Question

I am getting error "Cannot create TypedQuery for query with more than one return using requested result type" I tried with all columns value returning. That time the application hangs. I need to get list of Client, in arraylist. Please help, I am new to JPA.

@Override
    public ArrayList<Client> findAllClients() {
        EntityManager entity = this.emf.createEntityManager();
        List<Client> clients = entity.createQuery("select clientID,clientName from Client", Client.class).getResultList();
        return (ArrayList<Client>) clients;
    }

Client class is

package com.springmaven.models;

import java.util.Date;
import java.util.HashSet;
import java.util.Set;


import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;

@Entity
@Table(name="tblclient")
public class Client {
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY) @Column(name="ntClientID")
    private Long clientId;

    @Column(name="vcClientName")
    private String clientName;

    @Column(name="vcLocation")
    private String location;

    @Column(name="ofstTimeZone")
    private Date timeZone;

    @Column(name="vcCommunicationMode")
    private String communicationMode;

    @Column(name="vcContact")
    private String contact;

    @OneToMany(targetEntity=Project.class,mappedBy="client",
            cascade=CascadeType.ALL,fetch=FetchType.EAGER)
    private Set<Project> projects = new HashSet<Project>();

    public Set<Project> getProjects() {
        return projects;
    }

    public void setProjects(Set<Project> projects) {
        this.projects = projects;
    }

    public Long getClientId() {
        return clientId;
    }

    public void setClientId(Long clientId) {
        this.clientId = clientId;
    }

    public String getClientName() {
        return clientName;
    }

    public void setClientName(String clientName) {
        this.clientName = clientName;
    }

    public String getLocation() {
        return location;
    }

    public void setLocation(String location) {
        this.location = location;
    }

    public Date getTimeZone() {
        return timeZone;
    }

    public void setTimeZone(Date timeZone) {
        this.timeZone = timeZone;
    }

    public String getCommunicationMode() {
        return communicationMode;
    }

    public void setCommunicationMode(String communicationMode) {
        this.communicationMode = communicationMode;
    }

    public String getContact() {
        return contact;
    }

    public void setContact(String contact) {
        this.contact = contact;
    }

    public Client(){

    }
}

Answer 1

Usually on Hibernate you simply make selects of an specific entity, not necessarily defining what columns you want. Something like this:

List<Client> clients = entity.createQuery("select c from Client c", Client.class).getResultList();

You are getting the TypedQuery error because the EntityManager is waiting for a collection of Clients, but instead you are selecting two specific columns of it, which will make Hibernate unable to cast the results as a Client entity.

So in your case, use the query given above and everything should work fine.

Answer 2

您可以在（List＆lt; clients＆gt;）

中投射到您的结果中

List<Client> clients = (List<Client>) entity.createQuery("select clientID,clientName from Client", Client.class).getResultList();

Answer 3

这是对“客户端”的投影查询，仅返回clientID和clientName，而不是将整个对象加载到内存中。这种方法可以减少数据库服务器的网络流量并节省内存。因此，您可以使用下一个：

scala> var table1 = Seq((11, 25, 2, 0), (42, 20, 10, 0)).toDF("col_1", "col_2", "col_3", "col_4")
table1: org.apache.spark.sql.DataFrame = [col_1: int, col_2: int ... 2 more fields]

scala> table1.show()
+-----+-----+-----+-----+
|col_1|col_2|col_3|col_4|
+-----+-----+-----+-----+
|   11|   25|    2|    0|
|   42|   20|   10|    0|
+-----+-----+-----+-----+

scala> table1.createOrReplaceTempView("table1")


scala> val result = spark.sql(s""" select col_1,
     |                                    col_2,
     |                                    col_3,
     |                                    CASE WHEN col_1 > (col_2 + col_3)
     |                                           THEN 5
     |                                         ELSE   1
     |                                    END as col_4
     |                              from  table1 """)
result: org.apache.spark.sql.DataFrame = [col_1: int, col_2: int ... 2 more fields]


scala> result.show(false)
+-----+-----+-----+-----+
|col_1|col_2|col_3|col_4|
+-----+-----+-----+-----+
|11   |25   |2    |1    |
|42   |20   |10   |5    |
+-----+-----+-----+-----+

此结果集包含一个对象数组列表，每个数组代表一组属性，在这种情况下为clientID和clientName。现在您可以检索到此：

List<Object[]> results = 
entity.createQuery("select clientID, clientName from Client").getResultList();

Cannot create TypedQuery for query with more than one return using requested result type

3 个答案: