JavaWeb的jsp界面怎么显示Raise_application_error里的错误信息，在Servlet中怎么获取错误代码和内容？

Question

例如：触发器中当不满足某条件时执行Raise_application_error(-20000, '超量');
并在jsp界面中给出错误提示：超量
具体的怎么实现

Answer 1

自定义异常和Result工具类，如果不满足条件就抛异常，接受到异常就给Result工具类变量赋值，然后传递到jsp页面

/**
 * 
 * @ClassName: CMSException
 * @Description: 自定义异常
 * @author: dell
 * @date: 
 */
public class CMSException extends RuntimeException {

    /**
     * @fieldName: serialVersionUID
     * @fieldType: long
     * @Description: TODO
     */
    private static final long serialVersionUID = 1L;

    private String message;

    public CMSException() {
        super();

    }

    public CMSException(String message) {
        super();
        this.message = message;
    }

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }

/**
 * 
 * @ClassName: Result 
 * @Description: 返回统一的结果值
 * @author: charles
 * @date: 
 * @param <T>
 */
public class CMSResult<T> {
   //返回结果的状态
    private Integer code;
    private String  msg;
    private T data;
    public Integer getCode() {
        return code;
    }
    public void setCode(Integer code) {
        this.code = code;
    }
    public String getMsg() {
        return msg;
    }
    public void setMsg(String msg) {
        this.msg = msg;
    }
    public T getData() {
        return data;
    }
    public void setData(T data) {
        this.data = data;

    }

        try {

            //登录成功 存入session 普通用户
            session.setAttribute("user", u);
            result.setCode(200);
            result.setMsg("恭喜登录成功");
        } catch (CMSException e) {
            result.setCode(300);
            result.setMsg(e.getMessage());
        }catch (Exception e) {
            e.printStackTrace();
            result.setCode(500);
            result.setMsg("系统崩溃，请联系管理员");
        }