如何编写分页逻辑? [英] How to write Pagination logic?

问题描述

有人可以提供一些想法/逻辑来为我正在处理的搜索页面编写分页逻辑吗? 我所拥有的信息是该搜索的总页数-每页 10条记录我需要做的所有事情就是提取该信息并填充.我还获得了我所在页面的信息.我只能显示10个页面，如下所示:

Can anyone provide some idea/logic to write the pagination logic for the search page i am working on? The information i have is total number of pages for that search- 10 records per page also i am been sent the both the previous and next page number(no problem writing the logic all i need to do i pull that info and populate. I am also getting the info which page i am on. I can only display 10 pages like below

<previous 1 |2 |3 | 4| 5 | 6 | 7 | 8 | 9 | 10 next>

说如果总页数为15，并且当用户单击下一步"时，我需要这样显示

Say if total pages are 15 and when user click next then i need to display like this

<previous 2 |3 |4 |5 |6 |7 |8 |9 |10 |11 next>

任何时候我只需要在分页中显示10页即可.

At any time i just need to show then 10 pages in the pagination.

 #set($start = 1)
 #set($end = $Integer.parseInt($searchTO.getPagination().getNumberofPages()))
 #set($range = [$start..$end])

#set($iter = 1)
            #foreach($i in $range)
              #foreach($link in $searchTO.getPagination().getDirectPageLinks())
                    #if($i == $iter)
                        #if ($Integer.parseInt($searchTO.getPagination().getPageNumber())==$iter)
                            <a class="search_current" href="/?_page=SEARCH&_action=SEARCH$link">$i &nbsp|</a>
                        #else
                            <a href="/?_page=SEARCH&_action=SEARCH$link">$i &nbsp|</a>
                        #end
                        #set($iter = 1)
                        #break
                    #else
                        #set($iter=$iter+1)
                    #end

                 #end

            #end

推荐答案

这是我将如何实现它: 通常，创建一个Filter类来过滤数据并包含与分页有关的信息通常是一个好主意.我用这样的东西:

Here is how I would implement it: It is generally a good idea to create a Filter class that filters data and contains pagination related information for you. I use something like this:

public abstract class Filter{

     /** Member identifier for the current page number */
     private int currentPageNo;

     /** Member identifier for the current start page number in the page navigation */
     private int currentStartPageNo;

     /** Member identifier for the current end page number in the page navigation */
     private int currentEndPageNo;

     /** Member identifier for the number of elements on a page */
     private int elementsPerPage;

     /** Member identifier for the number of pages you have in the navigation (i.e 2 to  11 or 3 to 12 etc.) */      
     private int pageNumberInNavigation;

     public abstract Query createCountQuery();

     public abstract Query createQuery();

     public void setCurrentPageNo(){
         //Your code here
         //Validation, variable setup
     }

     public Long getAllElementsCount(){
          //Now this depends on the presistence framework you use, this code is
          //just for guidance and has Hibernate-like syntax
          Query query = createCountQuery();
          List list = query.list();
          return !list.isEmpty() && list.get(0) != null ? query.list().get(0) : 0;
     }

     public List getElements(){
          //Now this depends on the presistence framework you use, this code is
          //just for guidance and has Hibernate-like syntax
         Query query = createQuery();
         int from = ((currentPageNo - 1) * elementsPerPage);
         query.setFirstResult(from);
         query.setMaxResults(elementsPerPage);
         //If you use Hibernate, you don't need to worry for null check since if there are no results then an empty collection is returned
         return query.list();
     }

     public List getAllElements(){
         Query query = createQuery();
         return query.list();
     }

     public void refresh(){
         //Your code here
     }

     public List next(){
         //Move to the next page if exists
         setCurrentPageNo(getCurrentPageNo() + 1);
         getElements();
     }

     public List previoius(){
         //Move to the previous page if exists
         setCurrentPageNo(getCurrentPageNo() - 1);
         getElements();
     }

}

您可能有特殊的过滤器子类(取决于要检索的内容)，每个子类都将实现它的createCountQuery()和createQuery().

You could have special subclasses of filters (depending on what you want to retrieve) and and each subclass would implement it's createCountQuery() and createQuery().

然后将您的Filter放到Velocity上下文中，然后可以从此类中检索所需的所有信息.

You would put then your Filter to the Velocity context and you could retrieve all the information you need from this class.

在设置当前页面pf课程时，您会更新所需的所有其他信息(即currentStartPageNo，currentEndPageNo).

When you set the current page pf course you update all the other information that you need (i.e. currentStartPageNo, currentEndPageNo).

您还可以使用refresh()方法将过滤器放回其初始状态.

You could also have a refresh() method to put the filter back to its initial state.

当然，当用户在Filter所属的搜索页面上导航时，您应该在会话上保留相同过滤器的实例(我的意思是您的Web框架，如Struts，Turbine等).

And of course you should keep the instance of the same filter on the session (I mean you web framework like Struts, Turbine etc.) while the user navigates on the search page to which the Filter belongs.

这只是一个准则，一个想法，它不是完全编写的可执行代码，只是一个使您入门的示例.

This is just a guideline, an idea, it is not fully written executable code, just an example to get you started in a direction.

我还建议您使用一个具有分页支持的名为 jqGrid 的jQuery插件(尽管您必须有支持才能进行检索)数据)和许多更酷的东西. 您可以使用它在网格中显示数据.我将它与Velocity一起使用都没有问题.它具有很多非常好用的实用功能，例如过滤器工具栏，可编辑单元格，JSON，XML中的数据传输等. 老实说，我不知道它是否具有您所需的分页功能(我仅在导航中显示一个页面时使用它，并且不能单击页面，仅使用下一个上一个按钮即可导航)，但是它可能支持那个.

I would also recommend you a jQuery plugin called jqGrid that has pagination support (although you have to have a backed to retrieve data) and a lot more cool stuff. You can use it to display your data in a grid. I use it together with Velocity with no problem. It has a lot of very nice and useful features like filter toolbar, editable cells, data transfer in JSON, XML etc. Honestly, I do not know if it has pagination like you need (I use it with only one page displayed in the navigation and you can not click on a page just use the next a prev buttons to navigate), but it may have support for that.

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