14

我有对象列表。我需要做分页。
输入参数是每页的最大对象数和页码。

例如输入list = ("a", "b", "c", "d", "e", "f")
每页最大页数为 2 页数为 2 结果 = ("c", "d")

是否有任何现成的类（库）来做到这一点？比如Apache项目等等。

4

8 回答 8

23 
int sizePerPage=2;
int page=2;

int from = Math.max(0,page*sizePerPage);
int to = Math.min(list.size(),(page+1)*sizePerPage)

list.subList(from,to)
于 2012-08-31T12:03:07.467 回答
20

使用 Java 8 流：

list.stream()
  .skip(page * size)
  .limit(size)
  .collect(Collectors.toCollection(ArrayList::new));
于 2014-07-30T13:41:17.957 回答
3

尝试：

int page    = 1; // starts with 0, so we on the 2nd page
int perPage = 2;

String[] list    = new String[] {"a", "b", "c", "d", "e", "f"};
String[] subList = null;

int size = list.length;
int from = page * perPage;
int to   = (page + 1) * perPage;
    to   = to < size ? to : size;

if ( from < size ) {
    subList = Arrays.copyOfRange(list, from, to);
}
于 2012-08-31T12:02:32.837 回答
2

简单的方法

  public static <T> List<T> paginate(Page page, List<T> list) {
      int fromIndex = (page.getNumPage() - 1) * page.getLenght();
      int toIndex = fromIndex + page.getLenght();

      if (toIndex > list.size()) {
        toIndex = list.size();
      }

      if (fromIndex > toIndex) {
        fromIndex = toIndex;
      }

      return list.subList(fromIndex, toIndex);
  }
于 2015-10-14T20:05:07.130 回答
1

试试这个： 

int pagesize = 2;
int currentpage = 2;
list.subList(pagesize*(currentpage-1), pagesize*currentpage);

此代码返回一个列表，其中仅包含您想要的元素（页面）。

您还应该检查索引以避免 java.lang.IndexOutOfBoundsException。

于 2012-08-31T12:15:38.820 回答
0

根据您的问题，简单List.subList会给您预期的行为 size()/ 2= 页数

于 2012-08-31T12:02:38.327 回答
0

您可以使用List.subListusingMath.min来防范ArrayIndexOutOfBoundsException：

List<String> list = Arrays.asList("a", "b", "c", "d", "e");
int pageSize = 2;
for (int i=0; i < list.size(); i += pageSize) {
    System.out.println(list.subList(i, Math.min(list.size(), i + pageSize)));
}
于 2012-08-31T12:07:17.720 回答
0

使用 lombok 作为 getter 和 setter

@Getter
@Setter
public class Pagination<T> {
  
  private int pageSize = 10;
  private int paginationSize = 9;
  private int currentPage = 0;
  private List<T> list = null;

  public Pagination(List<T> list) {
    this.list = list;
  }
  
  public int getSize() {
    return list == null ? 0 : list.size();
  }
  
  public int getOffset() {
    return getPageSize() * getCurrentPage();
  }
  
  public int getLimit() {
    return getPageSize();
  }
  
  public List<T> getItemsOnCurrentPage() {
    return list.subList(getOffset(), getOffset()+getLimit());
  }
  
  public int getPageCount() {
    return getSize() / getPageSize() + (getSize() % getPageSize() > 0 ? 1 : 0);
  }
  
  public int getActualPaginationSize() {
    return Math.min(getPaginationSize(), getPageCount());
  }
  
  public int getLastPage() {
    return getPageCount() - 1;
  }
  
  public int getStartPage() {
    if (getPageCount() > getPaginationSize()
      && getCurrentPage() > getPaginationSize() / 2) {
      return Math.min(getCurrentPage() - getPaginationSize() / 2,
          getPageCount() - getActualPaginationSize());
    } else {
      return 0;
    }
  }
  
  public int getEndPage() {
    return Math.min(getStartPage() + getPaginationSize(), getPageCount()) - 1;
  }
  
  public void setCurrentPage(int newPage) {
    if (newPage < 0) newPage = 0;
    if (newPage > getLastPage()) newPage = getLastPage();
    this.currentPage = newPage;
  }
  
  public boolean isHasPreviousPage() {
    return hasPreviousPage();
  }
  
  public boolean hasPreviousPage() {
    return getStartPage() > 0;
  }
  
  public boolean isHasNextPage() {
    return hasNextPage();
  }
  
  public boolean hasNextPage() {
    return getEndPage() < getLastPage();
  }
  
  public boolean isShouldShowFirst() {
    return shouldShowFirst();
  }
  
  public boolean shouldShowFirst() {
    return getStartPage() - getPaginationSize() / 2 > 0;
  }
  
  public boolean isShouldShowLast() {
    return shouldShowLast();
  }
  
  public boolean shouldShowLast() {
    return getEndPage() + getPaginationSize() / 2 < getLastPage();
  }
  
}

页面索引从 0 开始，当前页面居中。

于 2021-12-05T16:10:25.010 回答