如何将HttpServletRequest
转换为String
?我需要解组HttpServletRequest
,但是当我尝试时,我的程序会抛出异常。
javax.xml.bind.UnmarshalException
- with linked exception:
[java.io.IOException: Stream closed]
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl.java:197)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(UnmarshallerImpl.java:168)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:137)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:184)
at com.orange.oapi.parser.XmlParsing.parse(XmlParsing.java:33)
我尝试使用以下代码来解组HttpServletRequest
。
InputStreamReader is =
new InputStreamReader(request.getInputStream());
InputStream isr = request.getInputStream();
ServletInputStream req = request.getInputStream();
我的解析器方法:
public root parse(InputStreamReader is) throws Exception {
root mc = null;
try {
JAXBContext context = JAXBContext.newInstance(root.class);
Unmarshaller um = context.createUnmarshaller();
mc = (root) um.unmarshal(is);
} catch (JAXBException je) {
je.printStackTrace();
}
return mc;
}
答案 0 :(得分:5)
在您处理完请求并回复客户端后,我的印象是您尝试从输入流中读取。你把代码放在哪里了?
如果要先处理请求,稍后再进行解组,则需要先将输入流读入String。如果您正在处理的请求很少,这样可以正常工作。
我建议使用像apache commons IOUtils这样的东西为你做这件事。
String marshalledXml = org.apache.commons.io.IOUtils.toString(request.getInputStream());
另请注意,您必须在request.getParameter(name)
和request.getInputStream
之间进行选择。你不能同时使用它们。
答案 1 :(得分:1)
String httpServletRequestToString(HttpServletRequest request) throws Exception {
ServletInputStream mServletInputStream = request.getInputStream();
byte[] httpInData = new byte[request.getContentLength()];
int retVal = -1;
StringBuilder stringBuilder = new StringBuilder();
while ((retVal = mServletInputStream.read(httpInData)) != -1) {
for (int i = 0; i < retVal; i++) {
stringBuilder.append(Character.toString((char) httpInData[i]));
}
}
return stringBuilder.toString();
}