從休眠數據庫獲取數據需要很長的時間

Question

我該如何提高性能？當從stackoverflow或gituhb rest api請求數據時，需要幾毫秒才能得到響應，但對於我的小應用程序，每個請求需要3-5秒。

例子：

@XmlRootElement 
@Entity 
@Table(name = "customers") 
public class Customer extends Model { 

@Id 
@GeneratedValue(strategy=GenerationType.AUTO) 
private int id; 

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

@Column(name = "email") 
private String email; 

@ManyToOne(fetch = FetchType.LAZY, cascade = {CascadeType.ALL}) 
@JoinColumn(name = "address_id") 
private Address address; 

@OneToMany(fetch = FetchType.LAZY, mappedBy = "customer") 
private List<Booking> bookings; 

/** 
* Constructor. 
* 
*/ 
public Customer() {} 

/** 
* Constructor. 
* 
* @param email Email of the customer. 
* @param address Address of the customer. 
* @param name Name of the customer. 
*/ 
public Customer(String email, Address address, String name) { 
    this.email = email; 
    this.address = address; 
    this.name = name; 
} 

/** 
* Set the id of the customer. 
* 
*/ 
public void setId(int id) { 
    this. id = id; 
} 

/** 
* Get the id of the customer. 
* 
* @return Id of the customer. 
*/ 
public int getId() { 
    return id; 
} 

/** 
* Get the name of the customer. 
* 
* @return Name of the customer. 
*/ 
public String getName() { 
    return name; 
} 

/** 
* Set the name of the customer. 
* 
* @param name New name of the customer. 
*/ 
public void setName(String name) { 
    this.name = name; 
} 

/** 
* Get the email of the customer 
* 
* @return Email of the customer. 
*/ 
public String getEmail() { 
    return email; 
} 

/** 
* Set the email of the customer 
* 
* @param email New email of the customer. 
*/ 
public void setEmail(String email) { 
    this.email = email; 
} 

/** 
* Get the address of the customer 
* 
* @return Address of the customer. 
*/ 
public Address getAddress() { 
    return address; 
} 

/** 
* Set the address of the customer 
* 
* @param address New address of the customer. 
*/ 
public void setAddress(Address address) { 
    this.address = address; 
} 

/** 
* Get the bookings of the customer. 
* 
* @return List of bookings. 
*/ 
@XmlTransient 
public List<Booking> getBookings() { 
    return bookings; 
} 

/** 
* Set the bookings of the customer. 
* 
* @param bookings List of bookings. 
*/ 
public void setBookings(List<Booking> bookings) { 
    this.bookings = bookings; 
} 

/** 
* Get all customers from the database. 
* 
* @return List of customers. 
*/ 
public static List<Customer> all() { 
    return (List<Customer>) Database.all(Customer.class); 
} 

/** 
* Find a customer by the id. 
* 
* @param id Id of the customer. 
* @return The customer object. Returns null if no customers were found. 
*/ 
public static Customer find(int id) { 
    return (Customer) Database.find(Customer.class, id); 
} 

/** 
* Check if a given customer id already exists. 
* 
* @param id Id of the customer. 
* @return Boolean. 
*/ 
public static boolean exists(int id) { 
    return Database.exists(Customer.class, id); 
} 

/** 
* Find customers by a specific column value. 
* 
* @param column Column of the database table. 
* @param value Value of the column. 
* @return List of customers. 
*/ 
public static List<Customer> where(String column, String value) { 
    return (List<Customer>) Database.where(Customer.class, column, value); 
} 

} 

數據庫類：

public class Database { 

/** 
* Get all objects of a model from the database. 
* 
* @param model Class type of the model. 
* @return List of all model objects. 
*/ 
public static List<? extends Model> all(Class<? extends Model> model) { 
    EntityManager entityManager = HibernateUtil.getEntityManagerFactory().createEntityManager(); 
    List<Model> list = (List<Model>) entityManager.createQuery("from " + model.getName(), model).getResultList(); 
    entityManager.close(); 
    return list; 
} 

/** 
* Find a model object by the id. 
* 
* @param model Class type of the model. 
* @param id Id of the model object. 
* @return The model object. Returns null if no objects were found. 
*/ 
public static Model find(Class<? extends Model> model, int id) { 
    EntityManager entityManager = HibernateUtil.getEntityManagerFactory().createEntityManager(); 
    Model m = entityManager.find(model, id); 
    entityManager.close(); 
    return m; 
} 

/** 
* Check if a given model object id already exists. 
* 
* @param model Class type of the model. 
* @param id Id of the model object. 
* @return Boolean. 
*/ 
public static boolean exists(Class<? extends Model> model, int id) { 
    return Database.find(model, id) != null; 
} 

/** 
* Find model objects by a specific column value. 
* 
* @param model Class type of the model. 
* @param column Column of the database table. 
* @param value Value of the column. 
* @return List of model objects. 
*/ 
public static List<? extends Model> where(Class<? extends Model> model, String column, String value) { 
    EntityManager entityManager = HibernateUtil.getEntityManagerFactory().createEntityManager(); 
    Query query = entityManager.createQuery("from " + model.getName() + " where " + column + "=:value"); 
    return (List<? extends Model>) query.setParameter("value", value).getResultList(); 
} 
} 

HibernateUtil類：

public class HibernateUtil { 

private static EntityManagerFactory entityManagerFactory; 

static { 

    entityManagerFactory = Persistence.createEntityManagerFactory("com.travelagency.jpa"); 

} 

/** 
* Get the EntityManager Factory object. 
* 
* @return EntityManagerFactory object. 
*/ 
public static EntityManagerFactory getEntityManagerFactory() { 
    return entityManagerFactory; 
} 
} 

謝謝

Answer 1

你的主要問題是，你使用Hibernate就好像它是ActiveRecord框架，這是錯誤的。但作爲一個快速的術語解決方案，您無需每次都重新創建EntityManager。這就是你需要這麼長時間。只創建一次，不要關閉它。

Answer 2

從你的代碼我可以看到你不應用分頁到你的查詢。這基本上意味着你將整個表加載到內存中。我不確定你的表有多大，但分頁可能是一個解決方案。

int offset = 0; 
int limit = 10; 

Query query = entityManager.createQuery("from " + model.getName() + " where " + column + "=:value"); 
query.setFirstResult(offset); 
query.setMaxResults(limit);