刷新页面后检查javascript保持复选框 [英] javascript keep checkbox checked after refresh page

问题描述

我是javascript新手。

hi i am javascript newbie.

我有一个slideDown其他输入的复选框，我想保持选中该复选框，并在刷新页面后显示其他输入选中输入复选框

I have checkbox that slideDown other input , i want to keep that checkbox checked and the other input showed after refreshing the page if the input checkbox checked

javascript：

function validateSitec(){
if (document.getElementById('sitecheck').checked){
$('#sitecheck').prop('checked', true);
$('#closedmsg').slideDown();
}else{
$('#closedmsg').slideUp();
$("#sitecheck").removeProp("checked").checkboxradio("refresh");
}
}

这是我的输入：

<label for="sitecheck">
<span style="font-weight:bold;">close site+ msg:</span>
</label>
<input name="" type="checkbox" id="sitecheck" onclick="validateSitec()" /><span style="font-weight:bold;">click to activate msg</span><br>
<input type="text" name="closedmsg" id="closedmsg" style="width:440px;height:120px;display:none;" value="enter closed msg.."/>

我希望如果选中保持检查..并且在刷新页面后不会更改，那么当取消选中时如此返回正常并在刷新页面时取消选中..

i want if checked stay checked.. and wont change after refreshing the page , then when unchecked so back to normal and be unchecked when refreshing the page..

我应该使用php使其在使用javascript检查后不会更改吗？

should i use php for making it not change after checking with javascript?

已编辑：

感谢所有人帮助
学分：leiyonglin。

Thanks to all for helping credit goes to : leiyonglin .

任何喜欢使用它的人的工作代码：

The working code for anyone who like to use it:

首先下载：
https://github.com/carhartl/jquery-cookie

然后使用此代码工作真棒：

then use this codes working awesome :

JavaScript：

JavaScript:

<script type="text/javascript">

function validateSitec(){
if (document.getElementById('sitecheck').checked){
    $('#sitecheck').prop('checked', true);
    $('#closedmsg').slideDown();
    $.cookie("cookieChecked", "#sitecheck");
}else{
    $('#closedmsg').slideUp();
    $("#sitecheck").removeProp("checked");
    $.cookie("cookieChecked","");
}
}


  $(function(){
      var cookieChecked = $.cookie("cookieChecked");
      if(cookieChecked){
          $('#sitecheck').prop('checked', true);
          $('#closedmsg').show();
      }else{
          $("#sitecheck").removeProp("checked");
          $('#closedmsg').hide();
      }
 })
    </script>

html输入：

<label for="sitecheck">
<span style="font-weight:bold;">close site temp:</span>
</label>
<input name="" type="checkbox" id="sitecheck" onclick="validateSitec()" /><span style="font-weight:bold;">close site and add message</span><br>
<input type="text" name="closedmsg" id="closedmsg" style="width:440px;height:120px;display:none;" value="<?php echo $data['csitemsg']; ?>" />

这个工作完美thx all。

This working perfect thx again all.

推荐答案

设置cookie：
将checked元素保存在cookieChecked中

Set cookies: Save the checked element in cookieChecked

function validateSitec(){
if (document.getElementById('sitecheck').checked){
    $('#sitecheck').prop('checked', true);
    $('#closedmsg').slideDown();
    $.cookie("cookieChecked", "#sitecheck");
}else if(document.getElementById('closedmsg').checked){
    $('#closedmsg').slideUp();
    $("#sitecheck").removeProp("checked").checkboxradio("refresh");
    $.cookie("cookieChecked", "#closedmsg");
} else {
    $.cookie("cookieChecked","");
}
}

页面上传时

 $(function(){
      var cookieChecked = $.cookie("cookieChecked");
      if(cookieChecked){
          $(cookieChecked).trigger("click");
      }
 })

我使用jquery cookie插件： https://github.com/carhartl/jquery-cookie

I use the jquery cookie plugin : https://github.com/carhartl/jquery-cookie

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