创建一个Servlet并重写doPost方法,然后新建一个form表单实现post形式的跳转;
时间: 2023-05-31 13:07:36 浏览: 185
servlet跳转页面的几种方法
好的,这是一个关于Java Servlet的问题,我可以回答。您可以参考以下代码实现:
```java
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
public class MyServlet extends HttpServlet {
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html;charset=utf-8");
PrintWriter out = response.getWriter();
out.println("<html>");
out.println("<head><title>Form doPost</title></head>");
out.println("<body>");
out.println("<h1>Hello World!</h1>");
out.println("</body>");
out.println("</html>");
out.close();
}
}
```
然后在web.xml文件中添加Servlet的映射,并在HTML页面中通过form表单实现post形式的跳转。如下:
```xml
<servlet>
<servlet-name>MyServlet</servlet-name>
<servlet-class>com.example.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>MyServlet</servlet-name>
<url-pattern>/MyServlet</url-pattern>
</servlet-mapping>
```
```html
<form action="MyServlet" method="post">
<input type="submit" value="Submit">
</form>
```
以上代码可以在Java Web容器中运行。希望可以帮助您解决问题。
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