Reading XML file from java web service
Hi
I am developing web service using Netbeans and the application server is glassfish.
but I am facing a problem
I have folder called "config" which will include the xml files. What I need to do setting path into that folder
but it will read from another location when I am deploying it
How to set our own path ?
Error
DPL5306:Servlet Web Service Endpoint [NewWebService] listening at address [http://kapila.epic.lk:8080/WebApplication2/NewWebServiceService]
deployed with moduleid = WebApplication2
java.io.FileNotFoundException: /root/MyWorks/glassfish-v2/domains/domain1/config/OSconfig/MConfig.xml (No such file or directory)
at java.io.FileInputStream.open(Native Method)
My "conifg" folder, i have put in to working directory but it will read from "/root/MyWorks/glassfish-v2/domains/domain1/config/OSconfig/MConfig.xm"
below method is used for getting its contain value
public String getMainCofig() {
String v ="";
try{
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document doc = docBuilder.parse(new File("OSconfig/MConfig.xml"));
doc.getDocumentElement().normalize();
NodeList listOfSPRMs = doc.getElementsByTagName("MainConfig");
Node firstSPRMNode = listOfSPRMs.item(0);
if (firstSPRMNode.getNodeType() == Node.ELEMENT_NODE) {
Element firstSPRAMElement = (Element) firstSPRMNode;
// Getting platform
NodeList server_port = firstSPRAMElement
.getElementsByTagName("RunningPlatform");
Element server_port_el = (Element) server_port.item(0);
NodeList server_port_List = server_port_el.getChildNodes();
v= ((Node) server_port_List.item(0))
.getNodeValue().trim();
}catch (Exception e){
e.printStackTrace();
return v;
Regards
I didn't quite get what the problem is but if the the desired file is in custom directory why don't you just use absolute path to reference it?
Like: "/home/my/OSconfig/MConfig.xml"
Similar Messages
-
How to read XML files from java
i need a sugession that how to read a xml file using java code
and i need to parse using some parsers and display attributes and entity seperately
as a string.......import org.dom4j.Document;
import org.dom4j.DocumentException;
import org.dom4j.io.SAXReader;
import java.io.File;
import java.text.AttributedCharacterIterator.Attribute;
import java.util.Iterator;
import java.util.StringTokenizer;
public class XmlParser
private String Result="";
private String Final="";
private String Delim="";
public void bar1(Document document) throws DocumentException
org.dom4j.Element root = document.getRootElement();
// System.out.println(root.getName());
bar2(root);
System.out.println(this.Result);
process();
public void bar2(org.dom4j.Element e)
for(Iterator i = e.elementIterator();i.hasNext();)
org.dom4j.Element Element = (org.dom4j.Element) i.next();
Result += Element.getName()+"\t"+Element.getText()+"\n";
bar2(Element);
public void process()
StringTokenizer Tokenizer = new StringTokenizer(this.Result,"\n");
String element;
while(Tokenizer.hasMoreTokens())
element = Tokenizer.nextToken();
StringTokenizer Tokenizer2 = new StringTokenizer(element,"\t");
// Do what ever String Process here Example
this.Final += element.getName();
this.Final += this.Delim;
System.out.println(this.Final);
public static void main(String s[])throws Exception
Document document = null;
SAXReader reader = new SAXReader();
File f1= new File("D:/Rajesh/EDI to XML/EDI.xml");
document = reader.read(f1);
Demo obj = new Demo();
obj.bar1(document);
i think this will hep full....... -
Reading an xml file in the web services java file
I want to read some data in the webservices file.so i am writing my own xml file through which i want to read data.but i am not able to read xml file from web services file.also , i am not able to put xml file in the .ear file,as there is no such tag in servicegen script through which i can add xml file in the .ear file.
i know in case of servlets,i will specify context-param attributes in web.xml,and i can access those in my servlet.can i do anything like this for webservices,and i am not using any servlets for web services.
kindly tell any solution.One possible option to parse an xml-file to a flex XML object:
public function parseConXML(source:String):void
xmlLoader = new URLLoader();
xmlLoader.load(new URLRequest(source));
// Eventlistener: if URL loaded --> onLoadComplete function
xmlLoader.addEventListener(Event.COMPLETE, xmlLoadComplete);
public function xmlLoadComplete(evt:Event):void{
var xml:XML = new XML();
// ignore comments in XML-File
XML.ignoreComments = true;
//ignore whitespaces in XML-File
XML.ignoreWhitespace = true;
// XML-Objekt erstellen
xml = new XML(evt.target.data);
//AFTERWARDS use your xml as your wish -
Open a local file from a web Service
Hello!
