar file and get an InputStream, do the following:
import java.io.*;
import java.util.Enumeration;
// jar stuff
import java.util.jar.*;
import java.util.zip.ZipEntry;
public class Jaris
   private JarFile jf;
   public Jaris(String jarFilename) throws Exception
       try
                       jf = new JarFile(jarFilename);
       catch (Exception e)
            jf=null;
            throw(e);
   public InputStream getInputStream(String matchString) throws Exception
       ZipEntry ze=null;
       try
          Enumeration resources = jf.entries();
          while ( resources.hasMoreElements() )
             JarEntry je = (JarEntry) resources.nextElement();
             // find a file that matches this string from anywhere in my jar file
             if ( je.getName().matches(".*\\"+matchString) )
                String filename=je.getName();
                ze=jf.getEntry(filename);
      catch (Exception e)
         throw(e);
      InputStream is = jf.getInputStream(ze);
      return is;
   // for testing the Jaris methods
   public static void main(String[] args)
      try
           String jarFilename=getJarfileName();
           Jaris jis = new Jaris(jarFilename); // this is the string we got from the other method listed above
           InputStream is=jis.getInputStream("myfile.txt"); // can be used for xml, xsl, etc
      catch (Exception e)
         // this is just a test, so we'll ignore the exception
}Happy coding! :)
Doc

How to read a Properties Resource File in KM with Web Dynpro Java

Hello all,
I would like to store a resource file in KM and have my web dynpro application read that file at runtime. Is this possible? Any examples of how to achieve this?
I hoping to be able to modify this resource file in KM to easily change certain behaviors in my webdynpro application.

Thanks Naga Raju Meesala.
How come all these methods are deprecated...getEP5User
Also, since I am building a weddynpro DC, what is the proper way to include these jar files as Used DCs?
Now proceed in the same way with the variable PORTAL_HOME and add the following .jar files:
u2022 \lib\prtapi.jar The portal runtime APIs
u2022 \portalapps\com.sap.portal.usermanagement\lib\com.sap.security.api.ep5.jar The user management APIs of the Enterprise Portal 5.0 are deprecated, but still in use in SAP NetWeaver 04
u2022 \portalapps\com.sap.netweaver.bc.rf\lib\bc.rf.framework_api.jar KM Repository Framework APIs
u2022 \portalapps\com.sap.netweaver.bc.rf.service\lib\bc.rf.global.service.urlgenerator_api.jar Repository Framework Utility: URL Generator
u2022 \portalapps\com.sap.netweaver.bc.sf\lib\bc.sf.framework_api.jar Repository Framework: Repository Services
u2022 \portalapps\com.sap.netweaver.bc.util\lib\bc.util.public_api.jar Repository Framework Utilities

Localization Issue - Application Doesn't Read Culture Specific Resource File

Dear Sirs,
i know there are lot of thread out there about localization, but most of them go over same rule, or are not enough complete that i be able to get them,...
Here's my issue, i use win form localization for UI, and every things is fine, then i pass the culture i want to application thread before start it, and every thing goes fine.
But when i want to handle error msg's... or Etc...i use Resource file exist over Resources, and my Resharper point to that, using command like
Resource.<ResourceName>;
So after doing every thing, i made an other resource file, copy exist text item in main resource file,
put a name like Resource.en-US.resx on it and then compile it, it generate a sub folder, or maybe two,
after it i run app, with passing culcure of en-US , and then the application went right, but when it come to show code resources, like what i talked about, it used the old default culture text messages...
what should i do? where do i go wrong?
I even tried to copy the resource dll file of en-US calture in exact bin folder, but it still didn't worked.
the thing i follow were on a thread of MSDN help.

