POST方法ddslickddslick下拉表單提交不工作
這在下拉jQuery的不工作的形式是:
<?php
if(isset($_POST['usern'])){
$value = $_POST['usern'];
if($value > 0)
{
echo "work";
}
else
{
echo "not work";
}
}
?>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script> <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.min.js"></script> <script type="text/javascript" src="http://web-0a54fa71-eb5f-4cdc-bfbe-142bcd20757b.runnablecodesnippets.com/jquery.ddslick.min.js"></script> <script type="text/javascript" src="http://web-0a54fa71-eb5f-4cdc-bfbe-142bcd20757b.runnablecodesnippets.com/script.js"></script>
<form action='users.php?j=60' method='post'>
<select id="demo-htmlselect-basic" form="post" name="usern">
<option value="0">Select</option>
<option data-imagesrc="image.jpg" value="1"> Part 1</option>
<option data-imagesrc="image.jpg" value="2"> Part 2</option>
<option data-imagesrc="image.jpg" value="1"> Part 3</option>
</select>
<input type='submit' value='Send'>
</form>
你可以發佈你的整個代碼,所以我們可以看到你是否複製了一切正確?編輯你的問題,然後將所有內容放入Javascript/HTML/css代碼片段工具中。看到這個圖片點擊什麼http://s32.postimg.org/x7oqo2bh1/post.png – mlegg