我正在尝试使用ajax删除mysql的记录。一切都很好。但我也想在点击垃圾按钮时添加加载到按钮。下面是我的按钮链接
<a href="#" id="delpost" data-id="'.$row['id'].'" class="btn btn-default pull-right" title="Delete"><i class="fa fa-trash-o"></i></a>
以下是我的ajax
$(function() {
$('body').on('click', '#delpost', function() {
var del_id = $(this).attr("data-id");
var info = 'id=' + del_id;
var parent = $(this).closest('li');
var btn = $(this);
btn.button('loading');
if (confirm("Sure you want to delete this Post? There is NO undo!")) {
$.ajax({
type: "POST",
url: "/delete-post.php",
data: info,
beforeSend: function() {
parent.animate({
'backgroundColor': '#fb6c6c'
}, 300);
},
success: function() {
btn.button('reset');
parent.slideUp(300, function() {
parent.remove();
});
}
});
}
return false;
});
});
答案 0 :(得分:0)
这是简单的jQuery解决方案,与bootstrap
无关
您可以添加一个隐藏的loading.gif图片,您可以从谷歌下载并调整其css,使其与删除链接显示在同一行:
<a href="#" id="delpost" data-id="'.$row['id'].'" class="btn btn-default pull-right" title="Delete"><i class="fa fa-trash-o"></i></a> <img src="path/to/your/loading.gif" style="display:none;" alt="loading" id="loading" />
修改你的jquery代码如下:
$(function() {
$('body').on('click','#delpost',function(){
var del_id = $(this).attr("data-id");
var info = 'id=' + del_id;
var parent = $(this).closest('li');
var btn=$(this);
btn.button('loading');
if(confirm("Sure you want to delete this Post? There is NO undo!"))
{
$.ajax({
type: "POST",
url: "/delete-post.php",
data: info,
beforeSend: function() {
parent.animate({'backgroundColor':'#fb6c6c'},300);
$('#loading').show();//shows loading gif image
},
success: function(){
$('#loading').hide();//hides loading gif image
btn.button('reset');
parent.slideUp(300,function() {
parent.remove();
});
}
});
}
return false;
});
});