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Re: Re[2]: [emf-dev] searching the object by id


The root of the object is "project".  Should the XPath look like this?:

Russell J. Nile

              Martin Zdila <zdila@xxxxxxxx>                                                                                 
                                                          To:                                         Frank Budinsky        
              Sent by: emf-dev-admin@xxxxxxxxxxx           <emf-dev@xxxxxxxxxxx>                                            
                                                          cc: l.fasianok@xxxxxxxxxxxxxxx                                    
                                                          Subject:   Re[2]: [emf-dev] searching the object by id            
              Wednesday January 14, 2004 11:35 AM                                                                           
              Please respond to emf-dev                                                                                     

Hello Frank

Sorry that I write here again (and I'll write to newsgroup too), but I
need to solve this problem as soon as possible. So I am asking for
help wherever I can.

Thanks for your useful reply. I've read the chapter 13, but although
I've found there lot of useful information, it has not helped me
solving my problem.

In that book is written that getURIFragment() uses XML-path like
locator. But it seems to be a little different to a W3C XPath, because
instead of node[0] there is @node.0 etc. And the locator written in
the following code is not working:

Java code:
Resource resource = new XMLResourceImpl(URI.createURI(fileName));
XMLResource.XMLMap xmlMap = new XMLMapImpl();
Map options = new HashMap();
options.put(XMLResource.OPTION_XML_MAP, xmlMap);

Project project = (Project) resource.getContents().get(0);
Sheet s = (Sheet) project.getSheets().getSheet().get(0);

// this will print:
// //@sheets/@sheet.0/@properties

// returns the 's' object
Object o1 = resource.getEObject("//@sheets/@sheet.0/@properties");

// this throws an exception :-( (instead of returning the 's' object)
Object o2 = resource.getEObject

XML file:
<?xml version="1.0" encoding="UTF-8"?>
<project id="String" version="String">
    <sheet id="sheet1074095906796">
        <name>New Sheet</name>
    <sheet id="sheet1074095906797">
        <name>New Sheet</name>

Please help me, how can I solve this problem. I don't want to iterate
through the list of object, because there will be lot of nested lists.

And before I've read about getEObject() method, I've used

Sheet s = (Sheet) PropertyUtils.getProperty(project,

This works exactly like that XPath-like selection - it uses only index
parameters too (sheet[0]).

2. Question: which of those two methots are more effective?

Thanks in advance

Best regards

Ing. Martin Zdila
EpiSoftware Slovakia Ltd.
Prazska 4, 040 11  Kosice
cellular: +421 908 363 848
phone: +421 55 643 9954
fax: +421 55 643 9954

Dňa 14. januára 2004, 14:53, Frank Budinsky napísal:

FB> Martin,

FB> Resource.getEObject() locates an object by URI fragment which can
either be
FB> a path or an ID. You can set ID's for objects using XMLResource.setID()
FB> you can set the isID attribute of a feature itself. This is all covered
FB> Chapter 13 of the EMF book.

FB> You should ask this kind of question on the EMF newsgroup instead of
FB> Actually, there have already been various discussions on the topic of
FB> there, which you should find helpful.

FB> Frank.

FB>                       Martin Zdila
FB>                       <zdila@xxxxxxxx>         To:
FB> emf-dev@xxxxxxxxxxx
FB>                       Sent by:                 cc:
FB> l.fasianok@xxxxxxxxxxxxxxx
FB>                       emf-dev-admin@ecl        Subject:
FB> [emf-dev] searching the object by id
FB>                       ipse.org

FB>                       01/14/2004 05:49
FB>                       AM
FB>                       Please respond to
FB>                       emf-dev

FB> Hello

FB> I am using EMF for serialization / deserialization of data holding
FB> objects into / from a XML file.

FB> Suppose, I have a XML file:

FB> <project>
FB>       <someItem id='345' value='a' />
FB>       <someItem id='666' value='b' />
FB>       <someItem id='12' value='c' />
FB> </project>

FB> I am possile to get the correct SomeItem object:

FB> Project project = (Project) resource.getContents().get(0);

FB> Iterator iter = project.iterator();
FB> while (iter.hasNext()) {
FB>       SomeItem item = (SomeItem) iter.next();

FB>       if (item.getId().equals("666")) {
FB>          System.out.println("Found: value = " + item.getValue());
FB>          break;
FB>       }
FB> }

FB> My answer is: it is possible to find value of that item if I know the
FB> id directly - something like this?:

FB> Project project = (Project) resource.getContents().get(0);
FB> SomeItem item = project.getItemById("666");
FB> ...

FB> I al looking for a method like getItemById - can EMF generate one for
FB> me - if yes, then how can I achieve this?
FB> Has an id some special meaning or is it jus a normal attribute? I am
FB> expecting that the id attribute has to be unique, of course.
FB> Or is there (somewhere) some another method which helps me in to find
FB> item?

FB> --
FB> Thanks in advance

FB> Martin Zdila
FB> Developer
FB> EpiSoftware Slovakia Ltd.
FB> Prazska 4, 040 11  Kosice
FB> cellular: +421 908 363 848
FB> phone: +421 55 643 9954
FB> fax: +421 55 643 9954
FB> mailto:m.zdila@xxxxxxxxxxxxxxx
FB> http://www.episoftware.com

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FB> emf-dev mailing list
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