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IF 结构设计有关问题,乱了

Jun 13, 2016 am 10:26 AM
input name quot type

IF 结构设计问题,乱了!
有如下需求,希望大侠指导
INPUT表单1
//用来输入文本1
//用来输入文本2
//用来提交输入的信息
我想点击了提交ida后界面输出INPUT表单2
//用来输入文本1
//用来输入文本2
//用来进一步提交输入的信息
//用来返回上一input表单1界面,但要把表单一输入的信息存储下来
当点击了提交idb后进入表单3
我想点击了提交ida后界面输出INPUT表单2
//用来输入文本1
//用来输入文本2
//用来再一次提交输入的信息
//用来返回上一input表单2界面,但要把表单二输入的信息存储下来

就输输入玩的信息,当在界面上点返回按钮时能把上一级的信息存下来,方便更改。求指导。。。。。。。。用IF结构设计的头有点大。。。

------解决方案--------------------
直接把参数通过url传啊
或者保存在session里。


url加个 step 来标识在第几步这样就行了。
------解决方案--------------------
这就是典型的ajax应用了.
A.将所有需要输入的部分 分别写在


B.默认从 第一个div 显示 当ajax提交之后 显示第二个div
C.类推B

需要掌握的内容

1.js/Jquery控制 div 的隐藏与显示
2.jquery 的 ajax 方法POST/GET

------解决方案--------------------
跟php没关系吧。
HTML code
<script>      function goBack(id,o){                document.getElementById(id).style.display='block';                o.parentNode.style.display='none';      }       function goNext(id,o){                document.getElementById(id).style.display='block';                o.parentNode.style.display='none';      }</script>
Copy after login

------解决方案--------------------
保存在session里, 外加个状态, 根据请求和当前状态做状态转移和数据的session记录.
------解决方案--------------------
看了半天,总算搞明白你的意思了
看看这个测试例,应该是你想要的
PHP code
Copy after login
= $mh) $val = array_merge($_POST['id'], $val); else $val = $_POST['id']; }else { $val = $_POST['id']; $h--; }}foreach($val as $k=>$v) { if(floor($k/$num) == $h) { $t = $k+1; echo "文本$t"; }else echo "";}echo ""; echo ""; echo "";if($h) echo "";echo "
h:$h mh:$mh val:" . join(',', $val);?>
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