帮我写一个代码,创建一个servlet,对其进行get请求时输出系统当前时间
时间: 2023-04-02 19:03:04 浏览: 54
好的,以下是创建一个servlet并输出系统当前时间的代码:
```java
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Date;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class TimeServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
out.println("<html><head><title>当前时间</title></head><body>");
out.println("<h2>当前时间是:" + new Date() + "</h2>");
out.println("</body></html>");
}
}
```
当你对这个servlet进行get请求时,它会输出当前系统时间。