Java 从另一个servlet表单标记调用servlet_Java_Servlets

Java 从另一个servlet表单标记调用servlet

java servlets

Java 从另一个servlet表单标记调用servlet,java,servlets,Java,Servlets,我有一个带有表单标签的Servlet。在这个表单标记中，我想调用另一个servlet out.println("<form id=\"myform\" action='/SubmitHome' method=\"post\">"); out.println（“”）；因此，当我单击“提交”按钮时： out.println("<input type=\"submit\" name=\"assignButton\" value=\"Assign\" /> out.pri

我有一个带有表单标签的Servlet。在这个表单标记中，我想调用另一个servlet

out.println("<form id=\"myform\" action='/SubmitHome' method=\"post\">");

out.println（“”）；

因此，当我单击“提交”按钮时：

out.println("<input type=\"submit\" name=\"assignButton\" value=\"Assign\" />

out.println（“

这不是所谓的“亚微米”

你知道为什么不打电话吗？

请检查你的HTML，你的标签是否正确关闭？ e、 g

Sytanx在下面的行中是错误的（您需要填写

println

大括号：

out.println（“

应该是

out.println（“”；

其他一切都很完美，工作正常：）

以下是完成任务的示例代码

步骤1：创建一个servlet

    package org.smarttechie;

    import java.io.IOException;
    import java.io.PrintWriter;

    import javax.servlet.ServletException;
    import javax.servlet.http.HttpServlet;
    import javax.servlet.http.HttpServletRequest;
    import javax.servlet.http.HttpServletResponse;

    /**
     * Servlet implementation class SampleServlet1
     */
    public class SampleServlet1 extends HttpServlet {
        private static final long serialVersionUID = 1L;

        /**
         * @see HttpServlet#HttpServlet()
         */
        public SampleServlet1() {
            super();
            // TODO Auto-generated constructor stub
        }

        /**
         * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
         */
        protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
            PrintWriter out = response.getWriter();
            out.println("<html><body>");
            out.println("<form action=\"SampleServlet2\" method=\"get\">");
            out.println("<input type=\"submit\" value=\"Submit\">");
            out.println("</form>");
            out.println("</body></html>");
        }
    }

步骤3:web.xml配置如下所示

<?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" id="WebApp_ID" version="2.5">
  <servlet>
    <description></description>
    <display-name>SampleServlet1</display-name>
    <servlet-name>SampleServlet1</servlet-name>
    <servlet-class>org.smarttechie.SampleServlet1</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>SampleServlet1</servlet-name>
    <url-pattern>/SampleServlet1</url-pattern>
  </servlet-mapping>
  <servlet>
    <description></description>
    <display-name>SampleServlet2</display-name>
    <servlet-name>SampleServlet2</servlet-name>
    <servlet-class>org.smarttechie.SampleServlet2</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>SampleServlet2</servlet-name>
    <url-pattern>/SampleServlet2</url-pattern>
  </servlet-mapping>
</web-app>


样本服务1
样本服务1
org.smarttechie.SampleServlet1
样本服务1
/样本服务1
样本服务2
样本服务2
org.smarttechie.SampleServlet2
样本服务2
/样本服务2

希望这能解决您的问题。

您应该检查web.xml中的url映射，这是表单未在所需页面上提交的唯一原因。

表单中是否显示了提交按钮？？表单

操作

属性与您当前所在的url主机相关。您的应用程序是否有上下文路径？另外，是否有y你关闭了你的表单？把HTML放在Java代码中是个坏主意。使用firebug，看看控制台上出现了什么错误？你在web.xml中定义了正确的映射吗？这是我在web.xml中的条目：submitome/submitome。实际上这个项目是一个基于restlet的web项目。所以我无法调用jsp文件。所以我创建了这些servlets。现在我无法将表单从第一个servlet提交到另一个servlet。@VikastjavaDeveloper当您单击“提交”按钮时会发生什么？您的输入标记不是自动关闭的

package org.smarttechie;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * Servlet implementation class SampleServlet2
 */
public class SampleServlet2 extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public SampleServlet2() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        System.out.println("The request is reached");
    }

}

<?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" id="WebApp_ID" version="2.5">
  <servlet>
    <description></description>
    <display-name>SampleServlet1</display-name>
    <servlet-name>SampleServlet1</servlet-name>
    <servlet-class>org.smarttechie.SampleServlet1</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>SampleServlet1</servlet-name>
    <url-pattern>/SampleServlet1</url-pattern>
  </servlet-mapping>
  <servlet>
    <description></description>
    <display-name>SampleServlet2</display-name>
    <servlet-name>SampleServlet2</servlet-name>
    <servlet-class>org.smarttechie.SampleServlet2</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>SampleServlet2</servlet-name>
    <url-pattern>/SampleServlet2</url-pattern>
  </servlet-mapping>
</web-app>