如何加入地圖

-2

如何加入newMap detals custMap。如何加入地圖

Map<String, Customer> custMap= new HashMap<String,Customer>(); 
Map<String, DoCustomer> newMap= new HashMap<String,DoCustomer>(); 
    for (Map.Entry<String, DoCustomer> cust: newMap.entrySet()) { 
    custMap.put(cust.getKey(),cust.getValue()); 
} 

public class DoCustomer { 
private Long id; 

private String custName; 

private String description; 
    private String status; 
    private List<DoCustomerBranch> doCustomerBranch=new ArrayList<DoCustomerBranch> 
public Long getId() { 
    return id; 
} 
public void setId(Long id) { 
    this.id = id; 
} 
public String getCustName() { 
    return custName; 
} 
public void setCustName(String custName) { 
    this.custName = custName; 
} 
public String getDescription() { 
    return description; 
} 
public void setDescription(String description) { 
    this.description = description; 
} 
public String getStatus() { 
    return status; 
} 
public void setStatus(String status) { 
    this.status = status; 
} 
getter/setters of doCustomerBranch 
} 

    @Entity 
    @Table(name = "CUSTOMER") 
    public class Customer implements Serializable{ 

private static final long serialVersionUID = 1L; 
private Long id; 

private String custName; 

private String description; 

private String createdBy; 
private Date createdOn; 

private String updatedBy; 
private Date updatedOn; 


private Set<CustomerBranch> customerBranch=new HashSet<CustomerBranch> 



@Id 
@GeneratedValue(generator = "CUSTOMER_SEQ") 
@SequenceGenerator(name = "CUSTOMER_SEQ", sequenceName = "CUSTOMERN_SEQ", allocationSize = 1) 
@Column(name = "ID") 
public Long getId() { 
    return id; 
} 

public void setId(Long id) { 
    this.id = id; 
} 


@Column(name = "CUST_NAME",nullable=false) 
public String getCustName() { 
    return custName; 
} 

public void setCustName(String custName) { 
    this.custName = custName; 
} 

@Column(name = "DESCRIPTION") 
public String getDescription() { 
    return description; 
} 

public void setDescription(String description) { 
    this.description = description; 
} 


@Column(name = "CREATED_BY", length = 50) 
public String getCreatedBy() { 
    return createdBy; 
} 

public void setCreatedBy(String createdBy) { 
    this.createdBy = createdBy; 
} 

@Temporal(TemporalType.TIMESTAMP) 
@Column(name = "CREATED_ON") 
public Date getCreatedOn() { 
    return createdOn; 
} 

public void setCreatedOn(Date createdOn) { 
    this.createdOn = createdOn; 
} 

@Column(name = "UPDATED_BY", length = 50) 
public String getUpdatedBy() { 
    return updatedBy; 
} 

public void setUpdatedBy(String updatedBy) { 
    this.updatedBy = updatedBy; 
} 

@Temporal(TemporalType.TIMESTAMP) 
@Column(name = "UPDATED_ON") 
public Date getUpdatedOn() { 
    return updatedOn; 
} 

public void setUpdatedOn(Date updatedOn) { 
    this.updatedOn = updatedOn; 
} 
    @OneToMany(cascade = { CascadeType.PERSIST, CascadeType.REMOVE }, fetch =   FetchType.LAZY, mappedBy = "customer") 
public Set<CustomerBranch> getCustomerBranch() { 
    return customerBranch; 
} 


public void setCustomerBranch(Set<CustomerBranch> customerBranch) { 
    this.customerBranch = customerBranch; 
} 

}

CustomerBranch

@Entity 
@Table(name = "CUSTOMER_BRANCH") 
public class CustomerBranch implements Serializable{ 

    @Id 
@GeneratedValue(generator = "CUSTOMER_BRANCH_SEQ") 
@SequenceGenerator(name = "CUSTOMER_BRANCH_SEQ", sequenceName = "CUSTOMER_BRANCH_SEQ", allocationSize = 1) 
@Column(name = "ID") 
private Long id; 
private String branchName; 

private String branchAddress; 

private Customer customer; 



public Long getId() { 
    return id; 
} 

public void setId(Long id) { 
    this.id = id; 
} 

@Column(name = "BRANCH_NAME",nullable=false) 
public String getBranchName() { 
    return branchName; 
} 



@ManyToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "MOBEE_CUSTOMER") 
public Customer getCustomer() { 
    return customer; 
} 

public void setCustomer(Customer customer) { 
    this.customer = customer; 
} 

    }

來源

2012-06-12 nag

你所說的'join'意思？你的示例代碼有什麼問題？它是做你想做的嗎？ –

我想遷移newMap的值到custMap.i我得到的錯誤是方法put（字符串，客戶）在類型地圖<字符串，客戶>不適用爲參數（字符串，DoCustomer） – nag

您可以張貼Customer和DoCustomer類的定義？ – tibtof

你的意思是：

custMap.putAll(newMap)

來源

2012-06-12 07:59:06 maks

-1這不起作用。「DoCustomer」不是「客戶」。 – Kai

雅我想要這個溶膠，但我得到錯誤是方法把（字符串，客戶）在類型映射<字符串，客戶>不適用爲參數（字符串，DoCustomer） – nag

從類'DoCustomer'我的名字假設它是'Customer'類的一個子類，但是我犯了一個錯誤。所以我提出的方法並不適合你的情況 – maks

與您的代碼的問題是，你想提出一個DoCustomer在Customer容器。它只適用於DoCustomer是Customer的子類。

編輯1：您可以使用BeanUtils將DoCustomer轉換爲Customer。 Here是一個很好的教程。

來源

2012-06-12 07:59:49 tibtof

謝謝DoCustomer不是客戶的子類，在這個例子中，客戶是實體，DoCustomer是客戶的特性，你能告訴我另一種方式解決這個問題嗎？可能使用BeanUtils ..... – nag

正如其他人所指出的，我們需要知道DoCustomer能夠提供哪些幫助。

但是，根據您給我們的建議，我建議將每個DoCustomer投射到客戶，或者更準確地說，從每個DoCustomer的字段中創建一個新客戶。

喜歡的東西：

custMap.put(cust.getKey(), new Customer(cust.getValue().getId(), cust.getValue().getCustName(), and so on..));

您的循環內。

我可以看到你所提供沒有一個構造函數中定義的客戶類，自然你就必須添加一個到它

來源

2012-06-12 10:04:39

謝謝邁克爾你寫，但在我的客戶實體中映射CustomerBranch在那段時間我是我pblm與構造器 – nag

你是指Set customerBranch？這與DoCustomer中的List doCustomerBranch相同嗎？我可以理解，在兩者之間轉換會使這個解決方案非常冗長。 –

好的，謝謝你，我會去替代 – nag

如何加入地圖

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