图像上传器。JQuery
我的任务有些问题。我创建了asp.NETMVC项目并将图像上传到服务器。对于上传,我使用swfupload。有创建上传程序的代码:图像上传器。JQuery,jquery,html,asp.net,Jquery,Html,Asp.net,我的任务有些问题。我创建了asp.NETMVC项目并将图像上传到服务器。对于上传,我使用swfupload。有创建上传程序的代码: <script> $(function() { $("#photo").makeAsyncUploader({ upload_url: "/Home/AsyncUpload/<%= ViewData["guid"]%>", flash_url: '/Scripts/swfupload.swf',
<script>
$(function() {
$("#photo").makeAsyncUploader({
upload_url: "/Home/AsyncUpload/<%= ViewData["guid"]%>",
flash_url: '/Scripts/swfupload.swf',
button_image_url: '/Content/blankButton.png',
disableDuringUpload: 'INPUT[type="submit"]'
});
});
</script>
$(函数(){
$(“#照片”).makeAsyncUploader({
上传url:“/Home/AsyncUpload/”,
flash_url:“/Scripts/swfupload.swf”,
按钮\图像\ url:'/Content/blankButton.png',
加载时禁用:“输入[type=“submit”]”
});
});
这是一个名为“照片”的按钮。但必须有更多名为“photo1”“photo2”的按钮。。。我为添加到div新输入创建了脚本:
<input type="button" onclick="add()" value="Add new button"/>
<script type="text/javascript">
function add() {
var newelement = document.createElement('p');
newelement.setAttribute("class", "file_input");
var x = ((document.getElementById('uploadImage').getElementsByTagName('p').length));
var fileinput = document.createElement('input');
fileinput.setAttribute("name", "photo" + x);
fileinput.setAttribute("type", "file");
newelement.appendChild(fileinput);
document.getElementById('uploadImage').appendChild(new_element);
}
</script>
函数add(){
var newelement=document.createElement('p');
setAttribute(“类”、“文件输入”);
var x=((document.getElementById('uploadImage').getElementsByTagName('p').length));
var fileinput=document.createElement('input');
setAttribute(“名称”、“照片”+x);
setAttribute(“类型”、“文件”);
newelement.appendChild(fileinput);
document.getElementById('uploadImage').appendChild(新的_元素);
}
它正在工作。它将添加按钮添加到我的div
<div id="uploadImage">
<p>Photo: <input type="file" id="photo" name="photo"/> </p>
</div>
照片:
我有一个问题,当我添加一个新按钮时,它将加载的脚本应该是什么。感谢并为我的英语感到抱歉。如果我正确理解了您的问题,您的add()函数应该如下所示:
<script>
var uploadButtonCount = 0;
function add() {
var uploadButtonId = "photo" + uploadButtonCount++;
var uploadInput = $('<input id="' + uploadButtonId + '" type="file" name="photo"/>');
$("#uploadImage").append(uploadInput);
uploadInput.makeAsyncUploader({
upload_url: "/Home/AsyncUpload/<%= ViewData["guid"]%>",
flash_url: '/Scripts/swfupload.swf',
button_image_url: '/Content/blankButton.png',
disableDuringUpload: 'INPUT[type="submit"]'
});
}
</script>
var uploadButtonCount=0;
函数add(){
var uploadButtonId=“photo”+uploadButtonCount++;
var uploadInput=$('');
$(“#uploadImage”).append(uploadInput);
uploadInput.makeAsyncUploader({
上传url:“/Home/AsyncUpload/”,
flash_url:“/Scripts/swfupload.swf”,
按钮\图像\ url:'/Content/blankButton.png',
加载时禁用:“输入[type=“submit”]”
});
}
您需要将新创建的具有唯一id的输入传递给插件。是。非常感谢你。