JavaWeb-分页查询

之前写的简易购物车系统,在显示图书列表的时候是全部显示出来,图书少的时候没有什么问题,但当图书数量变多的时候,一下子把全部图书显示出来会让界面看的不美观,所以就打算加个分页查询的功能

最终效果:

实现:

 分页查询有几个比较重要的参数,pageNum当前的页数,pageSize每页所展示的数据,totalRecord总共有多少数据,其他的数据可以根据这三个数据得出,有了这几个数据就可以分页查询了

public class Page {
    //当前的页数
    private int pageNum;
    //每页所展示的数据
    private int pageSize;
    //总共有多少数据
    private int totalRecord;
    //总页数
    private int totalPage;
    //查询数据的起始位置
    private int startIndex;
    //存放查询到的数据
    private List<Book> booklist;
    //相对的头页
    private int start;
    //相对的尾页
    private int end;

    public Page(int pageNum, int pageSize, int totalRecord) {
        this.pageNum = pageNum;
        this.pageSize = pageSize;
        this.totalRecord = totalRecord;

        if(totalRecord%pageSize == 0){
            //如果刚好能够整除,则totalPage = totalRecord/pageSize
            this.totalPage = totalRecord/pageSize;
        }else{
            //如歌不能整除,则需要多加一页显示余下的数据
            this.totalPage = (totalRecord/pageSize)+1;
        }
        this.startIndex = (pageNum-1)*pageSize;
        //初始头页
        this.start = 1;
        //初始尾页
        this.end = 5;
        if(totalPage<5){
            //如果总页数小于5页,则尾页为总页数
            this.end = this.totalPage;
        }else{
            this.start = this.pageNum-2;
            this.end = this.pageNum+2;
            if(this.start<0){
                this.start = 1;
                this.end = 5;
            }
            if(this.end>totalPage){
                this.end = totalPage;
                this.start = this.end-5;
            }
        }
    }

    //下面的为get/set方法
    public int getPageNum() {
        return pageNum;
    }

    public void setPageNum(int pageNum) {
        this.pageNum = pageNum;
    }

    public int getPageSize() {
        return pageSize;
    }

    public void setPageSize(int pageSize) {
        this.pageSize = pageSize;
    }

    public int getTotalRecord() {
        return totalRecord;
    }

    public void setTotalRecord(int totalRecord) {
        this.totalRecord = totalRecord;
    }

    public int getTotalPage() {
        return totalPage;
    }

    public void setTotalPage(int totalPage) {
        this.totalPage = totalPage;
    }

    public int getStartIndex() {
        return startIndex;
    }

    public void setStartIndex(int startIndex) {
        this.startIndex = startIndex;
    }

    public List<Book> getBooklist() {
        return booklist;
    }

    public void setBooklist(List<Book> booklist) {
        this.booklist = booklist;
    }

    public int getStart() {
        return start;
    }

    public void setStart(int start) {
        this.start = start;
    }

    public int getEnd() {
        return end;
    }

    public void setEnd(int end) {
        this.end = end;
    }
}

pageNum从请求中获得,pageSize自己设置,totalRecord通过mybatis查询获得,一开始登录的时候请求中没有pageNum,所以进行一个判断,如果没有pageNum则赋值为1

 //pageNum 
 String pageNum = request.getParameter("pageNum") == null?"1":request.getParameter("pageNum");
//pageSize 
   int pageSize = 7;
//totalRecord
int totalRecord = bookMapper.BookSize();
//mybatis
<select id="BookSize" resultType="int">
        select count(*) from book
 </select>

有了pageNum和startIndex就可以查询到特定范围的数据,通过使用mysql的limit函数,limit有两个参数,第一个参数为起始位置,即从那开始查,第二个参数为查询范围,即查多少个,如 limit 14,7 为从第14个数据开始往后查七个。

<select id="Books" parameterType="java.util.Map" resultType="book">
        select *from book  limit #{startIndex},#{pageSize}
    </select>
public static Page getPageUser(int pageNum,int pageSize){
        SqlSession sqlSession = GetSqlSession.getSqlSession();
        BookMapper bookMapper = sqlSession.getMapper(BookMapper.class);
        int totalRecord = bookMapper.BookSize();

        Page page = new Page(pageNum,pageSize ,totalRecord );
        int startIndex = page.getStartIndex();
        Map<String,Integer> map = new HashMap<>();
        map.put("startIndex",startIndex );
        map.put("pageSize",pageSize );
        page.setBooklist(bookMapper.Books(map));
        return page;
    }

jsp页面:

<a href="index?pageNum=1">首页</a>
            <%--当前页数为第一页--%>
        <c:if test="${sessionScope.page.pageNum == 1}">
            <c:forEach begin="${sessionScope.page.start}" end="${sessionScope.page.end}" step="1" var="i">
                <c:if test="${sessionScope.page.pageNum == i}">
                    <span  style="color: deeppink">${i}</span>
                </c:if>
                <c:if test="${sessionScope.page.pageNum != i}">
                    <a href="index?pageNum=${i}">${i}</a>
                </c:if>
            </c:forEach>
            <a href="index?pageNum=${sessionScope.page.pageNum+1}">下一页</a>
        </c:if>
        <%--当前页数为中间页数--%>
        <c:if test="${sessionScope.page.pageNum >1 && sessionScope.page.pageNum<sessionScope.page.totalPage}">
            <a href="index?pageNum=${sessionScope.page.pageNum-1}">上一页</a>
            <c:forEach begin="${sessionScope.page.start}" end="${sessionScope.page.end}" step="1" var="i">
                <c:if test="${sessionScope.page.pageNum == i}">
                    <span  style="color: deeppink">${i}</span>
                </c:if>
                <c:if test="${sessionScope.page.pageNum != i}">
                    <a href="index?pageNum=${i}">${i}</a>
                </c:if>
            </c:forEach>
            <a href="index?pageNum=${sessionScope.page.pageNum+1}">下一页</a>
        </c:if>
        <%--当前页数为最后一页--%>
        <c:if test="${sessionScope.page.pageNum == sessionScope.page.totalPage}">
            <a href="index?pageNum=${sessionScope.page.pageNum-1}">上一页</a>
            <c:forEach begin="${sessionScope.page.start}" end="${sessionScope.page.end}" step="1" var="i">
                <c:if test="${sessionScope.page.pageNum == i}">
                    <span  style="color: deeppink">${i}</span>
                </c:if>
                <c:if test="${sessionScope.page.pageNum != i}">
                    <a href="index?pageNum=${i}">${i}</a>
                </c:if>
            </c:forEach>
        </c:if>
        <a href="index?pageNum=${sessionScope.page.totalPage}">尾页</a>

其实这个用Ajax实现会更好,我这种写法每次都要请求整个页面,费时得很 以后时间再进行优化把哈哈哈

完整代码,在我的GitHub中:GitHub

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转载自blog.csdn.net/qq_40866897/article/details/88928446