用于过滤掉无效 xml 字符的 JAX WS SOAPHandler

Question

我有一个 Web 服务，其中客户端传递一个包含一些无效 xml 字符的字符串。在服务器端解析请求时，JAXWS 会抛出异常，因为它无法解析无效的 xml 字符。

为了解决这个问题，我尝试创建以下 SOAPHandler。在下面的 SOAPHandler 中，我尝试迭代子元素，这意味着我已经在解析 xml 并因此在处理程序本身中得到异常。

如何从邮件中删除无效的 xml 字符？

import java.util.Set;

import javax.xml.namespace.QName;
import javax.xml.soap.SOAPBodyElement;
import javax.xml.soap.SOAPException;
import javax.xml.soap.SOAPMessage;
import javax.xml.ws.handler.MessageContext;
import javax.xml.ws.handler.soap.SOAPHandler;
import javax.xml.ws.handler.soap.SOAPMessageContext;

import org.apache.commons.lang.StringUtils;
import org.apache.log4j.Logger;

public class InvalidXmlCharacterHandler implements SOAPHandler<SOAPMessageContext>{

     private static final Logger LOGGER = Logger.getLogger(InvalidXmlCharacterHandler.class);


    @Override
    public boolean handleMessage(SOAPMessageContext context) {

        System.out.println("Server : handleMessage()......");

        Boolean isRequest = (Boolean) context.get(MessageContext.MESSAGE_OUTBOUND_PROPERTY);

        //for request message only
        if(!isRequest){

            SOAPMessage soapMsg = context.getMessage();

            if (soapMsg != null) {
               try {

                   java.util.Iterator iterator = soapMsg.getSOAPBody().getChildElements();
                   while (iterator.hasNext()) {
                     SOAPBodyElement bodyElement = (SOAPBodyElement) iterator.next();
                     String val = bodyElement.getTextContent();
                     bodyElement.setTextContent(stripNonValidXMLCharacters(val));    
                     System.out.println("The Value is:" + val);
                   }


               } catch (SOAPException ex) {
                  LOGGER.error("Failed to get and set source", ex);
               }

            }

        }

        //continue other handler chain
        return true;
    }




     public static String stripNonValidXMLCharacters(String in) {
          StringBuffer out = new StringBuffer(); // Used to hold the output.
          char current; // Used to reference the current character.

          if (in == null || ("".equals(in))) return ""; // vacancy test.
          for (int i = 0; i < in.length(); i++) {
              current = in.charAt(i); // NOTE: No IndexOutOfBoundsException caught here; it should not happen.
              if ((current == 0x9) ||
                  (current == 0xA) ||
                  (current == 0xD) ||
                  ((current >= 0x20) && (current <= 0xD7FF)) ||
                  ((current >= 0xE000) && (current <= 0xFFFD)) ||
                  ((current >= 0x10000) && (current <= 0x10FFFF)))
                  out.append(current);
          }
          return out.toString();
      }   

    @Override
    public boolean handleFault(SOAPMessageContext context) {

        System.out.println("Server : handleFault()......");

        return true;
    }

    @Override
    public void close(MessageContext context) {
        System.out.println("Server : close()......");
    }

    @Override
    public Set<QName> getHeaders() {
        System.out.println("Server : getHeaders()......");
        return null;
    }

}

Answer 1

您需要做的是更改 XML 而不解析它，正如您已经知道的那样。这不一定是一个简单的问题，但至少更通用。

首先将您的消息作为裸字符串：

ByteArray OutputStream out = new ByteArrayOutputStream();
soapMsg.writeTo(out);
String messageAsString = new String(out.toByteArray());

然后只需在字符串上使用 stripNonValidXMLCharacters，并将结果用作 SOAPMessageContext::setMessage 的输入，在您的情况下似乎是 context.setMessage(...).

另请查看这些调整无效 XML 的更漂亮的方法：removing invalid XML characters from a string in java, and Parsing malformed/incomplete/invalid XML files。

用于过滤掉无效 xml 字符的 JAX WS SOAPHandler

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