如何基于重复的父对象合并列表中的子对象 [英] How to merge child objects in list based on duplicate parent object

问题描述

我有两个具有一对多关系的实体，这些实体通过复合主键连接在一起.由于Spring Data为oracle数据库生成了错误的计数独特查询，因此我具有带笛卡尔连接的SQL输出，这导致子对象的每一行都为父对象重复一行.我需要基于组合键找出不同的父对象，然后在列表中添加父对象的每个子对象并将其设置为父对象的属性

I have two entities with one to many relationship which are joined together by composite primary keys. Since Spring Data generates wrong count distinct query for oracle database , I have SQL output with cartesian join which leads to repeating row for parent object for every row of child object. I need to find out distinct parent objects based on composite keys and then add each child object for parent in a list and set it as property of parent object

我能够根据父对象的组合键找出不同的父对象.以下是相关代码

I am able to find out distinct parent object based on composite keys of the parent object. Following is the relevant code

private static <T> Predicate<T> distinctByKeys(Function<? super T, ?>... keyExtractors)
{
final Map<List<?>, Boolean> seen = new ConcurrentHashMap<>();
return t ->
{
final List<?> keys = Arrays.stream(keyExtractors)
.map(ke -> ke.apply(t))
collect(Collectors.toList());

return seen.putIfAbsent(keys, Boolean.TRUE) == null;
};
}

假设我有以下输入内容

list.add(new Book("Core Java", 200, new ArrayList(){{add("Page 1");}}));
list.add(new Book("Core Java", 200, new ArrayList(){{add("Page 2");}}));
list.add(new Book("Learning Freemarker", 150, new ArrayList(){{add("Page 15");}}));
list.add(new Book("Spring MVC", 300, new ArrayList(){{add("Page 16");}}));
list.add(new Book("Spring MVC", 300, new ArrayList(){{add("Page 17");}}));

我需要产生以下输出

Core Java,200, [Page 1, Page 2]
Learning Freemarker,150, [Page 15]
Spring MVC,300 , [Page 16, Page 17]

在这方面的任何帮助都将非常有帮助

Any help in this regard will be very helpful

推荐答案

一种选择是使用Collectors.toMap()，并使用title和pages值作为键.如果发现重复项，则可以合并两个列表:

One option is to use Collectors.toMap() using the title and pages value as key. If you find duplicates you can merge both lists:

Collection<Book> result = list.stream()
        .collect(Collectors.toMap(
                b -> Map.entry(b.getTitle(), b.getPages()),
                b -> new Book(b.getTitle(), b.getPages(), b.getList()),
                (b1, b2) -> new Book(b1.getTitle(), b1.getPages(),
                        Stream.concat(b1.getList().stream(), b2.getList().stream()).collect(Collectors.toList())),
                LinkedHashMap::new))
        .values();

或者，您可以使用Collectors.groupingBy():

List<Book> result = list.stream().collect(
        Collectors.groupingBy(b -> Map.entry(b.getTitle(), b.getPages()), LinkedHashMap::new,
                Collectors.flatMapping(b -> b.getList().stream(), Collectors.toList())))
        .entrySet().stream()
        .map(e -> new Book(e.getKey().getKey(), e.getKey().getValue(), e.getValue()))
        .collect(Collectors.toList());

第二种方法使用title和pages作为键来创建Map，并使用Collectors.flatMapping()合并所有书籍的列表.之后，将条目映射回Book对象.

The second approach creates a Map using title and pages as key and merging the lists of all books using Collectors.flatMapping(). After that you map the entries back to your Book object.

如果您不能使用Collectors.flatMapping()，请改为使用Collectors.mapping()`:

If you can not use Collectors.flatMapping() use Collectors.mapping()` instead:

List<Book> r = list.stream().collect(
        Collectors.groupingBy(b -> Map.entry(b.getTitle(), b.getPages()), LinkedHashMap::new,
                Collectors.mapping(Book::getList, Collectors.toList())))
        .entrySet().stream()
        .map(e -> new Book(e.getKey().getKey(), e.getKey().getValue(), 
                e.getValue().stream().flatMap(Collection::stream).collect(Collectors.toList())))
        .collect(Collectors.toList());

如果不能使用Map.entry()，请改为使用new AbstractMap.SimpleEntry<>().

If you can not use Map.entry() use new AbstractMap.SimpleEntry<>() instead.

两种情况下的结果都是这样:

The result in both cases will be this:

Book[title='Core Java', pages=200, list=[Page 1, Page 2]]
Book[title='Learning Freemarker', pages=150, list=[Page 15]]
Book[title='Spring MVC', pages=300, list=[Page 16, Page 17]]

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