我正在使用Spring 3.2.3,JPA 2.1,JUnit 4.11。我試圖運行一個junit測試,並且我一直得到抽象模式類型是未知錯誤。這裏是我的實體(截斷空間,它擁有所有的getter和setter):抽象模式類型未知
@Entity
@Table(name = "WEB_PROFILES")
public class TestWebProfile implements Serializable {
private static final long serialVersionUID = 1L;
@Transient
private String forward;
@Column(name = "ACCESS_FLAG")
private String accessFlag;
@Temporal(TemporalType.DATE)
@Column(name = "ACCESS_FLAG_UPD_DATE")
private Date accessFlagUpdDate;
@Column(name = "ACCESS_RESET_INTERVAL")
private BigDecimal accessResetInterval;
@Column(name = "ACCOUNT_TYPE")
private String accountType;
@Column(name = "CREATED_BY")
private String createdBy;
@Column(name = "E_MAIL")
private String eMail;
@Column(name = "FAILED_LOGIN_ATTEMPTS")
private BigDecimal failedLoginAttempts;
@Column(name = "FIRST_NAME")
private String firstName;
@Temporal(TemporalType.DATE)
@Column(name = "FROI_ACCESS_APPROVE_DENY_DATE")
private Date froiAccessApproveDenyDate;
@Column(name = "FROI_ACCESS_APPROVED_FLAG")
private String froiAccessApprovedFlag;
@Column(name = "FROI_ACCESS_REQUESTED")
private String froiAccessRequested;
@Column(name = "FROI_APPROVED_BY")
private String froiApprovedBy;
@Temporal(TemporalType.DATE)
@Column(name = "FROI_CONFIRM_EMAIL_SENT_DATE")
private Date froiConfirmEmailSentDate;
@Temporal(TemporalType.DATE)
@Column(name = "FROI_LETTER_SENT_DATE")
private Date froiLetterSentDate;
@Column(name = "LAST_LOGON_ADDR")
private String lastLogonAddr;
@Temporal(TemporalType.DATE)
@Column(name = "LAST_LOGON_DATE")
private Date lastLogonDate;
@Column(name = "LAST_NAME")
private String lastName;
@Column(name = "LAST_UPDATED_BY")
private String lastUpdatedBy;
@Column(name = "LAST_UPDATED_BY_NAME")
private String lastUpdatedByName;
@Column(name = "LAST_UPDATED_BY_SU_ID")
private BigDecimal lastUpdatedBySuId;
@Temporal(TemporalType.DATE)
@Column(name = "MAIL_SENT_DATE")
private Date mailSentDate;
@Temporal(TemporalType.DATE)
@Column(name = "MAINT_DATE")
private Date maintDate;
@Temporal(TemporalType.DATE)
@Column(name = "NEW_PIN_REQ_DATE")
private Date newPinReqDate;
@Column(name = "PASSWORD")
private String password;
@Transient
private String newPassword;
@Temporal(TemporalType.DATE)
@Column(name = "PASSWORD_UPD_DATE")
private Date passwordUpdDate;
@Column(name = "PHONE")
private String phone;
@Column(name = "PIN")
private String pin;
@Column(name = "POLICY_NUM")
private BigDecimal policyNo;
@Column(name = "PROFILE_CLASS_CODE")
private String profileClassCode;
@Temporal(TemporalType.DATE)
@Column(name = "PROFILE_REQ_DATE")
private Date profileReqDate;
@Temporal(TemporalType.DATE)
@Column(name = "PROFILE_UPDATE_DATE")
private Date profileUpdateDate;
@Column(name = "REMOTE_ADDR")
private String remoteAddr;
@Column(name = "SESSIONID")
private String sessionid;
@Column(name = "SUBSCRIBER_FLAG")
private String subscriberFlag;
@Column(name = "USER_ID")
private BigDecimal userId;
@Id
@Column(name = "USER_NO")
private BigDecimal userNo;
@Column(name = "USERNAME")
private String username;
我的JUnit測試:
@Test
public void testGetWebProfileByUsername() {
TestWebProfile wp = sso.getWebProfile("MARLENE");
System.out.println("name :" + wp.getFirstName());
System.out.println("last name :" + wp.getLastName());
}
我的DAO實現:
@Override
public TestWebProfile getWebProfile(String username) {
String sqlString = "select w from TestWebProfile w where w.username =:username";
return (TestWebProfile) getEntityManager()
.createQuery(sqlString, TestWebProfile.class)
.setParameter("username", username).getSingleResult();
}
谷歌搜索後在過去的一個小時裏,我發現似乎有意義的唯一罪魁禍首是沒有@Id和@Column註釋,但是我擁有那些關於userNo變量的註釋。任何可以提供的幫助將不勝感激!
謝謝@MGPJ,我看到了這篇文章。再看一遍後,我發現了這個問題。我們使用Spring的Java配置和entityManager bean,我需要添加一個額外的包來掃描。一旦我做到了,它就很好。 – NuAlphaMan