首页 \ 问答 \ servlet如何始终执行相同的任务?(How can a servlet always perform the same task?)

servlet如何始终执行相同的任务?(How can a servlet always perform the same task?)

我希望Servlet始终执行相同的任务。 无论是GET还是POST。 目前我只是从doPost()调用doGet() doPost() ,这很好。

然后我尝试重写service()方法,我认为它会以同样的方式工作。 但事实并非如此!

代码以某种方式执行,但响应不生成网页:

response.getWriter();
response.println(string); 

此代码适用于doGet / doPost方法,但不适用于该服务。 为什么?

Servlet的:

class MyWebServlet extends HttpServlet {
    @Override
    public void service(ServletRequest request, ServletResponse response) {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        String string = "teststring";
        out.println(string);
    }
}

I want a Servlet to perform always the same tasks. Regardless of if it is a GET or POST. At the moment I just call the doGet() from doPost(), which works fine.

Then I tried overriding the service() method, and I thought it would just work the same way. But it does not!

The code somehow gets executed, but the response does not generate the webpage:

response.getWriter();
response.println(string); 

This code works for the doGet/doPost methods, but not for the service. Why?

Servlet:

class MyWebServlet extends HttpServlet {
    @Override
    public void service(ServletRequest request, ServletResponse response) {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        String string = "teststring";
        out.println(string);
    }
}

原文:https://stackoverflow.com/questions/12637084
更新时间:2022-03-20 16:03

最满意答案

那篇文章已经过时了; 它引用了Web API的BETA版本。 如此处所述,已发布的版本中删除了通用HttpRequestMessage<T>类。 相反,您应该使用PostAsJsonAsync<T>扩展方法。 这是一个更新的教程 ,可以为您提供所需的信息。


That article is out of date; it is referencing the BETA version of Web API. The generic HttpRequestMessage<T> class was removed in the released version as explained here. Instead you should use the PostAsJsonAsync<T> extension method. Here is a more up-to-date tutorial that should give you what you need.

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