开发者

how to show the radio button is selected is still ticked?

my app has 2 radio buttons, when user ticks any of it, the value of the ticked radio button gets saved in the table via ajax, now the problem is, when user gets back to the page, the radio button is not ticked, how to make the selected radio button appear as ticked, so that the user knows which he selected before ?

here's my code

<ul>
    <li style="list-style-type: none;">    
<div align="center" class="radio_group">
    <input type="radio" id="gallerymenustyle1" class="element radio" name="gallerymenustyle[]" value="1" /> Gallery Link - In the navigation of my website, display one "gallery" link<br />
    <input type="radio" id="gallerymenustyle2" class="element radio" name="gallerymenustyle[]" value="2" /> Category Links - In the navigation of my websi开发者_运维技巧te, display a separate link to each category.
</div>
    </li>
</ul>


When you are rendering the page for the user in PHP you would need to do a query to where you have saved the selection (presumably a database). If the user has selected that particular radio button then you would display it as selected. Code would be something like:

// Do a database query or something to get the value that the user has stored before (if any)
<input type="radio" id="gallerymenustyle1" class="element radio" name="gallerymenustyle[]" value="1" <?php if ($gallerymenustyleFromDatabaseValue == 1){ echo 'selected'; }/> Gallery Link....<br />
<input type="radio" id="gallerymenustyle1" class="element radio" name="gallerymenustyle[]" value="2" <?php if ($gallerymenustyleFromDatabaseValue == 2){ echo 'selected'; }/> Category Link....


If the user is logged in you should query the database to get the value whether the button was ticked.

If the user isn't logged in you could use the session id. Although this only works until the user closes his browser / the session expires.

<?php
$checked = 0;
// this should come from db
$checked = 1;
?>

<ul>
  <li style="list-style-type: none;">    
    <div align="center" class="radio_group">
      <input type="radio" id="gallerymenustyle1" class="element radio" name="gallerymenustyle[]" value="1" <?php if ($checked == 1) print('checked="checked"') ?>/> Gallery Link - In the navigation of my website, display one "gallery" link<br />
      <input type="radio" id="gallerymenustyle2" class="element radio" name="gallerymenustyle[]" value="2" <?php if ($checked == 1) print('checked="checked"') ?>/> Category Links - In the navigation of my website, display a separate link to each category.
    </div>
  </li>
</ul>

Will check the first radio button

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