如何使用jQuery从PHP接收数据到弹出菜单?
<a href="#?da=<?php echo $fetch['da_ref'] ?>" class="big-link" data-reveal-id="myModal" data-animation="fade" style="text-decoration: none"><?php echo $fetch['da_ref'];?></a>
Perhaps you can use the method of retrieving the variable's value within your pop-up the same way you do with the attributes from your a tag. Analyse your script and see what is happening with your data-... attributes.
As far as I can see, you should add a new attribute to your a tag, something similar to data-variable="value".
A small snippet of code for this using jQuery, would be the following
The HTML:
<div>
<a href="#" class="big-link" data-reveal-id="myModal" data-variable="variable value " data-animation="fade">Click me</a>
</div>
The CSS:
a { display: inline-block; padding: 10px; color: #333; text-decoration: none; border: 1px solid #333; }
The JavaScript:
$(document).ready(function() {
$(".big-link").on("click", function() {
var variable = $(this).attr("data-variable");
$(this).parent().append(variable);
});
});
You can try this on http://jsfiddle.net/ .
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