在Java中生成XML时转义特殊字符

Question

I am trying to develop an XML export feature to give my application users to export their data in an XML format. I have got this feature ready and working until it started failing for some cases. Then I realized that it was because of some special characters that needs to be encoded. for example the data might contain & or ! or % or ' or # etc. etc. and this needs to be escaped properly. I was wondering if there is a generic utility available that can escape all of the special characters as per the XML specification. I couldn't find anything on Google.

Is there something like that already there? or Is there any other way to do it?

Here is the code I am using to generate XML

Document xmldoc = new DocumentImpl();
Element root = xmldoc.createElement("Report");

Element name= xmldoc.createElement((exportData.getChartName() == null) ? "Report" : exportData.getChartName());
if (exportData.getExportDataList().size() > 0
    && exportData.getExportDataList().get(0) instanceof Vector) {
    // First row is the HEADER, i.e name
    Vector name = exportData.getExportDataList().get(0);
    for (int i = 1; i  value = exportData.getExportDataList().get(i);
        Element sub_root = xmldoc.createElement("Data");
        //I had to remove a for loop from here. StackOverflow description field would not take that. :(
            // Insert header row
            Element node = xmldoc.createElementNS(null, replaceUnrecognizedChars(name.get(j)));
            Node node_value = xmldoc.createTextNode(value.get(j));
            node.appendChild(node_value);
            sub_root.appendChild(node);
            chartName.appendChild(sub_root);
        }
    }
}
root.appendChild(name);

// Prepare the DOM document for writing
Source source = new DOMSource(root);

// Prepare the output file
Result result = new StreamResult(file);

// Write the DOM document to the file
Transformer xformer = TransformerFactory.newInstance().newTransformer();
xformer.transform(source, result);`

Sample XML:

<Data>
    <TimeStamp>2010-08-31 00:00:00.0</TimeStamp>
    <[Name that needs to be encoded]>0.0</[Name that needs to be encoded]>
    <Group_Average>1860.0</Group_Average>
</Data>

Answer 1

You can use apache common lang library to escape a string.

org.apache.commons.lang.StringEscapeUtils

String escapedXml = StringEscapeUtils.escapeXml("the data might contain & or ! or % or ' or # etc");

But what you are looking for is a way to convert any string into a valid XML tag name . For ASCII characters, XML tag name must begin with one of _:a-zA-Z and followed by any number of character in _:a-zA-Z0-9.-

I surely believe there is no library to do this for you so you have to implement your own function to convert from any string to match this pattern or alternatively make it into a value of attritbue.

<property name="no more need to be encoded, it should be handled by XML library">0.0</property>

Answer 2

public class RssParser {
int length;
    URL url;
URLConnection urlConn;
NodeList nodeList;
Document doc;
Node node;
Element firstEle;
NodeList titleList;
Element ele;
NodeList txtEleList;
String retVal, urlStrToParse, rootNodeName;

public RssParser(String urlStrToParse, String rootNodeName){
    this.urlStrToParse = urlStrToParse;
    this.rootNodeName = rootNodeName;

    url=null;
    urlConn=null;
    nodeList=null;
    doc=null;
    node=null;
    firstEle=null;
    titleList=null;
    ele=null;
    txtEleList=null;
    retVal=null;
            doc = null;
    try {
        url = new URL(this.urlStrToParse);
                    // dis is path of url which v'll parse
        urlConn = url.openConnection();

                    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        DocumentBuilder db = dbf.newDocumentBuilder();

        String s = isToString(urlConn.getInputStream());
        s = s.replace("&", "&");
        StringBuilder sb =
                            new StringBuilder
                                    ("<?xml version=\"1.0\" encoding=\"utf-8\"?>");
        sb.append("\n"+s);
        System.out.println("STR: \n"+sb.toString());
        s = sb.toString();

        doc = db.parse(urlConn.getInputStream());
        nodeList = doc.getElementsByTagName(this.rootNodeName); 
        //  dis is d first node which
        //  contains other inner element-nodes
        length =nodeList.getLength();
        firstEle=doc.getDocumentElement();
    }
    catch (ParserConfigurationException pce) {
        System.out.println("Could not Parse XML: " + pce.getMessage());
    }
    catch (SAXException se) {
        System.out.println("Could not Parse XML: " + se.getMessage());
    }
    catch (IOException ioe) {
        System.out.println("Invalid XML: " + ioe.getMessage());
    }
    catch(Exception e){
        System.out.println("Error: "+e.toString());
    }
}


public String isToString(InputStream in) throws IOException {
    StringBuffer out = new StringBuffer();
    byte[] b = new byte[512];
    for (int i; (i = in.read(b)) != -1;) {
        out.append(new String(b, 0, i));
    }
    return out.toString();
}

public String getVal(int i, String param){
    node =nodeList.item(i);
    if(node.getNodeType() == Node.ELEMENT_NODE)
    {
        System.out.println("Param: "+param);
        titleList = firstEle.getElementsByTagName(param);
        if(firstEle.hasAttribute("id"))
        System.out.println("hasAttrib----------------");
        else System.out.println("Has NOTNOT      NOT");
        System.out.println("titleList: "+titleList.toString());
    ele = (Element)titleList.item(i);
    System.out.println("ele: "+ele);
        txtEleList = ele.getChildNodes();
    retVal=(((Node)txtEleList.item(0)).getNodeValue()).toString();
    if (retVal == null)
        return null;
            System.out.println("retVal: "+retVal);
    }
return retVal;
}
}

Answer 3

使用下面的代码使用XML来转义字符串中的字符.StringEscapeUtils在apche commons lang3 jar中可用

StringEscapeUtils.escapeXml11("String to be escaped");