Dear Sirs,
i know there are lot of thread out there about localization, but most of them go over same rule, or are not enough complete that i be able to get them,...
Here's my issue, i use win form localization for UI, and every things is fine, then i pass the culture i want to application thread before start it, and every thing goes fine.
But when i want to handle error msg's... or Etc...i use Resource file exist over Resources, and my Resharper point to that, using command like
Resource.<ResourceName>;
So after doing every thing, i made an other resource file, copy exist text item in main resource file,
put a name like Resource.en-US.resx on it and then compile it, it generate a sub folder, or maybe two,
after it i run app, with passing culcure of en-US , and then the application went right, but when it come to show code resources, like what i talked about, it used the old default culture text messages...
what should i do? where do i go wrong?
I even tried to copy the resource dll file of en-US calture in exact bin folder, but it still didn't worked.
the thing i follow were on a thread of MSDN help.
Hello,
I am afraid that we are not clear about what the error is and how you use these resources, I would recommend you share the detailed code with us, you could comment all sensitive data or if possible you could just create a simple winform project, and reproduce
that issue, then share it with us by uploading it to OneDrive.
You could share its link here that we could focus on that project to help you.
Regards.
Carl
We are trying to better understand customer views on social support experience, so your participation in this interview project would be greatly appreciated if you have time. Thanks for helping make community forums a great place.
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HERE to participate the survey.

Read a file resource in servlet

Hello,
I try to read a file ressource in a Servlet with following statements:
ServletContext ServCont = this.getServletContext();
String strConfFile = ServCont.getRealPath("/xml/config.xml");
With Tomcat5.5 it works, but with Bea 9.1 I get null.
I tried also read from a system ressource:
String strConfFile1 = System.getProperty("PTSHISTWeb.conFile");
but I don't know where to declare the property in Bea9.
Can anybody help?
Best regards
Peter

getRealPath can return null if the resource is in an archived WAR. Can you use getResourceAsStream instead?
You can declare system properties with -D flags on the java command-line you use to boot a server.
Of if you wanted to do it in code, you could run a startup class which is configured to run before application deployment and call System.setProperty there.
-- Rob
WLS Blog http://dev2dev.bea.com/blog/rwoollen/

Error reading configuration file

I've setup the Adobe Access 4.0 trial license server and when I run the Validator.bat with -g -r on the Tomcat install dir \licenseserver I get an 'Error reading configuration file' message. Here is the log dump:
[] 2012-12-20 22:46:32,176 INFO [[Partition(flashaccessserver)].com.adobe.flashaccess.server.license.context.SimpleContex tFactory] Creating class loader for partition 'flashaccessserver' with libraries '[file:/c:/Tomcat6/licenseserver/flashaccessserver/libs/, file:/c:/Tomcat6/licenseserver/flashaccessserver/libs/flashaccess-license-server-ext-samp le.jar]'
[] 2012-12-20 22:46:32,582 ERROR [[Partition(flashaccessserver)].com.adobe.flashaccess.server.license.tools.Validator] Failed to validate tenant deployment 'flashaccessserver/sampletenant'
com.adobe.flashaccess.server.common.configuration.ConfigurationException: Error reading configuration file
          at com.adobe.flashaccess.server.license.configuration.commonsadapter.Constants.parseTenantCo nfigurationStream(Constants.java:139)
          at com.adobe.flashaccess.server.license.configuration.commonsadapter.TenantConfigurationImpl .<init>(TenantConfigurationImpl.java:110)
          at com.adobe.flashaccess.server.license.configuration.commonsadapter.CommonsConfigurationBas edFactory.getTenantConfiguration(CommonsConfigurationBasedFactory.java:90)
          at com.adobe.flashaccess.server.license.tools.Validator.validateTenantDeployment(Validator.j ava:255)
          at com.adobe.flashaccess.server.license.tools.Validator.validatePartitionDeployment(Validato r.java:283)
          at com.adobe.flashaccess.server.license.tools.Validator.validateGlobalDeployment(Validator.j ava:301)
          at com.adobe.flashaccess.server.license.tools.Validator.process(Validator.java:173)
          at com.adobe.flashaccess.server.license.tools.Validator.main(Validator.java:117)
Caused by: org.apache.commons.configuration.ConfigurationException: Unable to load the configuration
          at org.apache.commons.configuration.XMLConfiguration.load(XMLConfiguration.java:863)
          at org.apache.commons.configuration.XMLConfiguration.load(XMLConfiguration.java:821)
          at com.adobe.flashaccess.server.license.configuration.commonsadapter.Constants.parseTenantCo nfigurationStream(Constants.java:134)
          ... 7 more
Caused by: org.xml.sax.SAXParseException; lineNumber: 121; columnNumber: 16; cvc-complex-type.2.4.b: The content of element 'KeyServer' is not complete. One of '{"http://licenseserver.flashaccess.adobe.com/tenant":File}' is expected.
          at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(Unkno wn Source)
          at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator$XSIErrorReporter.reportErro r(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.reportSchemaError(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.elementLocallyValidComplexT ype(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.elementLocallyValidType(Unk nown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.processElementContent(Unkno wn Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.handleEndElement(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaValidator.endElement(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanEndElement(Unk nown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDri ver.next(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl.next(Unknown Source)
          at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unkno wn Source)
          at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
          at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
          at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(Unknown Source)
          at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(Unknown Source)
          at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(Unknown Source)
          at org.apache.commons.configuration.XMLConfiguration.load(XMLConfiguration.java:855)
          ... 9 more
I'm using a relative path to my .pfx file for both the transportServerCredential and licenseServerCredential in the flashaccess-tenant.xml and my password has been scrambled using the Scrambler.bat.
Also, when I verify setup using http://<LicenseServer>:8080/flashaccessserver/flashaccess/license/v2 I get the message 'License server is setup correctly.'
Any ideas of why the Validator.bat can't read the configuration file?

