Creating Pagination in Spring Data JPA

Question

I am trying to implement pagination feature in Spring Data JPA. I am referring this Blog My Controller contains following code :

 @RequestMapping(value="/organizationData", method = RequestMethod.GET)
  public String list(Pageable pageable, Model model){
    Page<Organization> members = this.OrganizationRepository.findAll(pageable);
    model.addAttribute("members", members.getContent());
    float nrOfPages = members.getTotalPages();
    model.addAttribute("maxPages", nrOfPages);
    return "members/list"; 
  }

My DAO is following :

@Query(value="select m from Member m", countQuery="select count(m) from Member m")
  Page<Organization> findMembers(Pageable pageable);

I am able to show first 20 records, how do I show next 20??? Is there any other pagination example that I can refer??

The blog you refer to also has a section which allows you to specify the page number. I would imagine that would be what you would want to use. — Makoto, Nov 21 '16 at 07:18
Yes, I saw that but, it is Spring MVC, I want to implement same in Spring Data JPA. How do i modify it to Spring data JPA?/ — Tejas, Nov 21 '16 at 07:21
...If you've been following the entire tutorial, which was geared towards Spring MVC using Spring Data, then why not at least give it a try? Or did you give it a try already? If you did, then could show the code you attempted, what you expected to happen, and what did happen? — Makoto, Nov 21 '16 at 07:23
Just pas a request parameter with the name `page`. that will be bound to the `Pageable` object in your request method. — M. Deinum, Nov 21 '16 at 07:24
Actually the project that I am working on is in Spring Data JPA, so i need to implement same in Spring Data JPA. — Tejas, Nov 21 '16 at 07:26
@M.Deinum I am also thinking the same, but I dont know where exactly should I pass the parameter. — Tejas, Nov 21 '16 at 07:29
The URL you are calling. As explained [here](http://docs.spring.io/spring-data/jpa/docs/current/reference/html/#core.web.basic.paging-and-sorting) in the reference guide. However if you follow the tutorial that will also (in the end) do exactly the same... — M. Deinum, Nov 21 '16 at 07:29
As mentioned a request parameter named page with the page number you want to see. — M. Deinum, Nov 21 '16 at 07:42
@Tejas Either you can specify PageRequest as specified in the answer it will work but it is redundant since by passing "url?page=1&size=20" Pageable will automatically wire the page & size property and will fetch 2nd page (Since page number starts with 0) with 20Items. — VelNaga, Nov 21 '16 at 16:09

score 46 · Answer 1 · answered Feb 20 '19 at 09:29

46

The constructors of Pageable are deprecated, use of() instead:

Pageable pageable = PageRequest.of(0, 20);

answered Feb 20 '19 at 09:29

snowfox

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This doesn't exist in my version of spring, how can I create a peageable object to limit the query to 1? – Dimitri Kopriwa May 25 '20 at 14:21

xenteros · Accepted Answer · 2019-10-17T10:46:35.490

34

I've seen similar problem last week, but can't find it so I'll answer directly.

Your problem is that you specify the parameters too late. Pageable works the following way: you create Pageable object with certain properties. You can at least specify:

Page size,
Page number,
Sorting.

So let's assume that we have:

PageRequest p = new PageRequest(2, 20);

the above passed to the query will filter the results so only results from 21th to 40th will be returned.

You don't apply Pageable on result. You pass it with the query.

Edit:

Constructors of PageRequest are deprecated. Use Pageable pageable = PageRequest.of(2, 20);

edited Oct 17 '19 at 10:46

answered Nov 21 '16 at 07:46

xenteros

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1

Note: Index is zero-based page index. – rbawaskar Feb 20 '19 at 10:23
2

new PageRequest() constructor is deprecated – J3ernhard Oct 17 '19 at 10:44
@J3ernhard it wasn't when this answer was prepared. – xenteros Oct 17 '19 at 10:45
@xenteros that's right, further information about PageRequest: https://stackoverflow.com/q/44848653/4121281 – J3ernhard Oct 17 '19 at 10:55
how to set page size unlimited? – Sandeep Kumar Dec 23 '19 at 17:50
@SandeepKumar what do you mean, by setting the unlimited page size? – xenteros Dec 23 '19 at 20:21
I was writing a wrapper on top of an existing paginated API to fetch all record, I did it by passing Integer.Max_Value. – Sandeep Kumar Dec 24 '19 at 04:08
`PageRequest p = new PageRequest(2, 20);` can't be passed to spring data – Dimitri Kopriwa May 25 '20 at 14:21
1

Use `PageRequest.of` instead of constructor – masterxilo Jun 09 '21 at 14:22

Tiago PC · Answer 3 · 2022-04-28T19:36:33.567

You can use Page, List or Slice.

If you dont need the number of pages, and only need to know if the next page exists, use Slice, since it does not do the "count" query:

  for (int i = 0;  ; i++) { 
     Slice<Organization> pageOrganization = organizationRepository.find(PageRequest.of(0, 100));        
    
    
     List<Organization> organizationList = pageOrganization.getContent();
     for (Organization org : organizationList) {
           // code
     }
    
    if (!pageOrganization.hasNext()) {
        break;
    }   
}

score 0 · Answer 4 · answered Feb 24 '22 at 04:14

0

Pageable object by default comes with size 20, page 0, and unsorted So if you want the next page in front end the url can be sent with query params page, size,sort and these u can test it on postman.

answered Feb 24 '22 at 04:14

blessy kiran00

1
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Creating Pagination in Spring Data JPA

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