Likely this is a trivial question, but I need to open a local file from a Web Service and i don't know how to get the file path. I mean that I have a WS source code placed in a proper package (my.ws.package) with an additional configuration XML file placed in the same directory. I need to read it, but if I use merely
File conf = new File("Conf.xml");it can't find it. I've tryed with:
MyWS.class.getClass().getProtectionDomain().getCodeSource().getLocation().getPath()in order to get the file path, but a
java.security.AccessControlException access denied (java.lang.RuntimePermission getProtectionDomain)arises...
I don't know...
Suggestions?
Thanks.
AnhurHello!
Likely this is a trivial question, but I need to open a local file from a Web Service and i don't know how to get the file path. I mean that I have a WS source code placed in a proper package (my.ws.package) with an additional configuration XML file placed in the same directory. I need to read it, but if I use merely
File conf = new File("Conf.xml");it can't find it. I've tryed with:
MyWS.class.getClass().getProtectionDomain().getCodeSource().getLocation().getPath()in order to get the file path, but a
java.security.AccessControlException access denied (java.lang.RuntimePermission getProtectionDomain)arises...
I don't know...
Suggestions?
Thanks.
Anhur -
Generation of xml file from java code
hi,
I want to manipulate data in a xml file with java code.I have read data from xml file and also changed it. But i am unable to covert it again in xml file from java code. Can you please tell me how i can do this?Let me know which parser are you using currently for reading xml files so that i assist you. For now, you can refer to STAX Parser API under this link
http://java.sun.com/webservices/docs/1.6/tutorial/doc/SJSXP3.html -
Send an event from Java Web Service to BPEL
Hi,
I have a requirement where from an Async BPEL service I have to call a Java Web Service. After the completion of its task Java Web Service will fire an event that has to be consumed by the BPEL to initiate further process. Can anybody help me with how to send an event from Java Web Service and at the same time consume it in BPEL?
Thanks,
AnujSee the following posts for your answer
http://blogs.oracle.com/soabpm/entry/event_delivery_network_chapter
http://blogs.oracle.com/soabpm/entry/using_the_event_api_to_publish
http://guidoschmutz.wordpress.com/2010/01/12/using-the-event-api-to-publish-an-event-to-the-event-delivery-network-edn-the-spring-way/ -
Is there any way to read XML directly from a Web Page ??
i have a url, which on sending request, shows XML in browser.
Now i need to read this XML in browser and then manipulate it according to my need and display it on another page.
actually the process is. :
1) i have to first retrieve an xml from other site. (XML will only be shown in browser)
2.)then i have to read the Xml and show it in according to my requirements.
Is there any way to read XML directly from a Web Page ??
is their logic to accomplish this.
e.g in Servlet i can do somewhat like this :
String wholeXml=Somemethod(url);
Please Advicethe average Java XML parser will accept an InputStream, so just open an URLConnection to the webpage, get the inputstream from it and feed that inputstream to the XML parser. If the URL has valid XML data, it will get parsed without problems.
-
How to edit the existing data in the XML file from java programming.
Hi all
i am able to create XML file with the sample data as below from java programming.
i need sample code on how to edit the existing data in the XML file?
for example
<?xml version="1.0"?>
<mydata>
<data1>
<key1>467</key1>
<name1>Paul</name1>
<id1>123</id1>
</data1>
<data2>
<key2>467</key2>
<name2>Paul</name2>
<id2>123</id2>
</data2>
</mydata>
i am able to insert the data in the XML.
now i need sample code on how to modify the data in the above XML file from the java programming for only key2,name2,id2 tags only. the remaining tags data in the XML file i want to keep same data except for key2,name2,id2 which are i want to modify from java code
Regards
Sunil
[points will be always rewardable]hi
u need a parser or validate the xml file for to read the xml file from java coding u need for this
xml4j.jar u can download this file from here
http://www.alphaworks.ibm.com/tech/xml4j
or we can use the SAX(simple API for XML)
some sample applications for this
http://www.java-tips.org/java-se-tips/javax.xml.parsers/how-to-read-xml-file-in-java.html
http://www.developertutorials.com/tutorials/java/read-xml-file-in-java-050611/page1.html
http://www.xml-training-guide.com/e-xml44.html
let me know u need any other info
bvr -
Creating XML file from Java Bean
Hi
Are there any standard methods in Java 1.5 to create XML file from java bean,
i can use JAXB or castor to do so,
But i would like to know if there is any thing in java core classes,
I have seen XMLEncoder, but this is not what i want.
