[英]Java 8 sorting Arraylist of object based on multiple conditions and date
我有一个对象关系列表,里面有一个名为 Contact 的对象,它将包含 elecType 对象或 postType
**Relationships:**
Relationship:
date :15/10/20
Contact :
code : 1
usageCode : 1
elecType(object) :(ecode : 1, detail : ssss )
Relationship:
date :14/10/20
Contact :
code : 1
usageCode : 2
elecType(object) :(ecode : 2, detail :yyy )
Relationship:
date :10/10/20
Contact :
code : 1
usageCode : 2
elecType(object) :(ecode : 2, detail :eee )
Relationship:
date :13/10/20
Contact :
code : 1
usageCode : 2
elecType(object) :(ecode : 1, detail :zzz )
Relationship:
date :15/10/20
Contact :
code : 1
usageCode : 1
elecType(object) :(ecode : 2, detail :ttt )
Relationship:
date:12/10/20
Contact :
code : 1
postType(object) : ( detail :xxx )
Relationship:
date:11/10/20
Contact :
code : 2
postType(object) : (detail :yyy )
Relationship:
date:13/10/20
Contact :
code : 2
postType(object) : (detail :zzz )
如果代码为 2,我需要根据以下条件对关系、联系人对象进行排序,我需要从每个具有不同代码的联系人中获取最新日期的关系对象,即:将是上述示例的最终输出
Relationship:
date:12/10/20
Contact :
code : 1
postType(object) : (detail :xxx )
Relationship:
date:13/10/20
Contact :
code : 2
postType(object) : (detail :zzz )
同样,如果代码为 1 ,我需要从每个具有不同使用代码、编码的联系人记录中获取最新的关系
即:从上面的数据,输出将是
Relationship:
date :15/10/20
Contact :
code : 1
usageCode : 1
elecType(object) :(ecode : 1, detail : ssss )
Relationship:
date :15/10/20
Contact :
code : 1
usageCode : 1
elecType(object) :(ecode : 2, detail :ttt )
Relationship:
date :13/10/20
Contact :
code : 1
usageCode : 2
elecType(object) :(ecode : 1, detail :zzz )
Relationship:
date :14/10/20
Contact :
code : 1
usageCode : 2
elecType(object) :(ecode : 2, detail :yyy )
public class Relationship {
private Date date;
private Contact contact;
}
public class Contact{
private Integer code;
private Integer usageCode;
private PostalType postalType;
private ElecType elecType;
}
public class PostalType{
private String detail;
}
public class ElecType{
private String detail;
private Integer eCode
}
在 Java 8 或更高版本中实现这一点的最佳方法是什么(是否可以使用 lambda 和流实现)
您可以在您的类关系中实现 Comparable 并对列表进行排序
Collections.sort(testList);
如果要过滤列表,可以使用带过滤器的流。
// filters a List with relationships with code 1
relationships.stream().filter(r -> r.getContacts().getCode() == 1).collect(Collectors.asList());
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