Question

如何在下面的代码中获取唯一的对象数组列表：

    import java.util.*;

    class t1 {
        public static void main(String[] args) {

            Object[] o1 = new Object[] { null, "abc@aaa.com", "ENG", "775778435", 1};
            Object[] o2 = new Object[] { null, "abc@aaa.com", "ENG", "775778435", 1};

            List<Object[]> result = new ArrayList<Object[]>(); 
            result.add(o1); 
            result.add(o2); 

            // The above result list is coming from some DB & I cannot change the structure of above script. 
            // Now, I need to remove the duplicates from this result list meaning I need to find duplicates from the objects within the list.
            // I tried below code but it still prints duplicates. Any help???

            Set<User> setResult = new HashSet<User>();
            User userInfo = null;

            for (Object[] userData : result) {
                userInfo = new User((String)userData[0], (String)userData[1], (String)userData[2], (String)userData[3], (Integer) userData[4]);
                setResult.add(userInfo);
            }

            Iterator it = setResult.iterator();
            while (it.hasNext()) {
                Object o = it.next();

                User u = (User) o;
                System.out.println("non-duplicate = " + u.getEmail());
            }

            // Expected result: non-duplicate = abc@aaa.com
            // Actual   result: non-duplicate = abc@aaa.com getting printed twice i.e. duplicate not getting removed!
        }
    }

    class User {
        public String firstName;
        public String email;
        public String language;
        public String productCode;
        public int status;

        public User() {         
        }

        public User(String fName, String userId, String lang, String productCode, int status) {
            this.firstName = fName;
            this.email = userId;
            this.language = lang;
            this.productCode = productCode;
            this.status = status;
        }   

        public String getFirstName() {
            return firstName;
        }

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

        public String getEmail() {
            return email;
        }

        public void setEmail(String email) {
            this.email = email;
        }

        public String getLanguage() {
            return language;
        }

        public void setLanguage(String language) {
            this.language = language;
        }

        public String getProductCode() {
            return productCode;
        }

        public void setProductCode(String productCode) {
            this.productCode = productCode;
        }

        public int getStatus() {
            return status;
        }

        public void setStatus(int status) {
            this.status = status;
        }

        @Override
        public int hashCode() {
            int fNameHash = 0;
            int lNameHash = 0;
            int emailHash = 0;
            int langHash = 0;
            int productCodeHash = 0;

            if (this.firstName != null) {
                fNameHash = this.firstName.hashCode();
            }

            if (this.email != null) {
                emailHash = this.email.hashCode();
            }

            if (this.language != null) {
                langHash = this.language.hashCode();
            }

            if (this.productCode != null) {
                productCodeHash = this.productCode.hashCode();
            }

            return (fNameHash + lNameHash + emailHash + langHash + productCodeHash + this.status);
        }

        @Override
        public boolean equals(Object obj) {
            if(obj != null && obj instanceof User) {
                User temp = (User) obj;

                if (this.firstName != null && temp.firstName != null && this.firstName.equalsIgnoreCase(temp.firstName) 
                   && this.email != null && temp.email != null && this.email.equalsIgnoreCase(temp.email) 
                   && this.language != null && temp.language != null && this.language.equalsIgnoreCase(temp.language)
                   && this.productCode != null && temp.productCode != null && this.productCode.equalsIgnoreCase(temp.productCode)
                   && this.status == temp.status) {             
                    return true;
                }
            }
            return false;
        }   
    }

我的预期结果是打印abc@aaa.com一次，但打印两次!!

任何人都可以告诉我如何更正此代码吗？

谢谢！

Answer 1

问题是您的User.equals会预测会返回false，因为这两个实例都有null firstName s：

if (this.firstName != null && temp.firstName != null && this.firstName.equalsIgnoreCase(temp.firstName)

因此，即使它们具有相同的hashCode，它也仅仅被视为碰撞而不是匹配，因为equals已被破坏。

尝试纠正:-)用......之类的东西替换逻辑。

firstName == temp.firstName || firstName != null && firstName.equalsIgnoreCase(temp.firstName)

......应该有效; - ）

您可以看到完整的版本here。

顺便说一句，您忘了在lNameHash中分配User.hashCode。

Answer 2

您的equals方法和hashcode方法都未正确实施。

当两个对象都有equals个成员时，您的false方法会返回null。
hashCode的实现可以为某些比较相等的对象返回不同的哈希码。

您必须提供equals和hashCode的正确实施才能在HashSet中使用您的对象。

Answer 3

问题在于firstName值对于两个对象都为null。因此，equals（）方法返回false，Set将您的对象视为不同。

尝试使用以下内容更改您的equals和hashcode：

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (!(o instanceof User)) return false;

    User user = (User) o;

    if (status != user.status) return false;
    if (email != null ? !email.equals(user.email) : user.email != null) return false;
    if (firstName != null ? !firstName.equals(user.firstName) : user.firstName != null) return false;
    if (language != null ? !language.equals(user.language) : user.language != null) return false;
    if (productCode != null ? !productCode.equals(user.productCode) : user.productCode != null) return false;

    return true;
}

@Override
public int hashCode() {
    int result = firstName != null ? firstName.hashCode() : 0;
    result = 31 * result + (email != null ? email.hashCode() : 0);
    result = 31 * result + (language != null ? language.hashCode() : 0);
    result = 31 * result + (productCode != null ? productCode.hashCode() : 0);
    result = 31 * result + status;
    return result;
}

最好的方法是按照另一个答案中的建议，从IDE生成等号和哈希码。

获取对象数组的唯一列表

3 个答案: