如何将List<Schedule>合并为MyPojo列表？JPA数据合并处理

现有数据库表数据如下：

Id name descr type freq
 1  A   desc   SMS  daily
 1  A   desc   Push weekly
 1  A   desc   InAp custom
 2  B   desc1  SMS  weekly
 2  B   desc1  Push daily

通过JPA将该表映射为包含id、name、desc、type、freq字段的List<Schedule>集合。需要对该集合做以下处理：

• 当name与desc相同时，合并为一个MyPojo对象：公共字段直接赋值，不同的type和freq分别收集为列表；
• 若name或desc不同，则单独作为列表项。

MyPojo类定义如下：

class MyPojo{
  private int id;
  private String name;
  private String desc;
  private List<String> type;
  private List<String> frequency;
  //setters and getters
}

预期输出：

O/P= [{"id":"1", "name":"A", "desc":"desc", "type":["SMS","Push","InApp"], "freq":["daily","weekly","custom"]}, {"id":"2", "name":"B", "desc":"desc1", "type":["SMS","Push"], "freq":["daily","weekly"]}]

可以通过Java Stream的分组+映射实现需求，具体代码如下：

完整实现代码 #

import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import java.util.AbstractMap.SimpleEntry;

// 假设Schedule实体类结构（需与JPA映射一致）
class Schedule {
    private int id;
    private String name;
    private String desc;
    private String type;
    private String freq;

    // 构造器、getter方法
    public Schedule(int id, String name, String desc, String type, String freq) {
        this.id = id;
        this.name = name;
        this.desc = desc;
        this.type = type;
        this.freq = freq;
    }

    public int getId() { return id; }
    public String getName() { return name; }
    public String getDesc() { return desc; }
    public String getType() { return type; }
    public String getFreq() { return freq; }
}

class MyPojo{
    private int id;
    private String name;
    private String desc;
    private List<String> type;
    private List<String> frequency;

    // 构造器、getter/setter方法
    public MyPojo(int id, String name, String desc, List<String> type, List<String> frequency) {
        this.id = id;
        this.name = name;
        this.desc = desc;
        this.type = type;
        this.frequency = frequency;
    }

    public int getId() { return id; }
    public String getName() { return name; }
    public String getDesc() { return desc; }
    public List<String> getType() { return type; }
    public List<String> getFrequency() { return frequency; }
}

public class ScheduleMerger {
    public static void main(String[] args) {
        // 模拟JPA查询得到的Schedule集合
        List<Schedule> schedules = List.of(
                new Schedule(1, "A", "desc", "SMS", "daily"),
                new Schedule(1, "A", "desc", "Push", "weekly"),
                new Schedule(1, "A", "desc", "InAp", "custom"),
                new Schedule(2, "B", "desc1", "SMS", "weekly"),
                new Schedule(2, "B", "desc1", "Push", "daily")
        );

        // 核心处理逻辑
        List<MyPojo> mergedResult = schedules.stream()
                // 按name+desc的组合分组，相同组合的Schedule归为一组
                .collect(Collectors.groupingBy(
                        sch -> new SimpleEntry<>(sch.getName(), sch.getDesc()),
                        Collectors.toList()
                ))
                .values()
                .stream()
                // 将每组转换为MyPojo对象
                .map(group -> {
                    // 取组内第一个元素的公共字段值（name/desc/id相同）
                    Schedule firstItem = group.get(0);
                    // 收集组内所有type和freq
                    List<String> types = group.stream().map(Schedule::getType).collect(Collectors.toList());
                    List<String> freqs = group.stream().map(Schedule::getFreq).collect(Collectors.toList());
                    return new MyPojo(firstItem.getId(), firstItem.getName(), firstItem.getDesc(), types, freqs);
                })
                .collect(Collectors.toList());

        // 打印验证结果
        mergedResult.forEach(pojo -> {
            System.out.println(String.format(
                    "{\"id\":\"%d\", \"name\":\"%s\", \"desc\":\"%s\", \"type\":%s, \"freq\":%s}",
                    pojo.getId(), pojo.getName(), pojo.getDesc(), pojo.getType(), pojo.getFrequency()
            ).replace("[", "[\"").replace("]", "\"]").replace(", ", "\", \""));
        });
    }
}

代码说明 #

1. 分组逻辑：用SimpleEntry<String, String>作为分组键，将name和desc的组合作为唯一标识，把同组合的Schedule分到一组；
2. 对象转换：对每个分组，取第一个元素的id、name、desc作为公共值，再收集组内所有type和freq组成列表，最终实例化MyPojo；
3. 自定义分组键：如果不想用SimpleEntry，也可以自定义一个包含name和desc的类，重写equals和hashCode方法作为分组键，效果一致。

内容的提问来源于stack exchange，提问作者vyas