Any ideas
AshishMarshall JavaBean to an XML document with JAXB or XMLBeans.
-
How to modify an existing xml file from java code.
Hi
I have worked on creating a new xml file from java code using xmlbeans.But if i try to modify an already existing file using java code I am unable to get errorfree xmlfile.
For example if xml file(studlist.xml) is as below:
<?xml version="1.0" encoding="UTF-8"?>
<StudentList xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="D:\kchaitanya\xmlprac1\abc\Studlist.xsd">
<Student>
<Name>ram</Name>
<Age>27</Age>
</Student>
<Student>
<Name>sham</Name>
<Age>26</Age>
</Student>
</StudentList>
Now suppose i have set name to victor using student.setName,
and set age to 20 using setAge from javacode,
the new xml file is as follows:
<?xml version="1.0" encoding="UTF-8"?>
<StudentList xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="D:\kchaitanya\xmlprac1\abc\Studlist.xsd">
<Student>
<Name>ram</Name>
<Age>27</Age>
</Student>
<Student>
<Name>sham</Name>
<Age>26</Age>
</Student>
</StudentList>
<Student>
<Name>victor</Name>
<Age>20</Age>
</Student>
As observed this is not a valid xml file.But how can i modify without any errors?I know it's an old post, but I found this while doing a google search for something else, and don't like to leave it un-aswered
Just in case anyone has a similar problem... In this case the new elements have been appended outside of the root element
What you need to do is first get the root element and then append the new children to that, there are several ways of getting the root element, which depend on what you want to do with the elements you get back here's a simple (incomplete) way.
// gets the root element of the specified file (code not shown)
Element rootElement= new SAXReader().read(file).getRootElement();Then just append the new elements as below (this is non-generic code and would need to be modified for your situation)
// write a new student element
Element student = document.createElement("Student"); // creates the new student
rootElement.appendChild(student); // ***appends it to the root element***
Element name = document.createElement("Name"); // creates the name element
name.appendChild(document.createTextNode("Fred")); // adds the name text to the name element
student.appendChild(name); // appends the name to the student
Element age= document.createElement("Age"); // creates the age element
age.appendChild(document.createTextNode("26")); // adds the age text to the age element
student.appendChild(age); // appends the name to the studentThen flush ya buffers or whatever and write the file
Edited by: Dream-Scourge on Apr 23, 2008 11:10 AM -
FM to read XML files from Application server in ECC5.0
Hi All,
We need to pick up an XML file from Application server/FTP server. The requirement is to parse the XML file and process it to create material master. SAP provides standard function modules to read XML files.
Now we need to read the XML file contents of MM01 and upload into SAP Data Base through BAPI
I need to know about the Function modules to read XML files from Application Server and also about the FM's that will update the Date base tables with the data obtained form XML files.
Regards
PrathimaParsing XML data:
http://help.sap.com/saphelp_nw04/helpdata/en/86/8280ba12d511d5991b00508b6b8b11/frameset.htm
or alternatively check out ABAP online help for "CALL TRANSFORMATION".
For creating the material master look at BAPI_STANDARDMATERIAL_CREATE.
Thomas -
Reading XML file from application server and put into internal table-4.6C
Dear All,
Is there any way of reading XML file from application server to SAP? I am using 4.6C. Function module SCMS_STRING_TO_XSTRING function module is not available. Please suggest.
Thanks and regards,
AtanuHi Atanu!
Simply use the XSLT transformation 'ID'.
FIELD-SYMBOLS <ls_result> TYPE ANY.
CREATE DATA lref_data TYPE (your_structure).
ASSIGN lref_data->* TO <ls_result>.
CALL TRANSFORMATION id
SOURCE XML xmlstr
RESULT result = <ls_result>.
"xmlstr" contains your XML file. Just read it into it via standard I/O operations. "<ls_result>" will contain your DDIC formatted content.
Best regards
Torsten -
Read XML file from presentation server
Hi All,
I want read XML file from presentation server currently i am using GUI_UPLOAD fm . but it is reading some junk data.
DATA : BEGIN OF upl OCCURS 0,
f(255) TYPE c,
END OF upl.
CALL FUNCTION 'GUI_UPLOAD'
EXPORTING
filename = D:\XX.XML'
filetype = 'BIN'
TABLES
data_tab = upl.
is there any other alternative.
Thanks
Swarup,Hi Swarup,
Use method IMPORT_FROM_FILE of class CL_XML_DOCUMENT.
A sample code snippet :-
PARAMETERS: p_filnam TYPE localfile OBLIGATORY
DEFAULT 'C:\Documents and Settings\ssaha\Desktop\test.xml'.
