【发布时间】:2015-08-23 16:51:29
【问题描述】:
如何从 html 调用 servlet 文件。我试过了,但是当我点击提交时,它没有采取任何行动,我也没有收到任何错误。我想在数据库中提交数据。操作事件不起作用。请帮助我
HTML 代码
<form name="form" method="post" action="NewServlet" >
<label for='name' ><font size="2">Your Full Name*: </label><br/>
<input type='text' name='name' id='name' maxlength="50" style="height:30px; width :250;" placeholder="Enter Full Name"/><br/><br/>
<label for='email' >Email Address*:</label><br/>
<input type='text' name='email' id='email' maxlength="50" style="height:30px; width :250;" placeholder="Enter your Email"/><br/><br/>
<label for='phone' >Phone Number*:</label><br/>
<input type='text' name='phone' id='phone' maxlength="15" style="height:30px; width :250;" placeholder="Enter Phone Number"/><br/><br/>
<label for='Reason' >Reason*:</label><br/>
<select name="reason" style="height:30px; width :250;">
<option>Select</option>
<option>Enquiry</option>
<option>Complain</option>
<option>Order</option>
</select>
</br>
</br>
</br>
<label for='message' >Address Or Message:</label><br/>
<textarea style="height:100px; width :400;" name='message' id='message' placeholder="Enter Address or Message"></textarea></p>
</b>
</size>
<%--
<input type='submit' name='btnSubmit' value="Submit"/>
--%>
<img src="Image/submit1.png" style="width:150px; height:70px;top:50%px; " onmouseover="this.src='Image/submit2.png'" onmouseout="this.src='Image/submit1.png'"/>
</form>
Servlet 代码
public class NewServlet extends HttpServlet {
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
String driver= "com.microsoft.sqlserver.jdbc.SQLServerDriver";
String url="jdbc:sqlserver://localhost:1433;databaseName=wristwatch; username=sa; password=abc@123;";
try {
//int id=Integer.parseInt(request.getParameter("txt_id"));
String nm=request.getParameter("name");
String email=request.getParameter("email");
int phone=Integer.parseInt(request.getParameter("phone"));
String reason=request.getParameter("reason");
String add=request.getParameter("Address");
Class.forName(driver);
Connection c=DriverManager.getConnection(url);
out.println("Data Inserted");
}
catch(Exception e)
{
System.out.print(e);
}
finally {
out.close();
}
}
web.xml
<servlet>
<servlet-name>NewServlet</servlet-name>
<servlet-class>NewServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>NewServlet</servlet-name>
<url-pattern>/NewServlet</url-pattern>
</servlet-mapping>
【问题讨论】:
-
请同时发布您的
web.xml文件,因为我们可能需要它来回答您的问题。 -
如果您使用的是 Apache TomCat,您是否在 web.xml 文件中添加了有关您的 servlet 的信息?
-
web.xml
NewServlet NewServlet NewServlet /NewServlet -
将其添加(编辑)到您的问题并正确格式化。
-
您的
doPost()方法在 servlet 中在哪里?