如何在java中比较行并找到常用值和不常用值（使用逗号）

这是我的桌子

表

Id      Town     City     Country  
1        a         b         c  
1        a         b         d  
1        a         b         e  
2        f         g         h  
2        f         g         i  
2        f         g         j  
3        k         l         m  

在我的jsp页面中，这就是我需要如何显示表中的数据

          Title

Id      Town     City     Country  
1        a         b       c,d,e  
2        f         g       h,i,j  
3        k         l         m  

我运行查询，从表中选择id、城镇、城市和国家。然后

public List<Place> getAddressChangeView () 
{
    List <Place> countriesList = new ArrayList < Place> ();

    List <Object[]> z = query.getResultList();

    List<String> countries = new ArrayList String;
    String delim ="," ;

    Place places = null;
    for (Object[] j: z)
    {

        places = new Place();
        int id = ((String) obj[0]);
        String town=((String) obj1);
        String city=((String) obj[2]);
        String country=((String) obj[3]);

        if (! id.equals(id1) && ! town.equals(town1) && ! city.equals(city1) && ! country.equals(country1)) 
        {
            id1= id;
            town1= town;
            city1=city;
            country1=country;

            places.setId(id);
            places.setTown(town);
            places.setCity(city);

        }

        countries.add(country);

        StringBuilder sb = new StringBuilder();

        for (String s: countries)
        {
            sb.append(s).append(delim);

        }
        places.setCountry(sb.toString());

        countriesList.add(places);

    }

    return countriesList;

}

---

实现分组操作的常用方法是使用映射。换句话说，您可以创建从每个组到该组中对象列表的映射。然后，您可以遍历映射来处理每个组。在您的情况下，看起来您只是按照ID进行分组，尽管这可能只是您提供的示例数据

使用java 8流，它看起来像：

List<Place> places = table.stream()
    .map(Place::new).collect(Collectors.toList());
Map<String, List<Place>> grouping = places.stream()
    .collect(Collectors.groupingBy(Place::getID));

然后分组语句变为：

Map<Group, List<Place>> grouping = places.stream()
    .collect(Collectors.groupingBy(Group::forPlace));

---

由于地图的键是now Group，您可以在遍历条目集时获得该组的城镇和城市。

您可以在SQL中使用CTEs one recursive和rank函数在查询中这样做：

with Countries(Id, Town, City, Country, Seq) as (
  select Id, Town, City, Country,
    row_number() over(partition by Id order by Country)
  from TableA
), Maxes(Id, Seq) as (
  select Id, max(Seq) from Countries group by Id
), Recursed(Id, Town, City, Country, Seq) as (
  select Maxes.Id, Town, City, Country, Maxes.Seq
  from Countries, Maxes
  where Countries.Id = Maxes.Id and Countries.Seq = Maxes.Seq
  union all
  select c.Id, c.Town, c.City, c.Country || ',' || r.Country, c.Seq
  from Countries c, Recursed r
  where c.Id = r.Id and c.Seq = r.Seq - 1
)
select Id, Town, City, Country
from Recursed
where Seq = 1
order by Id

假设字符串串联运算符| |，这将产生：

Id   Town  City  Country  
1     a      b    c,d,e  
2     f      g    h,i,j  
3     k      l      m  

with Countries(Id, Town, City, Country, Seq) as (
  select Id, Town, City, Country,
    row_number() over(partition by Id order by Country)
  from TableA
), Maxes(Id, Seq) as (
  select Id, max(Seq) from Countries group by Id
), Recursed(Id, Town, City, Country, Seq) as (
  select Maxes.Id, Town, City, Country, Maxes.Seq
  from Countries, Maxes
  where Countries.Id = Maxes.Id and Countries.Seq = Maxes.Seq
  union all
  select c.Id, c.Town, c.City, c.Country || ',' || r.Country, c.Seq
  from Countries c, Recursed r
  where c.Id = r.Id and c.Seq = r.Seq - 1
)
select Id, Town, City, Country
from Recursed
where Seq = 1
order by Id

Id   Town  City  Country  
1     a      b    c,d,e  
2     f      g    h,i,j  
3     k      l      m