在Servlet 中的获取所有请求参数

Myth丶恋晨 2023-09-25 12:16 65阅读 0赞

在本教程中，我们将向您展示如何在 Java Servlet 中获取所有请求参数。这是开发 Servelt 应用程序时必须考虑的最基本步骤，因为 HTTP 主要基于参数交换。

基本上，为了获取 Servlet 中的所有请求参数，应该执行以下步骤：

让我们看看下面的简单代码片段：

package com.javacodegeeks.snippets.enterprise;
    
    import java.io.IOException;
    import java.io.PrintWriter;
    import java.util.Enumeration;
    
    import javax.servlet.http.HttpServlet;
    import javax.servlet.http.HttpServletRequest;
    import javax.servlet.http.HttpServletResponse;
    
    public class GetAllRequestParametersInServlet extends HttpServlet {
    
        private static final long serialVersionUID = -2128122335811219481L;
    
        public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException {
            handleRequest(req, res);
        }
    
        public void doPost(HttpServletRequest req, HttpServletResponse res) throws IOException {
            handleRequest(req, res);
        }
    
        public void handleRequest(HttpServletRequest req, HttpServletResponse res) throws IOException {
    
            PrintWriter out = res.getWriter();
            res.setContentType("text/plain");
    
            Enumeration<String> parameterNames = req.getParameterNames();
    
            while (parameterNames.hasMoreElements()) {
    
                String paramName = parameterNames.nextElement();
                out.write(paramName);
                out.write("\n");
    
                String[] paramValues = req.getParameterValues(paramName);
                for (int i = 0; i < paramValues.length; i++) {
                    String paramValue = paramValues[i];
                    out.write("\t" + paramValue);
                    out.write("\n");
                }
    
            }
    
            out.close();
    
        }
    
    }

**web.xml**

<?xml version="1.0" encoding="UTF-8"?>
    
    <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
             xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
             version="2.5">
    
        <display-name>JCG Snippets Web Project</display-name>
    
        <servlet>
            <servlet-name>JCG Snippets Application</servlet-name>
            <servlet-class>com.javacodegeeks.snippets.enterprise.GetAllRequestParametersInServlet</servlet-class>
        </servlet>
    
        <servlet-mapping>
            <servlet-name>JCG Snippets Application</servlet-name>
            <url-pattern>/jcgservlet</url-pattern>
        </servlet-mapping>
    
    </web-app>

**网址：**

http://localhost:8080/jcgsnippets/jcgservlet?param1=paramvalue1&param2=paramvalue2a&param2=paramvalue2b

**输出：**

param2
    	paramvalue2a
    	paramvalue2b
    param1
    	paramvalue1

这是一个关于如何在 Servlet 中获取所有请求参数的示例。