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ext中ComboBox+Ajax+分页联动代码和注释

今天花了1下午时间研究出了ComboBox+Ajax+分页联动效果.ext学习初步见效下面代码直接就能运行,用list模拟了数据库中取数据以下是目录结构以下是代码test1.jsp<%@pagelanguage="java"import="java.util.*"pageEncoding="UTF-8"%></span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:28:19</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/15076" title="attachMovie外部swf中的元素" target="_blank" onclick="return recordId('',)">attachMovie外部swf中的元素 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/15076" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/3.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/15076';}else{window.open('/article/15076');}" style="cursor: pointer;">[1b]作者的个人站点:[/1b][url=http://jim.blueidea.com/]http://jim.blueidea.com/[/url]相信大家都用过attachMovie这个语句，可以方便的绑定库中带Linkage输出的元素，在一些游戏以及交互程度比较高的网站中经常会用到。那么这里就遇到一个问题，假如我在sub.fla库里面用linkage设定了几个元素，然后想在main.fla里面attachMovie这几个元素，该如何处理尼？[1b</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:23:49</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/14722" title="html4 的默认样式问题" target="_blank" onclick="return recordId('',)">html4 的默认样式问题 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/14722" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/10.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/14722';}else{window.open('/article/14722');}" style="cursor: pointer;">写div的时候遇到一个问题,一个div里面嵌套了一个h3但是h3会跑到div的外面去.试了各种方法之后,都不行,原因就是h3默认有自己的margin:html4的默认样式列表:http://www.w3cn.org/article/tips/2005/116.htmlh3有默认的样式,直接用的话会发生问.下面是默认的h3默认的样式:h3{font-size:1.17em;margin:.83em0}有默认的外边距的,如果不去改掉的话,就会错位..一般建项目</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:19:21</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/13523" title="MTK一些特殊特效的制作" target="_blank" onclick="return recordId('',)">MTK一些特殊特效的制作 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/13523" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/16.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/13523';}else{window.open('/article/13523');}" style="cursor: pointer;">使用刷屏函数可以制作简单的特效,但特效方方正正,中矩而不中规,过于单调.使用拷屏函数可以做许多单靠刷屏无法实现的函数,但要制作三维特效,不规则区域的特效,这些方法就有些不够使用了.更复杂的特效,一是使用3D模,二是使用数学函数.当然前提是你必须对lcd_layer_struct层的定义十分熟悉,对DM管理机制也比较熟悉.要不然你会不知道数学函数或者3D建立的模型怎么使用,或者写出来对不知道怎么在系统中调用.常用数学函数公式:.1,直线:(直线方程如下图)有</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:04:52</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/13314" title="汉诺塔递归解法" target="_blank" onclick="return recordId('',)">汉诺塔递归解法 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/13314" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/9.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/13314';}else{window.open('/article/13314');}" style="cursor: pointer;">汉诺塔的递归算法假设有3个分别命名为X、Y和Z的塔座，在塔座X上插有n个直径大小各不相同、依小到大编号为1、2…n的圆盘。现要求将X轴上的n个圆盘移至塔座Z上并仍按同样顺序叠排，圆盘移动时必须遵守下列规则：1、每次只能移动一个圆盘；2、圆盘可以插在X、Y和Z中任一塔座上；3、任何时刻都不能将一个较大的圆盘压在较小的圆盘之上。递归算法的思路：当n=1时，问题比较简单，只要将编号为1的圆盘从塔座X直接移至塔座Z上即可。当n>1时，需利用塔座Y作辅助塔座，若能设</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:02:00</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/12156" title="二叉树遍历--递归实现" target="_blank" onclick="return recordId('',)">二叉树遍历--递归实现 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/12156" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/15.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/12156';}else{window.open('/article/12156');}" style="cursor: pointer;">递归这东西真是抽象，我看着看着算法，就囫囵吞枣地的写了下，写得囧了···这次先用递归实现先序，中序，后序遍历算法。先大概说下原理：我输入一大串字符，中间#就是代表了空，基本的储存结构就是二叉链表。主要就是二叉树的创建和三种顺序的遍历。二叉树的创建通过从左孩子开始创建不断递归，知道读取了#，开始创建对应的右孩子，继续递归。访问的时候对于三种顺序不过就是对于操作的顺序改变而已。对于下面的程序，按照图里面的二叉树建立方式：输入ABD#G###CE##FH###就</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 09:30:11</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2187</span></span>  </p> </article><div class="pagination" style="display:;"> <ul><li><a href="/search/web/1/p/1">首页</a></li> <li class="prev-page"><a href="/search/web/1/p/2465">上一页</a></li> <li><a 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