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_filnam.
DATA: l_v_fieldname TYPE dynfnam.
l_v_fieldname = p_filnam.
CALL FUNCTION 'F4_FILENAME'
EXPORTING
program_name = syst-cprog
dynpro_number = syst-dynnr
field_name = l_v_fieldname
IMPORTING
file_name = p_filnam.
START-OF-SELECTION.
TYPES:
BEGIN OF ty_tab,
name TYPE string,
value TYPE string,
END OF ty_tab.
DATA:
lcl_xml_doc TYPE REF TO cl_xml_document,
v_subrc TYPE sysubrc,
v_node TYPE REF TO if_ixml_node,
v_child_node TYPE REF TO if_ixml_node,
v_root TYPE REF TO if_ixml_node,
v_iterator TYPE REF TO if_ixml_node_iterator,
v_nodemap TYPE REF TO if_ixml_named_node_map,
v_count TYPE i,
v_index TYPE i,
v_attr TYPE REF TO if_ixml_node,
v_name TYPE string,
v_prefix TYPE string,
v_value TYPE string,
v_char TYPE char2.
DATA:
itab TYPE STANDARD TABLE OF ty_tab,
wa TYPE ty_tab.
CREATE OBJECT lcl_xml_doc.
CALL METHOD lcl_xml_doc->import_from_file
EXPORTING
filename = p_filnam
RECEIVING
retcode = v_subrc.
CHECK v_subrc = 0.
v_node = lcl_xml_doc->m_document.
CHECK NOT v_node IS INITIAL.
v_iterator = v_node->create_iterator( ).
v_node = v_iterator->get_next( ).
WHILE NOT v_node IS INITIAL.
CASE v_node->get_type( ).
WHEN if_ixml_node=>co_node_element.
v_name = v_node->get_name( ).
v_nodemap = v_node->get_attributes( ).
IF NOT v_nodemap IS INITIAL
* attributes
v_count = v_nodemap->get_length( ).
DO v_count TIMES.
v_index = sy-index - 1.
v_attr = v_nodemap->get_item( v_index ).
v_name = v_attr->get_name( ).
v_prefix = v_attr->get_namespace_prefix( ).
v_value = v_attr->get_value( ).
ENDDO.
ENDIF.
WHEN if_ixml_node=>co_node_text OR
if_ixml_node=>co_node_cdata_section.
* text node
v_value = v_node->get_value( ).
MOVE v_value TO v_char.
IF v_char <> cl_abap_char_utilities=>cr_lf.
wa-name = v_name.
wa-value = v_value.
APPEND wa TO itab.
CLEAR wa.
ENDIF.
ENDCASE.
* advance to next node
v_node = v_iterator->get_next( ).
ENDWHILE.
LOOP AT itab INTO wa.
ENDLOOP.
Regards
Abhii -
How I can create a XML file from java Aplication
How I can create a XML file from java Aplication
whith have a the following structure
<users>
<user>
<login>anyName</login>
<password>xxxx</password>
</user>
</users>
the password label must be encripted
accept any suggestionLet us assume you have all the data from the jsp form in an java bean object..
Now you want a xml file. This can be acheived in 2 ways
1. Write it into a file using java.io classes. Say you have a class with name
write("<name>"+obj.getName+</name>);
bingo you have a flat file with the xml
2. Use data binding to do the trick
will recommend JiBx and Castor for the 2nd option
Regards,
Rajagopal -
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.
Maybe you are looking for
-
How can I implement external authentication module for Oracle database? We have a homegrown SSO system, and we need database users to authenticate directly against it.
-
Issues with iTunes 11.1.5 and Windows 8.1
I recently upgraded my iTunes to 11.1.5 on my Windows 8.1 computer, since the upgrade iTunes has become unusable. Before the upgrade I had an occasional error, device not recognized or a download error, sometimes i had to unplug and then plug
-
Stuck on 'Fixed' ADSL and no help from overseas ca...
After trying 3 times to get some sense from an overseas call centre, I am now trying this forum in the hopes that a moderator can help as I have heard great things about them! I am obviously trying to get the fastest broadband speed possible for my l
-
3.5 Weeks ago i sent my Thinkpad Yoga (12.5'') to repair because of several problems. I never heard anything from the company, until 1.5 Weeks later i phoned them, they told me they were waiting for a replacement mainboard. Since then i phoned more t
-
Hi Vishal Badiani I think the problem is you dont create net 80 configurations to run any product with Oracle 8i but you must check the following: - 1- Create net80 from start programs oracle-orahome 81 network administration Net8 Configuration Assis