Question

我在使用Hibernate 3.5.1-Final作为提供商的JPA 2.0 Criteria API中获得了非常奇怪的行为。

我正在尝试在JPQL中构建一个看起来像这样的动态查询：

 SELECT e FROM Employee e WHERE lower(e.firstName) like lower(:employeeName) OR lower(e.lastName) like lower(:employeeName)

但不断收到错误

java.lang.IllegalArgumentException: org.hibernate.QueryException: Not all named parameters have been set: [param0] [select generatedAlias0 from com.company.model.Employee as generatedAlias0 where ( lower(generatedAlias0.firstName) like lower(:param0) ) or ( lower(generatedAlias0.lastName) like lower(:param1) ) order by generatedAlias0.firstName asc]

如果我取走lastName的路径，它就可以了。难道我做错了什么？以下是涉及的类和查询。谢谢！

AbstractEntity.java：

@MappedSuperclass
@SuppressWarnings("serial")
public abstract class AbstractEntity<ID extends Serializable> extends AbstractBaseEntity
    implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Basic(optional = false)
@Column(nullable = false)
protected ID id;

/**
 * Gets the identifier of the entity.
 * 
 * @since 0.0.1
 * @return the identifier
 */
public ID getId() {
    return this.id;
}

/**
 * Sets the identifier of the entity
 *
 * @since 0.0.1
 * @param id the identifier
 */
public void setId(ID id) {
    this.id = id;
}

/**
 * Determines if the entity is new by checking if the Id is null
 * 
 * @since 0.0.1
 * @return true if id == null
 */
public boolean isNew() {
    return id == null;
}

@Override
public int hashCode() {
    int hash = 0;
    hash += (id != null ? id.hashCode() : 0);
    return hash;
}

@Override
public String toString() {
    return this.getClass().toString() + "[id=" + id + "]";
}

}

Employee.java：

@Entity
public class Employee  {

@Basic(optional = false)
@Column(nullable = false, length = 50)
private String firstName;

@Basic(optional = false)
@Column(nullable = false, length = 50)
private String lastName;

public Employee() {
}

public Employee(Integer id) {
    this.id = id;
}

public Employee(Integer id, String firstName, String lastName) {
    this.id = id;
    this.firstName = firstName;
    this.lastName = lastName;
}

public Employee(String firstName, String lastName) {
    this.firstName = firstName;
    this.lastName = lastName;
}

public String getFirstName() {
    return firstName;
}

public void setFirstName(String firstName) {
    this.firstName = firstName;
}

public String getLastName() {
    return lastName;
}

public void setLastName(String lastName) {
    this.lastName = lastName;
}

@Override
public boolean equals(Object object) {
    // TODO: Warning - this method won't work in the case the id fields are not set
    if (!(object instanceof Employee)) {
        return false;
    }
    Employee other = (Employee) object;
    if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
        return false;
    }
    return true;
}

}

EmployeeService.java方法：

public List<Employee> findByCriteria(String employeeName, String officeName,
        int pageNumber, int pageSize) {
    CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
    Root<Employee> emp = c.from(Employee.class);
    c.select(emp);
    c.orderBy(cb.asc(emp.get("firstName")));

    List<Predicate> criteria = new ArrayList<Predicate>();
    ParameterExpression<String> paramEmployeeName = cb.parameter(String.class);
    //Build the criteria with parameters
    if (!StringUtils.isEmpty(employeeName)) {

        criteria.add(cb.or(
                cb.like(cb.lower(emp.<String>get("firstName")), cb.lower(paramEmployeeName)),
                cb.like(cb.lower(emp.<String>get("lastName")), cb.lower(paramEmployeeName))));
    }
    //Add the criteria to the CriteriaQuery
    c.where(criteria.toArray(new Predicate[0]));

    TypedQuery<Employee> query = entityManager.createQuery(c);
    if (!StringUtils.isEmpty(employeeName)) {
        query.setParameter(paramEmployeeName, "%" + employeeName + "%");
    }

    page(query, pageNumber, pageSize);

    return query.getResultList();
}

Answer 1

这可能会使用位置参数。如果将:employeeName替换为？在这两种情况下然后使用：

 query.setParameter(0, "name");
 query.setParameter(1, "name");

Answer 2

您需要执行以下操作

SELECT e FROM Employee e WHERE lower(e.firstName) like :employeeName

以及您设置员工姓名的任何地方

执行以下操作

String employeename = employeename.toLowerCase();

org.hibernate.QueryException：并未设置所有命名参数

2 个答案: