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java - 使用 Java HttpServlet 解码 Soap 信封

转载 作者:行者123 更新时间:2023-12-02 03:37:23 25 4
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我正在尝试使用一个简单的 HttpServlet 来管理来自另一台服务器的 SOAP 请求。该请求只有一个byte[]类型的参数(一个简单的字符串)。

相关代码是:

@Override
protected void service(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
try {
InputStream is = req.getInputStream();
byte[] body = IOUtils.toByteArray(is);
String stringRequest = new String(body);
log.info("Request -> "+stringRequest );
}catch(Exception){log.error(e);}

我收到请求,如果我打印它,它会以这种方式显示:

    <?xml version="1.0" encoding="UTF-8"?>
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<soapenv:Body>
<fixedResearch soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/">
<MYPARAMETER xsi:type="xsd:hexBinary">
*****bytearray******
</MYPARAMETER>
</fixedResearch>
</soapenv:Body>
</soapenv:Envelope>

我需要获取 MYPARAMETER 标记内的值(它是一个 byte[])。有一种聪明的方法,也许使用 Axis1 的一些 utils 类(我不能使用 Axis2)来净化传入的请求?

最佳答案

我不确定您可以使用什么,但“手动”方式是使用 XPath。下面的代码我运行了一次,它似乎有效。它没有利用命名空间,但它是一个开始。您需要优化它 - 这只是一个示例,但我已经有代码可以完成 99% 的工作。

package tld.domainname.stuff;


import java.io.IOException;
import java.io.StringReader;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;


public class XPathTest {
public static void main(String[] argv) {
String xmlString = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
"<soapenv:Envelope xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\">\n" +
" <soapenv:Body>\n" +
" <fixedResearch soapenv:encodingStyle=\"http://schemas.xmlsoap.org/soap/encoding/\">\n" +
" <MYPARAMETER xsi:type=\"xsd:hexBinary\">\n" +
" *****bytearray******\n" +
" </MYPARAMETER>\n" +
" </fixedResearch>\n" +
" </soapenv:Body>\n" +
"</soapenv:Envelope>";

String value = null;

XPath xpath = XPathFactory.newInstance().newXPath();
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = null;
Document doc = null;
try {
builder = factory.newDocumentBuilder();
} catch (ParserConfigurationException e) {
e.printStackTrace();
}
InputSource is = new InputSource(new StringReader(xmlString));
try {
doc = builder.parse(is);
} catch (SAXException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}

if (doc == null) {
System.out.println("can't parse doc");
return;
}

Node parentNode = doc.getDocumentElement();

String path = "Body/fixedResearch/MYPARAMETER";

NodeList nodeList;
try {
nodeList = (NodeList) xpath.evaluate(path, parentNode, XPathConstants.NODESET);
} catch (XPathExpressionException xpe) {
throw new IllegalArgumentException("Cannot evaluate xpath with path \"" + path + "\"", xpe);
}

if ((nodeList == null) || (nodeList.getLength() == 0)) {
System.out.println("found nothing");
return;
}

if (nodeList.getLength() > 1)
System.out.println("found " + nodeList.getLength() + " nodes in the path \"" + path + "\" - using only the first");

Node nextNode = nodeList.item(0);

if (nextNode == null) {
System.out.println("found nothing");
return;
}

if (nextNode.hasChildNodes()) {
Node child = nextNode.getFirstChild();
value = child.getNodeValue();
}

System.out.println("found value of \"" + value + "\"");
}
}

关于java - 使用 Java HttpServlet 解码 Soap 信封,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37282671/

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