I created xml file getFilesDir().getAbsolutePath()+ File.separator + test.xml in this path, i want to

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Asked: May 23, 20262026-05-23T00:35:53+00:00 2026-05-23T00:35:53+00:00

I created xml file getFilesDir().getAbsolutePath()+ File.separator + test.xml in this path, i want to

0

I created xml file

getFilesDir().getAbsolutePath()+ File.separator + "test.xml"

in this path, i want to read xml values from this path how can i do it. please send me code for accessing file on that path. this xml is created in this path in DDMS (data/data/<packagename>/files/xmlname.xml) please help me.

Thank You,

TextView  txtoldpwd=(TextView)findViewById(R.id.txtOldPwd);
TextView txtnewpwd=(TextView)findViewById(R.id.TxtNewPwd);
File objFile=new File(getFilesDir().getAbsolutePath()+ File.separator + "test.xml");
try {
    if (objFile.exists()){
        objFile.delete();
    }
    objFile.createNewFile();
    FileOutputStream  objFileStream=new FileOutputStream(objFile);

    XmlSerializer objXmlSrl=Xml.newSerializer();
    objXmlSrl.setOutput(objFileStream,"UTF-8");
    objXmlSrl.startDocument(null, true);
    objXmlSrl.setFeature("http://xmlpull.org/v1/doc/features.html#indent-output", true);
    objXmlSrl.startTag(null, "ChangePassword");
    objXmlSrl.startTag(null, "Oldpassword");
    objXmlSrl.text(txtoldpwd.getText().toString());
    objXmlSrl.endTag(null, "Oldpassword");
    objXmlSrl.startTag(null, "Newpassword");
    objXmlSrl.text(txtnewpwd.getText().toString());
    objXmlSrl.endTag(null, "Newpassword");
    objXmlSrl.endTag(null, "ChangePassword"); 
    objXmlSrl.endDocument();
    objFileStream.close();
} catch (Exception e) {
    // TODO: handle exception
}