英文:
Get page number of specific item in spring data rest
问题
我正在使用spring-data-rest构建一个网络应用程序。
我想在我的前端以分页表格的形式显示来自大型数据库表的数据。该表格会异步加载来自API的当前页面数据。通过只需拥有一个类似于以下的存储库，一切都可以正常运行：
public interface KeywordRepository extends JpaRepository<Keyword, String>, QuerydslPredicateExecutor<Keyword> {
}
现在，我想实现一种功能，使我能够在分页表格中跳转到包含特定项目的正确页面。问题是，我不知道特定项目位于哪个页面。
我需要一种类似于findPageOfItemById(Long id, Pageable pageable)的端点，以根据当前的筛选和排序参数告诉我特定项目（按ID）所在的页面号。
我该如何实现这个？
由于表格相当大，我不想将整个内容保存在内存中。
英文:
I'm building a web application with spring-data-rest.
I want to show data from a big db-table in a paginated table in my frontend. The table loads asynchronically just the current page from the API. That works all fine and dandy out of the box by just having a repository like
public interface KeywordRepository extends JpaRepository<Keyword, String>, QuerydslPredicateExecutor<Keyword> {
}
Now I want to implement a functionality to jump in my paginated table to the correct page where a certain item is.
Problem is that I don't know on what page that specific item is.
I need some kind of endpoint to tell me the page number of a specific item (by id) according to the current filter- and sorting-parameters. Basically findPageOfItemById(Long id, Pageable pageable).
How can I get this?
Since the table is quite big, I don't want to have the whole content in memory.
答案1
得分: 0
为了保证完整性，我会回答自己的问题。
这个方法运行得相当不错，而且不需要将整个列表加载到内存中，就像所需的那样。
但是我仍然希望能找到更多关于spring-data-rest的答案。
@GetMapping("/getPositionOfItem/{id}")
public long getPositionOfItem(@PathVariable String id, @QuerydslPredicate(root = SomeEntity.class) Predicate predicate, Pageable pageable) {
Iterable<SomeEntity> elements = someEntityRepository.findAll(predicate, pageable.getSort());
return findFirst(elements.iterator(), id);
}
private long findFirst(Iterator<SomeEntity> iterator, String id) {
long index = 0;
while (iterator.hasNext()) {
if (iterator.next().getId().equals(id)) {
return index;
}
index++;
}
return -1;
}
注意，我计算了项目的位置。要获取元素所在的页数，我们需要除以页面大小。
英文:
For the sake of completeness, I'll answer my own question.
It works quite nice and without loading the whole list into memory as reqested.
But I was still hoping to find a little more spring-data-resty answer.
@GetMapping("/getPositionOfItem/{id}")
public long getPositionOfItem(@PathVariable String id, @QuerydslPredicate(root = SomeEntity.class) Predicate predicate, Pageable pageable) {
Iterable<SomeEntity> elements = someEntityRepository.findAll(predicate, pageable.getSort());
return findFirst(elements.iterator(), id);
}
private long findFirst(Iterator<SomeEntity> iterator, String id) {
long index = 0;
while (iterator.hasNext()) {
if (iterator.next().getId().equals(id)) {
return index;
}
index++;
}
return -1;
}
Note that I calculate the position of the item. To get the page of the element, we need to divide by the pagesize.
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