这是执行删除操作的代码。出现四个图像,但是通过提供onload功能,不会显示删除操作的警告框。请指导我......这是代码。
// script for deletedelete operation
$(document).ready(function(){
$('a.delete').on('click',function(e){
e.preventDefault();
imageID = $(this).closest('.image')[0].id;
alert('Now deleting "'+imageID+'"');
$(this).closest('.image')
.fadeTo(300,0,function(){
$(this)
.animate({width:0},200,function(){
$(this)
.remove();
});
});
});
});
HTML
//four images being given with delete link
<div id="container">
<div class="image" id="image1" style="background-image:url(http://lorempixel.com/100/100/abstract);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image2" style="background-image:url(http://lorempixel.com/100/100/food);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image3" style="background-image:url(http://lorempixel.com/100/100/people);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image4" style="background-image:url(http://lorempixel.com/100/100/technics);">
<a href="#" class="delete">Delete</a>
</div>
</div>
答案 0 :(得分:0)
一切正常......让我在这里拍摄一下。您可能没有在代码中包含Jquery文件???
您是否已加入此标记
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
此FIDDLE只是您粘贴的代码副本。没有一个改变
HTML
<div id="container">
<div class="image" id="image1" style="background-image:url(http://lorempixel.com/100/100/abstract);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image2" style="background-image:url(http://lorempixel.com/100/100/food);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image3" style="background-image:url(http://lorempixel.com/100/100/people);">
<a href="#" class="delete">Delete</a>
</div>
<div class="image" id="image4" style="background-image:url(http://lorempixel.com/100/100/technics);">
<a href="#" class="delete">Delete</a>
</div>
</div>
JQUERY
$('a.delete').on('click',function(e){
e.preventDefault();
imageID = $(this).closest('.image')[0].id;
alert('Now deleting "'+imageID+'"');
$(this).closest('.image')
.fadeTo(300,0,function(){
$(this)
.animate({width:0},200,function(){
$(this)
.remove();
});
});
});
答案 1 :(得分:0)
代码工作正常......如果你没有包含Jquery文件,请包含它..
&LT; script src =“http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js”&gt;
&LT; /脚本&GT;
&LT;脚本&GT; $(文件)。就绪(函数(){
$('a.delete').on('click',function(e){
e.preventDefault();
imageID = $(this).closest('.image')[0].id;
alert('Now deleting "'+imageID+'"');
$(this).closest('.image')
.fadeTo(300,0,function(){
$(this)
.animate({width:0},200,function(){
$(this)
.remove();
});
});
});
});
&LT; /脚本&GT;