I agree with you that if I can successfully playback content the license serve is setup successfully. Here is what I've tried and how I've set things up:
I have my license server setup (its the Protected Streaming version).
I'm using Adobe Media Server as my content/packaging server. I successfully served Vanilla PHLS sample streams to the sample Adobe Access player on iOS devices.
I have configured the Adobe Media Server to point to the license server and certificates as per the documenation and this Adobe Dev article (http://www.adobe.com/devnet/adobe-media-server/articles/content-protection-using-phds-phls .html). Since I am using a trial version of Adobe Access my .der files are the same for transport and packaging.
I am using the local key mode to remove one more variable (I have setup a remote key server as well).
I have placed a copy of the 'vod-policy.pol' policy file from the reference license server resources directory in my Adobe Media Server 'creds' directory and used the relative path '..creds/vod-policy.pol' in the httpd.conf file.
When I attempt to load the sample stream http://<mymediaserver>/hls-vod/sample2_1000kbps.f4v.m3u8 using the Adobe Access sample player on my iOS device I receive the following errors in the player:
DRM error Major[3363] minor:[0] NSError:(null)]
From my knowledge of working with the Adobe Access Objective C library to create a PhoneGap plugin it appears that a decrypted playlist is not being returned by the Adobe Media Server. Additionally, I can find no information in the Adobe Media Server or Adobe Access logs that pertain to my setup.
I would love to have someone from Adobe speak to the problems I am having. I find the documentation to be hit and miss and most of my successful results during this proofing process have been from piecing together disparate pieces of information and trial and error. Information on using the Adobe Media Server with Adobe Access is limited to one Devnet article (above) and the help in Adobe Media Server that doesn't explain pathing to the license server or Java policy files in any detail.
Colour me frustrated

Failed to read wsdl file from url

Hi all,
I am struggling with WL 9.2, consuming a WSRP enabled portlet at the following point:
Having created a proxy portlet, refering to the WSDL of the WSRP producer, I did successfully deploy the portal. When accessing the portal and consuming the portlet the following exception is thrown:
com.bea.wsrp.faults.TransportException: Failed to read wsdl file from url due to -- java.net.ConnectException: Tried all: '8' addresses, but could not connect over HTTP to server: 'www.w3.org', port: '80';
The WSDL includes a schema (http://www.w3.org/2001/XMLSchema). The URL is definetely accessible by the server.
I am open for any kind of support ;) Thanks in advance!
Kind regards,
Matthias

Yes, they are all local schemas. The way ALSB works, even if the import path in the wsdl or parent schema doesn't exist, relative to the current directory, you can specify which schema resolves that reference. My WSDL imports several schemas, all of which import others. When I was done importing all the schemas and resolving references, all the references in WSDLs and XSDs were assigned and valid. However, after all of those imports and reference resolution, it still says the WSDL is invalid, with the following odd error message:
The WSDL is not semantically valid: Failed to read wsdl file from url due to -- java.net.MalformedURLException: no protocol: /XMLSchema/PaymentServices/Resources/PaymentServices-200802.
That path listed refers to one of the schemas imported by the WSDL. The reference to that schema in the WSDL was resolved, and it doesn't complain about that reference.

