javascript - 你们怎样实现前端分页的?
滿天的星座
滿天的星座 2017-07-05 10:58:25
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后端的nodejs我能够实现分页但是前端的代码如何写呀?

滿天的星座
滿天的星座

全部回复(4)
学霸

估计题主需要的是前端分页脚本吧:

var getPageList = function(options){
    if(typeof options != "object" || !('pageId' in options) || !('pageRecord' in options)){
        throw Error("options Error! eg:{pageId:1, pageRecord:12, pageUrlTemplate : '/page/{PAGE}/'}");
    };
    options.pageId = parseInt(options.pageId);
    options.pageRecord = parseInt(options.pageRecord);
    options.pageSize = options.pageSize || 10;
    
    options.pageUrlTemplate = options.pageUrlTemplate || "?page={PAGE}";
    options.pageCount = parseInt(( options.pageRecord - 1 ) / options.pageSize ) + 1;
    
    var page = [];
    var firstPage = parseInt(( options.pageId - 1 ) / options.pageSize ) * options.pageSize + 1;
    
    options.getLink = options.getLink || function(pageId){
        return options.pageUrlTemplate.replace("{PAGE}", pageId);
    };
    
    page.push({
        name    : '首页',
        style   : options.pageId == 1 ? "disabled" : "",
        link    : options.getLink(1)
    });
    page.push({
        name    : '上一页',
        style   : options.pageId == 1 ? "disabled" : "",
        link    : options.getLink(options.pageId - 1)
    });
    for( var pageId = firstPage; pageId < firstPage + 10; pageId ++){
        if( pageId >= 1 && pageId <= options.pageCount ){
            page.push({
                name    : pageId,
                link    : options.getLink(pageId),
                style   : pageId == options.pageId ? "active" : ""
            });
        }
    }
    page.push({
        name    : '下一页',
        style   : options.pageId == options.pageCount ? "disabled" : "",
        link    : options.getLink(options.pageId + 1)
    });
    page.push({
        name    : '尾页',
        style   : options.pageId == options.pageCount ? "disabled" : "",
        link    : options.getLink(options.pageCount)
    });
    page.toString = function(){
        return page.map(function(item){
            return '<a href="' + item.link + '" class="' + item.style + '">' + item.name + '</a>';
        }).join("");
    };
    return page;
};

getPageList({pageId:1,pageRecord:1200});
/*
    [
        {"name":"首页","style":"disabled","link":"?page=1"},
        {"name":"上一页","style":"disabled","link":"?page=0"},
        {"name":1,"link":"?page=1","style":"active"},
        {"name":2,"link":"?page=2","style":""},
        {"name":3,"link":"?page=3","style":""},
        {"name":4,"link":"?page=4","style":""},
        {"name":5,"link":"?page=5","style":""},
        {"name":6,"link":"?page=6","style":""},
        {"name":7,"link":"?page=7","style":""},
        {"name":8,"link":"?page=8","style":""},
        {"name":9,"link":"?page=9","style":""},
        {"name":10,"link":"?page=10","style":""},
        {"name":"下一页","style":"","link":"?page=2"},
        {"name":"尾页","style":"","link":"?page=120"}
    ]
*/
'' + getPageList({pageId:1,pageRecord:1200,pageUrlTemplate:'/category/{PAGE}/view'});
/*
    <a href="/category/1/view" class="disabled">首页</a>
    <a href="/category/0/view" class="disabled">上一页</a>
    <a href="/category/1/view" class="active">1</a>
    <a href="/category/2/view" class="">2</a>
    <a href="/category/3/view" class="">3</a>
    <a href="/category/4/view" class="">4</a>
    <a href="/category/5/view" class="">5</a>
    <a href="/category/6/view" class="">6</a>
    <a href="/category/7/view" class="">7</a>
    <a href="/category/8/view" class="">8</a>
    <a href="/category/9/view" class="">9</a>
    <a href="/category/10/view" class="">10</a>
    <a href="/category/2/view" class="">下一页</a>
    <a href="/category/120/view" class="">尾页</a>
*/

如果提主恰巧使用的是 AngularJS,可以下载直接使用我的开源小项目:ng-pagination。

滿天的星座

前端直接写啊
1,2,3,4,5....
点击的时候 把相应的数字如‘4’,发送到后端
后端写sql语句如:select * from tab limit a*10,b

这个sql应该能懂把,

注: 如果数据库数据很多的话 使用这种sql效率不高

给我你的怀抱

前端根据后端需要的参数来发送请求啊,发什么参数取到对应的数据,就实现分页了,就是每次翻页都是需要发送ajax的

伊谢尔伦

比如。后台进行200条每页的数据分页来传给前端,并传回来总页数。前端就可以做每页10、20、50、100等不同条数的展示。前端做个计算,把前端的页码和后台分页页码相对应。每页数据就直接在200条里面取就好了。像angular,ng-repeat="item in items.slice(page*10, 10)"。封装个指令,屡试不爽

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