| 久丰学习Jsp&Servelet的笔记(写出一个 servelet 程序) |
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| 日期：2007-5-25 11:44:35 人气：198 [大 中 小] |
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//Begin assembling the HTML content out.println("<html><head>");
out.println("<title>Help Page</title></head><body>");
out.println("<h2>Please submit your information</h2>");
//make sure method="post" so that the servlet service method
//calls doPost in the response to this form submit
out.println( "<form method="post" action ="" + request.getContextPath( ) + "/firstservlet" >");
out.println("<table border="0"><tr><td valign="top">");
out.println("Your first name: </td> <td valign="top">");
out.println("<input type="text" name="firstname" size="20">");
out.println("</td></tr><tr><td valign="top">");
out.println("Your last name: </td> <td valign="top">");
out.println("<input type="text" name="lastname" size="20">");
out.println("</td></tr><tr><td valign="top">");
out.println("Your email: </td> <td valign="top">");
out.println("<input type="text" name="email" size="20">");
out.println("</td></tr><tr><td valign="top">");
out.println("<input type="submit" value="Submit Info"></td></tr>");
out.println("</table></form>"); out.println("</body></html>");
}
//doGet
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, java.io.IOException ...{
//display the parameter names and values
Enumeration paramNames = request.getParameterNames( );
String parName;
//this will hold the name of the parameter |
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| 出处：本站原创 作者：佚名 |
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