为什么我使用String.replace和HashMap的方法不能替换字符串？

Question

I am attempting to write a little class to escape characters in an XML document. I am using xpath to get the nodes of the XML document, and passing each node to my class. However, it is not working. I want to change:

"I would like a burger & fries."

to

"I would like a burger & fries."

Here is the code for my class:

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

public class MyReplace{
    private static final HashMap<String,String> xmlCharactersToBeEscaped;
    private Iterator iterator;
    private String newNode;
    private String mapKey;
    private String mapValue;

    static {
        xmlCharactersToBeEscaped = new HashMap<String,String>();
        xmlCharactersToBeEscaped.put("\"","&quot;");
        xmlCharactersToBeEscaped.put("'","&apos;");
        xmlCharactersToBeEscaped.put("<","&lt;");
        xmlCharactersToBeEscaped.put(">","&gt;");
        xmlCharactersToBeEscaped.put("&","&amp;");
    }

    public String replaceSpecialChar(String node){
        if(node != null){
            newNode = node;
            iterator = xmlCharactersToBeEscaped.entrySet().iterator();
            while(iterator.hasNext()){
                Map.Entry mapEntry = (Map.Entry) iterator.next();
                mapKey = mapEntry.getKey().toString();
                mapValue = mapEntry.getValue().toString();

                if(newNode.contains(mapKey)){
                    newNode = newNode.replace(mapKey,mapValue);
                }
            }
            return newNode;
        } else {
            return node;
        }
    }
}

What is happening is that it is replacing the first special character in the map, the quote, and skipping everything else.

Answer 1

Your solution is over complicated.

Use StringEscapeUtils (Part of the Commons Lang library). This has a built in feature to escape and unescape XML, HTML and much more. Commons lang is very easy to import and the following examples are from the latest stable release (3.4). Previous versions use different methods, look up the Java doc dependant on your version. It's very flexible so you can do a lot more with it than just simple escapes and unescapes.

String convertedString = StringEscapeUtils.escapeXml11(inputString);

If you're using XML 1.0 they also offer the following

String convertedString10 = StringEscapeUtils.escapeXml10(inputString);

Get it here: https://commons.apache.org/proper/commons-lang/

Java docs here (3.4): https://commons.apache.org/proper/commons-lang/javadocs/api-3.4/org/apache/commons/lang3/StringEscapeUtils.html

Answer 2

Make it simpler (and see comment below):

xmlCharactersToBeEscaped = new HashMap<String,String>();
xmlCharactersToBeEscaped.put("\"","&quot;");
xmlCharactersToBeEscaped.put("'","&apos;");
xmlCharactersToBeEscaped.put("<","&lt;");
xmlCharactersToBeEscaped.put(">","&gt;");
/* xmlCharactersToBeEscaped.put("&","&amp;"); <-- don't add this to the map */

//...
public String replaceSpecialChars(String node) {
    if (node != null) {
        String newNode = node.replace("&", "&amp;"); 
        for (Map.Entry<String, String> e : xmlCharactersToBeEscaped.entrySet()) {              
             newNode = newNode.replace(e.getKey(), e.getValue());
        }
        return newNode;
    } else {
        return null;
    }
}

or better use StringEscapeUtils for such stuff.