How to add resource files to a new folder to a project in JDeveloper 11g

Currently, new folder can only be created when creating an item within JDeveloper via setting the destination folder.
I have some resource files that are binary files and created outside JDeveloper.
How can I add those resource files to a specific folder under an existing project so that we can do versioning control on those files using SVN within JDeveloper?
Or I have to manage the versioning control of those files outside JDeveloper?

Go to project properties - >Project content - > resources and add the directory.

Unable to read Properties file from Java code

Hi,
The directory structure of my application is as follows:-
My App
++++++ src
++++++++com
++++++++++readProp.java
++++++++resource
++++++++++message.properties
I am trying to read the file as follows:-
<code>
public Static final string FilePath="resource.message.properties"
Here the code to read the file. I tried using the following two techniques but to no use... :(
File accountPropertiesFile = new File(FacesContext.getCurrentInstance()
.getExternalContext().getRequestContextPath()
+ FilePath);
properties.load(externalContext.getResourceAsStream(FilePath));
</code>
But none yeild any sucess while reading through the Bean class. please help...

Your source code isn't there at run time, at least it shouldn't be, so the directory structure you've shown is irrelevant.
If the file is in the corresponding position under WEB-INF/classes at runtime, the second technique should work, but you need to fix FilePath to agree with what it says under Class.getResource() and friends.

Problem while reading XML file from Aplication server(Al11)

Hi Experts
I am facing a problem while reading XML file from Aplication server using open data set.
OPEN DATASET v_dsn IN BINARY MODE FOR INPUT.
IF sy-subrc <> 0.
EXIT.
ENDIF.
READ DATASET v_dsn INTO v_rec.
WHILE sy-subrc <> 0.
ENDWHILE.
CLOSE DATASET v_dsn.
The XML file contains the details from an IDOC number , the expected output is XML file giving all the segments details in a single page and send the user in lotus note as an attachment, But in the present output after opening the attachment i am getting a single XML file which contains most of the segments ,but in the bottom part it is giving the below error .
- <E1EDT13 SEGMENT="1">
<QUALF>001</QUALF>
<NTANF>20110803</NTANF>
<NTANZ>080000</NTANZ>
<NTEND>20110803<The XML page cannot be displayed
Cannot view XML input using XSL style sheet. Please correct the error and then click the Refresh button, or try again later.
Invalid at the top level of the document. Error processing resource 'file:///C:/TEMP/notesD52F4D/SHPORD_0080005842.xml'.
/SPAN></NTEND>
<NTENZ>000000</NTENZ>
for all the xml its giving the error in bottom part , but once we open the source code and if we saved in system without changing anything the file giving the xml file without any error in that .
could any one can help to solve this issue .

Hi Oliver
Thanx for your reply.
see the latest output
- <E1EDT13 SEGMENT="1">
<QUALF>003</QUALF>
<NTANF>20110803</NTANF>
<NTANZ>080000</NTANZ>
<NTEND>20110803</NTEND>
<NTENZ>000000</NTENZ>
<ISDD>00000000</ISDD>
<ISDZ>000000</ISDZ>
<IEDD>00000000</IEDD>
<IEDZ>000000</IEDZ>
</E1EDT13>
- <E1EDT13 SEGMENT="1">
<QUALF>001</QUALF>
<NTANF>20110803</NTANF>
<NTANZ>080000</NTANZ>
<NTEND>20110803<The XML page cannot be displayed
Cannot view XML input using XSL style sheet. Please correct the error and then click the Refresh button, or try again later.
Invalid at the top level of the document. Error processing resource 'file:///C:/TEMP/notesD52F4D/~1922011.xml'.
/SPAN></NTEND>
<NTENZ>000000</NTENZ>
E1EDT13 with QUALF>003 and <E1EDT13 SEGMENT="1">
with <QUALF>001 having almost same segment data . but E1EDT13 with QUALF>003 is populating all segment data
properly ,but E1EDT13 with QUALF>001 is giving in between.

Reading resource files

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