【发布时间】:2018-12-01 08:32:21
【问题描述】:
我正在使用 JavaEE 构建一个非常简单的 Java 全栈应用程序,因此我使用带有 Prime Faces 6.2 的 JSF 在前端渲染 xthml,在后端渲染 EJB、JPA 和 Hibernate 和 postgresql,但是,当我从 Prime Faces 的 dataTable 组件中设置了 rowKey="#{person.id}"。抛出下一个异常。
"00:13:36,307 错误 [io.undertow.request](默认任务 55)UT005023:对 /javaee-app/faces/listPersons.xhtml 的异常处理请求:javax.servlet.ServletException ... 由:java.lang.NullPointerException"
listPersons.xhtml
Prime Faces DataTable Component(opening tag and its attributes)
<p:dataTable id="persons"
value="#{personBean.persons}"
var="person"
editable="true"
rowKey="#{person.id}"
selection="#{personBean.personSelected}"
selectionMode="single">
The exception thrown that appears when trying to render the page is this.
但是,如果我设置 rowKey="#{person.name}" 甚至 rowKey="#{person.email}" 而不是 rowKey="#{person.id}" 问题就会消失并且 xthml 页面是正确渲染。
<p:dataTable id="persons"
value="#{personBean.persons}"
var="person"
editable="true"
rowKey="#{person.name}"
selection="#{personBean.personSelected}"
selectionMode="single">
with rowKey="#{person.name}" or rowKey="#{person.email}" xhtml page is rendered correctly"
模型/实体
@Entity
@Table(name = "person")
@NamedQueries({
@NamedQuery(name = "Person.findAll", query = "SELECT p FROM Person p"),
@NamedQuery(name = "Person.findById", query = "SELECT p FROM Person p
WHERE p.id = :id")
, @NamedQuery(name = "Person.findByName", query = "SELECT p FROM
Person p WHERE p.name = :person_name")
, @NamedQuery(name = "Person.findByLastname", query = "SELECT p FROM
Person p WHERE p.lastname = :lastname")
, @NamedQuery(name = "Person.findByEmail", query = "SELECT p
FROM Person p WHERE p.email = :email")
, @NamedQuery(name = "Person.findByPhone", query = "SELECT p
FROM Person p WHERE p.phone = :phone")})
public class Person implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 60)
@Column(name = "person_name")
private String name;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 60)
private String lastname;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 60)
private String email;
@Size(max = 60)
private String phone;
@OneToMany(mappedBy = "person", fetch = FetchType.EAGER)
private List<Users> users;
public Person() { }
public Person(Integer id) {
this.id = id;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getLastname() {
return lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPhone() {
return phone;
}
public void setPhone(String phone) {
this.phone = phone;
}
public List<Users> getUsers() {
return users;
}
public void setUsers(List<Users> users) {
this.users = users;
}
@Override
public int hashCode() {
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
}
@Override
public boolean equals(Object object) {
if (!(object instanceof Person)) {
return false;
}
Person other = (Person) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
return false;
}
return true;
}
@Override
public String toString() {
return "Person [id = " + id + ", name=" + name
+ ", lastName=" + lastname + " email=" + email + ", phone=" + phone + "]";
}
有解决这个问题的想法吗?提前致谢
我还添加了应用服务器中生成的错误图像,在我的例子中是 Wildfly 8.2
【问题讨论】:
-
person 实体是如何填充的?我看到 5 个构造函数,一个没有 id 参数。你确定那不是用来构造实体的吗?为什么你还需要这么多?
-
您是如何填充列表的?错误
Caused by:java.lang.NullPointerException表示至少一个人的 ID 为空。默认情况下,JPA 调用空构造函数,而该构造函数恰好将 ID 保留为null。也许在您支持的某个时刻,您向列表中添加了一个新人(尚未在数据库中持久化),而这个人恰好有一个空 ID? -
@kTest 感谢您的关注,我删除了不必要的构造函数,但是,问题仍然存在,关于空构造函数,这是必要的,因为它是默认构造函数。
-
@Al1 也谢谢你,你说的有道理,但是,dataTable 根本没有呈现。如果您检查它是从名为 person 的表中填充的实体,我添加了一个包含表当前内容的屏幕截图。当要呈现 xhtml 时,它会带来数据库中的人员,因此它应该可以正常工作,但我还没有得到它。
-
你能把 getId() 改成返回 Integer 而不是 int 吗?
标签: java hibernate jpa jsf jakarta-ee