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up vote 2 down vote favorite

I put my employeedetailxml into a folder named "res\raw". When I attempt to open it by specifying the file and folder name, I get the error "file not found".

How can I specify the path to my file?

I would much prefer to be able to pass 'R.raw.employeedetailxml' to FileInputStream to specify that it open that file. Is that possible? (I tried it, and got an error.)

Can FileInputStream take a Resource ID as parameter?

try{
   SAXParserFactory spf=SAXParserFactory.newInstance();
   SAXParser sp=spf.newSAXParser();
   XMLReader xr=sp.getXMLReader();

   EmployeeDetailHandler empDetailHandler=new EmployeeDetailHandler();
   xr.setContentHandler(empDetailHandler);

   xr.parse(new InputSource(new FileInputStream("\\res\\raw\\employeedetailxml.xml")));
}
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3 Answers

up vote 6 down vote accepted

Yes, you can use something similar to passing R.raw.employeedetailxml. You just need to fetch the resource XML file using that resource name, like so:

InputStream ins = getResources().openRawResource(R.raw.file_name);
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Thank you It works fine – mahe madhi Oct 7 '11 at 6:24
up vote 4 down vote

this.getResources().openRawResource() returns an InputStream that will probably help you to parse the xml.

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up vote 5 down vote

Try this:

InputStream inputStream =
    getResources().openRawResource(R.raw.employeedetailxml);
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Thanks dude It works 100% – mahe madhi Oct 7 '11 at 6:18

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