0

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

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

所以当我点击提交按钮时：

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

它没有调用“/SubmitHome”。

知道为什么这不叫吗？

4

4 回答 4

0

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。

于 2013-09-05T07:29:29.980 回答
0

以下行的 Sytanx 错误（您需要完成println大括号：

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

应该

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

其他一切都很完美，工作正常:)

于 2013-09-05T05:53:30.997 回答
0

这是实现您的任务的示例代码。

第一步：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 2：创建另一个 servlet。这将从表单操作标签中调用。如果您观察到表单操作标记值，即指向不包括“/”的“SampleServlet2”url 模式。如果要给出绝对路径（包含'/'），我们需要指定上下文路径。

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");
    }

}

Step3：web.xml配置如下。

<?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>

希望这能解决您的问题。

于 2013-09-05T06:24:26.687 回答
0

请检查您的 HTML，您的标签是否正确关闭？例如

<form action="" method="post">
<input type="submit" value="Submit">
</form>
于 2013-09-05T05:44:30.823 回答