HTTP状态 500 -内部服务器错误
来源:2-12 自由编程
weixin_慕的地0390393
2022-10-14 20:05:07
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 SampleServlet extends HttpServlet {
@Override
protected void service(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
String number = req.getParameter(“n”);
int n = Integer.parseInt(number);
int add = 0;
for(int i = 1; i <= n;i++){
add += i;
}
PrintWriter out = resp.getWriter();
out.println(“累加和为:”+add);
}
}
second
SecondServlet
second
/hi
sample
SampleServlet
sample
/sample
4.0.0
<groupId>org.example</groupId>
<artifactId>SecondSerclet</artifactId>
<version>1.0-SNAPSHOT</version>
<dependencies>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>javax.servlet-api</artifactId>
<version>3.1.0</version>
</dependency>
</dependencies>
HTTP状态 500 -内部服务器错误
java.lang.NumberFormatException: null
java.lang.Integer.parseInt(Integer.java:542)
java.lang.Integer.parseInt(Integer.java:615)
SampleServlet.service(SampleServlet.java:12)
javax.servlet.http.HttpServlet.service(HttpServlet.java:765)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
1回答
好帮手慕小蓝
2022-10-15
同学你好,报错信息中描述了后端代码抛出了NumberFormatException并且是因为值为null引起的,发生的位置为int n = Integer.parseInt(number);语句。
猜测同学访问时使用的url中没有拼接请求参数“?n=5”。
建议同学在请求后拼接参数再尝试一下。
祝学习愉快~
相似问题
回答 1
回答 1
回答 2
回答 1
回答 1