关于java：org.springframework.orm.hibernate3.HibernateQueryException – HibernateTemplate

org.springframework.orm.hibernate3.HibernateQueryException - HibernateTemplate

我对HibernateTemplate有问题，不知道哪里出了问题。我使用的是Hibernate3和Tomcat6。在我的DAO中，我有一些函数尝试使用字符串查询数据库，它们看起来工作正常。这样地

1
2
3
4
5
6
7
8
9
10
11

public AppUser getUserByUserName(String username){
HibernateTemplate template=new HibernateTemplate(sessionFactory);
try{
List<AppUser> appList = template.find(" from AppUser as au where au.username=?", username);
tempUser = appList.get(0);
return tempUser;
} catch(Exception e){
System.out.println("Problem in AppUserDao--get byUsername:" + e.toString());
return null;
}
}

然而，当我尝试使用整数进行查询时。像：

1
2
3
4
5
6
7
8
9
10
11
12
13

public List<AppUser> getAllMerchants(){
HibernateTemplate template=new HibernateTemplate(sessionFactory);
try{
List<AppUser> appList = template.find(" from appuser as au where au.securityLevel!=?", 112);
if(appList.size() > 0)
return appList;
else
return null;
} catch(Exception e){
System.out.println("Problem in AppUserDao--getAllMerchants:" + e.toString());
return null;
}
}

号

我得到这个错误：

1	org.springframework.orm.hibernate3.HibernateQueryException: appuser is not mapped [ from appuser as au where au.securityLevel!=?]; nested exception is org.hibernate.hql.ast.QuerySyntaxException: appuser is not mapped [ from appuser as au where au.securityLevel!=?]

。

我的实体似乎有必要的注释。因为它适用于第一个函数，所以我不明白为什么它不适用于第二个函数。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

@Entity
@Table(name="appuser")
public class AppUser {
private int id;
private String username;
private String password;
private String firstName;
private String secondName;
private int state;
private int securityLevel;

@Id
@GeneratedValue(strategy = GenerationType.AUTO, generator="idSeq")
@SequenceGenerator(name="idSeq",sequenceName="app_user_seq_id")
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;
}

@Column(name="password_2")
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getSecondName() {
return secondName;
}
public void setSecondName(String secondName) {
this.secondName = secondName;
}
public int getState() {
return state;
}
public void setState(int state) {
this.state = state;
}
public int getSecurityLevel() {
return securityLevel;
}
public void setSecurityLevel(int securityLevel) {
this.securityLevel = securityLevel;
}

}

。

你的代码好像有错别字。在第一个函数中，您使用"from appuser"，在第二个函数中使用"from appuser"。尝试将第二个查询更改为"from appuser"。