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java - 如何从 HttpServletRequest (Tomcat 9.0) 获取查询参数?

转载 作者:塔克拉玛干 更新时间:2023-11-03 03:36:43 29 4
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服务器接收来自两个客户端的请求——Raspberry Pi 和 Android 应用程序,它们都使用 HttpURLConnection 发送请求。我需要通过这些请求传递参数,例如:

http://192.168.0.10:8080/MyProject/MyServer/rpi/checktask?rpi="rpi"

这样做:

String requestUrl = "http://192.168.0.10:8080/MyProject/MyServer/rpi";
String query = String.format("/checktask?rpi=%s",
URLEncoder.encode("rpi", "UTF-8"));
URL url = new URL(requestUrl + query);
conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setDoInput(true);
conn.setRequestProperty("Accept-Charset", "UTF-8");
conn.connect();

Servlet 有注解:

@WebServlet(name = "MyServer", urlPatterns = { "/MyServer/rpi/*", "/MyServer/app/*"})

但是当 Servlet 收到上述请求时,会发生以下情况:

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

String path = request.getRequestURI(); // /MyProject/MyServer/rpi/*
String query = request.getQueryString(); // null
String context = request.getContextPath(); // /MyProject
String servlet = request.getServletPath(); // /MyServer/rpi
String info = request.getPathInfo(); // /*
}

虽然根据那些答案: How to use @WebServlet to accept arguments (in a RESTFul way)?How come request.getPathInfo() in service method returns null?

它应该是这样的:

String path = request.getRequestURI();     // /MyProject/MyServer/rpi//checktask?rpi="rpi"
String query = request.getQueryString(); // rpi="rpi"
String context = request.getContextPath(); // /MyProject
String servlet = request.getServletPath(); // /MyServer/rpi
String info = request.getPathInfo(); // /checktask?rpi="rpi"

我做错了什么?

最佳答案

您的网址字符串是

http://192.168.0.10:8080/MyProject/MyServer/rpi/checktask?rpi="rpi" 

上述String中的参数名称为“rpi”。

下面的代码将为您提供参数“rpi”的所需值。

String rpi = request.getParameter("rpi");

关于java - 如何从 HttpServletRequest (Tomcat 9.0) 获取查询参数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39509892/

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