如何将HttpServletRequest转换为String？ [英] How to convert an HttpServletRequest to String?

问题描述

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。

How can I convert an HttpServletRequest to String? I need to unmarshall the HttpServletRequest but when I try to, my program throws an exception.

 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)

我试过以下代码解组 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;
    }

推荐答案

我的印象是你'在处理完请求并响应客户端后，尝试读取输入流。你把代码放在哪里了？

I have the impression you're trying to read from the input stream after you've handled the request and responded to your client. Where did you put your code?

如果你想先处理请求，然后再进行解组，你需要先将输入流读入一个字符串。如果你正在处理小请求，这样可以正常工作。

If you want to handle the request first, and do the unmarshalling later, you need to read the inputstream into a String first. This works fine if it's small requests you're handling.

我建议使用像apache commons IOUtils这样的东西为你做这个。

I suggest using something like apache commons IOUtils to do this for you.

String marshalledXml = org.apache.commons.io.IOUtils.toString(request.getInputStream());

另请注意，您必须在 request.getParameter（名称）之间进行选择）和 request.getInputStream 。你不能同时使用它们。

Also keep in mind that you have to choose between request.getParameter(name) and request.getInputStream. You can't use both.

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