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BOM?  org.xml.sax.SAXParseException: Content is not allowed in prolog.

Posted on 2006-08-24
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Last Modified: 2012-06-27
I read in a SOAP http response which is an xml string and try to build a DOM object out of it for parsing but I think a BOM (byte order mark) is causiing this exception...


2006-08-24 18:21:28,894 INFO  [STDOUT] Exception: Content is not allowed in prolog.
2006-08-24 18:21:28,904 INFO  [STDOUT] org.xml.sax.SAXParseException: Content is not allowed in prolog.
2006-08-24 18:21:28,904 INFO  [STDOUT]       at org.apache.xerces.parsers.DOMParser.parse(Unknown Source)
2006-08-24 18:21:28,904 INFO  [STDOUT]       at org.apache.xerces.jaxp.DocumentBuilderImpl.parse(Unknown Source)
2006-08-24 18:21:28,904 INFO  [STDOUT]       at javax.xml.parsers.DocumentBuilder.parse(Unknown Source)


I googled around for a solution but no success? Anyone now how to strip away a BOM?



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Question by:defa0009
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CEHJ earned 180 total points
ID: 17387139
Try something like:


try {
      PushbackReader r = new PushbackReader(new InputStreamReader(new URL(url).openStream()));
      int ch = -1;
      // scan for the < character
      while((ch = r.read()) != '<');
      if( ch == '<')
            r.unread('<') ;
      builder.parse(new InputSource(r));
      System.out.println("ok");
} catch (Exception e) {
      e.printStackTrace();
}
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by:defa0009
ID: 17388717
Thanks..
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by:CEHJ
ID: 17388772
:-)
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