Question

我有一个接受HttpServletRequest的servlet类，我需要将请求传递给另一个方法。我怎样才能以另一种方法获得请求。

的Servlet

     public void doPost(final HttpServletRequest request,
                               final HttpServletResponse response)
      XmlParser xmlParser = new XmlParser();
      xmlParser.getRequest(request);

XmlParser类：

public class XmlParser{
    public void getRequest () //How can i get the request parameter from the servlet. 
    {
    }
}

Answer 1

将其传递给构造函数：

public class XmlParser{
    final private HttpServletRequest request;
    public XmlParser(HttpServletRequest request)
    {
        this.request = request;
    }
    // use it in othe methods...
}

Answer 2

public class XmlParser {
    private final HttpServletRequest request = null;

    public XmlParser(HttpServletRequest request) {
        this.request = request;
    }

    ... other methods

Answer 3

在XmlParser类中存储对请求的引用。当您需要从此请求中获取参数时，请在XmlParser方法中使用它。

Answer 4

public class XmlParser{
HttpServletRequest request;
public XmlParser(HttpServletRequest request){
this.request=request;
}
public void getRequest () 
{}
}

Answer 5

所有其他答案都有效，但我建议您不要将XmlParser与HttpServletRequest结合使用。从HttpServletRequest（一个读取正文内容的InputStream？）中检索所需的所有数据，并将其传递给XmlParser。

将httpservletrequest传递给方法

5 个答案: