使用jQuery将事件绑定到动态创建的元素

使用jQuery将事件绑定到动态创建的元素

本文介绍了使用jQuery将事件绑定到动态创建的元素的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在上传多张图片,并将其放置在某些div上,这些div必须切换某些类。我是否必须放置在ajax成功函数中添加onclick事件的部分?非常感谢!

我正在使用jqueryon,但似乎不起作用。可能我错过了一些东西



以下是我的代码:



Javascript:


'b $ b

  $(。box-picture-selected> div> div)。 (this).toggleClass('image-selected'); 
});

$('#uploadForm')。submit(function(e){
e.preventDefault();

var form = new FormData($(' (var i = 0; i< files.length; i ++){
form。
var files = document.getElementsByClassName('pics');
append(files [+ i +],files [i] [0]); //添加收据以形成
}
form.append('action','upload-photos' ); //指定动作
form.append('csrfmiddlewaretoken','{{csrf_token()}}');
$ .ajax({{b $ url:'{{url( / b / b)}}',
type:'POST',
data:form,
cache:false,
processData:false,
contentType: false,
success:function(data){
$ .each($(data),function(key,value){
var displayDiv = document.getElementById(displayPics);
var grid = document.createElement(div);
grid.setAttribute(class, 'col-md-3 col-sm-4 col-xs-6 grid-changes image-selected');
var picDiv = document.createElement(div);
picDiv.setAttribute(class,'col-xs-12 images-box');
picDiv.setAttribute(style,'background-image:url({{url('fit /?image ='。urlencode(asset('filestorage / temp'))}}'+'%2F '+ value +');');
displayDiv.appendChild(grid);
grid.appendChild(picDiv);
});
},
error:function(xhr,desc,err){
//我在这里有一些错误处理逻辑
}
});
});

HTML:

 < div id =displayPicsclass =col-xs-12 grid-4-picture> 
@if(isset($ files)&&!empty($ files))
@foreach($ files as $ photo)
< div class =col-md- 3 col-sm-4 col-xs-6 grid-changes>

< input type =hiddenvalue ={{url('fit /?image ='。urlencode(asset('filestorage / temp /'.$ photo)))}}>
< / div>
@endforeach
@endif
< / div>


解决方案

对于动态创建的元素,您必须使用然而, live()已被弃用在1.7中有利于 on(),并在 1.9 中完全删除。 live()签名:



如果您有更大版本的 jQuery than 1.9 您可以使用 .on 下面是.on函数的签名

  $(document).on(eventName,selector,function(){}) ; 
$ b $(body)。on(click,#YOUR_DYNAMICALLY_CREATED_ELEMENT,function(event){
// Do Some stuff
});

解决的版本:

 ('click','.box-picture-selected> div> div',function(event)
{
$(this ).toggleClass('image-selected');
});


I am uploading multiple images and placing them inside some divs which onclick have to toggle some class. Do I have to place the part in which I add the onclick event inside the ajax success function? Thanks a lot!

I am using the jquery "on", but doesn't seem to work . Probably I'm missing something

Here is my code:

Javascript:

$(".box-picture-selected > div > div").on( 'click' , function () {
    $(this).toggleClass('image-selected');
});

$('#uploadForm').submit(function (e) {
    e.preventDefault();

    var form = new FormData($('form')[0]);
    var files = document.getElementsByClassName('pics');
    for (var i=0; i<files.length; i++) {
        form.append("files[" + i + "]", files[i][0]); // add receipt to form
    }
    form.append('action', 'upload-photos'); // specify action
    form.append('csrfmiddlewaretoken', '{{ csrf_token() }}');
    $.ajax({
        url: '{{url("/photos/device")}}',
        type: 'POST',
        data: form,
        cache: false,
        processData: false,
        contentType: false,
        success:function(data) {
            $.each($(data), function(key, value) {
                var displayDiv = document.getElementById("displayPics");
                var grid = document.createElement("div");
                grid.setAttribute("class", 'col-md-3 col-sm-4 col-xs-6 grid-changes image-selected');
                var picDiv = document.createElement("div");
                picDiv.setAttribute("class" , 'col-xs-12 images-box');
                picDiv.setAttribute("style", 'background-image: url({{ url('fit/?image='.urlencode(asset('filestorage/temp'))) }}' + '%2F' + value + ');'  );
                displayDiv.appendChild(grid);
                grid.appendChild(picDiv);
            });
        },
        error: function(xhr, desc, err) {
            // I have some error handling logic here
        }
    });
});

HTML:

<div id="displayPics" class="col-xs-12 grid-4-picture">
      @if (isset($files) && !empty($files))
          @foreach ($files as $photo)
              <div class="col-md-3 col-sm-4 col-xs-6 grid-changes">
                  <div class="col-xs-12 images-box" style="background-image: url({{ url('fit/?image='.urlencode(asset('filestorage/temp/'.$photo))) }});"></div>
                  <input type="hidden" value="{{ url('fit/?image='.urlencode(asset('filestorage/temp/'.$photo))) }}">
             </div>
          @endforeach
       @endif
    </div>

For dynamically created element you have to use .live() However, live() was deprecated in 1.7 in favour of on(), and completely removed in 1.9. The live() signature:

If you have greater version of jQuery than 1.9 you can use jQuery.fn.on

I would recommend to use .on below is a signature of .on function

$(document).on( eventName, selector, function(){} );

$("body").on("click", "#YOUR_DYNAMICALLY_CREATED_ELEMENT", function(event){
    //Do Some stuff
});

Solved version:

$("body").on('click', '.box-picture-selected > div > div', function(event)
{
    $(this).toggleClass('image-selected');
});

这篇关于使用jQuery将事件绑定到动态创建的元素的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-06 15:22