exception-新手學hibernate簡單登錄遇到未知錯誤

新手學hibernate簡單登錄遇到未知錯誤

package com.gloomy.rigister.model;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;

@Entity
public class user {

private int id;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}
public String getUsername() {
    return username;
}
public void setUsername(String username) {
    this.username = username;
}
public String getPassword() {
    return password;
}
public void setPassword(String password) {
    this.password = password;
}
private String username;
private String password;

}
錯誤信息
嚴重: Developer Notification (set struts.devMode to false to disable this message):
Unexpected Exception caught setting 'passrword' on 'class com.gloomy.rigister.action.userAction: Error setting expression 'passrword' with value ['25', ]
九月 27, 2015 10:18:34 上午 com.opensymphony.xwork2.interceptor.ParametersInterceptor error
嚴重: Developer Notification (set struts.devMode to false to disable this message):
Unexpected Exception caught setting 'password1' on 'class com.gloomy.rigister.action.userAction: Error setting expression 'password1' with value ['25', ]
九月 27, 2015 10:18:34 上午 com.opensymphony.xwork2.interceptor.ParametersInterceptor error
嚴重: Developer Notification (set struts.devMode to false to disable this message):
Unexpected Exception caught setting 'username' on 'class com.gloomy.rigister.action.userAction: Error setting expression 'username' with value ['he', ]

最佳回答：

我自己粗心把index.jsp中form表單中的password寫成passrword了