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我試圖上傳我的多個文件時遇到問題。我有兩種方法通過ajax發送到一個頁面。一個是dropzone,另一個是我的XMLHttpRequest
。每次我通過單擊按鈕發送時,它只會複製我的請求,只使用第一種方法,即dropzone並忽略我的XMLHttpRequest
。所以我試圖找出將它綁定到formData的方式,使它只有1個文件一起請求。JQuery:如何使用dropzone.js將多個文件放入formData並通過ajax發送
我的形式
<form style="border:3px dashed #D9EDF7;" action="UploadImage"
class="dropzone dz-clickable"
id="my-awesome-dropzone" enctype="multipart/form-data">
<div class="dz-message">Drop files here or click to upload.
<br> <span class="note">(Selected files are not
automatically uploaded.)</span>
</div>
<div class="fallback">
<input name="file" id="filez" type="file" multiple/>
</div>
</form>
我的js
Dropzone.autoDiscover = false;
var myDropzone = new Dropzone('#my-awesome-dropzone', {
url : '../ajax/ajax_add/ajax_addNEWContestant.php?
multipleImage=multiple_image&event_id='+event_id,
autoProcessQueue : false
});
$(document).on('click','#addnewContestant', function(e){
myDropzone.processQueue();
var formdata = new FormData();
formdata.append('file_addnew', file_addnew.files[0]);
//If there is a way, I just want to bind all the files from dropzone into formData
var data = new FormData();
for(var b=0; b<imageContainer.length; b++){
formdata.append('All_files_from_dropzone[], input_file.files[b]);
}
var param = "?event_id="+encodeURIComponent(event_id)+
"&contestant_name="+encodeURIComponent(contestant_name)+
"&contestant_lastName="+encodeURIComponent(contestant_lastName)+
"&conAge="+encodeURIComponent(conAge)+
"&hAddress="+encodeURIComponent(hAddress)+
"&email_add="+encodeURIComponent(email_add)+
"&conContactNum="+encodeURIComponent(conContactNum)+
"&conDesc="+encodeURIComponent(conDesc)+
"&conId_hidden="+encodeURIComponent(conId_hidden)+
"&hidden_gender="+encodeURIComponent(hidden_gender)+
"&random_gender="+encodeURIComponent(random_gender)+
"&multipleImage="+encodeURIComponent(multipleImage);
beforeSend();
xhr = new XMLHttpRequest();
var url = '../ajax/ajax_add/ajax_addNEWContestant.php';
xhr.open('post', url+param, true);
});