CODEWITHSUNDEEP
javaxmlsaxparsersaxparseexception
Watz
Watz

Reputation: 859

SAXParseException occurrs in my program

I've written a program to read a set of source files and convert them into XML files using SrcML tool. Basically the procedure as follows.

for (------------------) {
    -------------------
    String xmlUri = GetXmlFile(sourceFileUri); // create xml file and get its uri
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();          
    Document doc = dBuilder.parse(xmlUri);
    -------------------
}

For each source file the program creates a XML file in the same location (by overriding the previously created file) and read the XML file. For some source files this procedure works fine. But most of them it gives some SAX Parse Exceptions as follows:

The SrcML tool creates valid XML documents. When I check the XML file for some of these exception it doesn't show anything wrong with the format. All exceptions pointed out to the same line in the code which is:

 "Document doc = dBuilder.parse(xmlUri);"

I have gone through number of discussions related to this topic in stack over flow as well as in other forums. Neither provides me a clue to overcome this problem.

I really appreciate if someone can help me to solve this problem. Thank you.

Here's the source code written to read XML file:

private static Document GetXmlDom(String xmlFilePath)
    throws SAXException, ParserConfigurationException, IOException {

File tempFile;
try {
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();          
    Document doc = dBuilder.parse(xmlFilePath);
    if (doc.hasChildNodes()) {
        return doc;
    }
}
catch (IOException e) {
    e.printStackTrace();
    throw e;
}
catch (SAXParseException e) {
    e.printStackTrace();
    throw e;
}

return null;
}

private static String GetXmlFile(String inputFile) throws IOException {
if (new File(inputFile).isFile()) {
    String outFile = FileNameHandler.GetNextNumberedFileName(FileNameHandler.getXmlFlePath(), "outFile.xml");
    Process process = new ProcessBuilder("srcML\\src2srcml.exe", inputFile,
                                        "-o", outFile).start();

    return outFile;
}
else {
    System.out.println("\nNo XML file is created. File does not exist: " + inputFile);
}   
return null;
}

public static List<Tag> SourceToXML(String inputFile) 
    throws SAXException, ParserConfigurationException, IOException {

List<Tag> tagList = new LinkedList<Tag>();
String xmlUri = GetXmlFile(inputFile);
Document doc = GetXmlDom(xmlUri);
if (doc != null) {
    LinkedList<Integer> id = new LinkedList<Integer>();
    id.add(1);
    TagHierarchy.CreateStructuredDom(new TagId(id), doc.getFirstChild(), tagList);
    tagList.get(0).setAncestor(null);
    TagHierarchy.SetTagHierarchy(tagList);
}

return tagList;
}

Here's the exception thrown:

[Fatal Error] outFile.xml:461:300: The element type "argcl" must be terminated by the matching end-tag "". org.xml.sax.SAXParseException; systemId: file:/E:/srcML/Output/outFile.xml; lineNumber: 461; columnNumber: 300; The element type "argcl" must be terminated by the matching end-tag "". at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(Unknown Source) at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(Unknown Source) at javax.xml.parsers.DocumentBuilder.parse(Unknown Source) at vocab.util.file.FileConverter.SourceToXML(FileConverter.java:188) at vocab.CodeVocabulary.Create(CodeVocabulary.java:59) at vocab.CodeVocabulary.(CodeVocabulary.java:53) at vocab.util.DataAcccessUtil.GetCodeVocabularies(DataAcccessUtil.java:331) at vocab.TestMain.main(TestMain.java:57)

Upvotes: 0

Views: 988

Answers (1)

JB Nizet
JB Nizet

Reputation: 692231

It seems like you're starting a process which generates an XML file, and read the generated file directly after. This means that the parser will read the file while the process is running and writing to the same file. So the parser will not see the complete generated file.

You should wait for the process to finish before reading the file it generates.

You should also respect the Java naming conventions: methods start with a lowercase letter.

Upvotes: 1

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