我正在使用codiegniter通过Ajax上传文件，但我也想从数据中发送，并在控制器上获取数据并执行。
文件上载正在工作，但无法获取其他表单数据。
我还序列化了表单数据，但没有成功。
Html代码：
<form action="<?php echo site_url("admin_panel/upload_project_image") ?>" id="form-upload">
<input type="hidden" name="project_id" id="project_id" value="<?=$project_details->ID;?>">
<div class="fileinput fileinput-new input-group" data-provides="fileinput">
<div class="form-control" data-trigger="fileinput" style="margin:3px"> <i class="glyphicon glyphicon-file fileinput-exists"></i>
<span class="fileinput-filename"></span>
</div>
<span class="input-group-addon btn btn-default btn-file" style="margin:3px">
<span class="fileinput-new"> <i class="glyphicon glyphicon-paperclip"></i>
Select file
</span>
<span class="fileinput-exists">
<i class="glyphicon glyphicon-repeat"></i>
Change
</span>
<input type="file" name="file[]" multiple id="file"></span>
<a href="#" class="input-group-addon btn btn-default fileinput-exists" data-dismiss="fileinput" style="margin:3px">
<i class="glyphicon glyphicon-remove"></i>
Remove
</a>
<a href="#" id="upload-btn" class="input-group-addon btn btn-success fileinput-exists" style="margin:3px">
<i class="glyphicon glyphicon-open"></i>
Upload
</a>
</div>
</form>
Jquery代码：
$('#upload-btn').on('click', function(event) {
var filesToUpload = inputFile[0].files;
var projectID = $('#project_id').val();
if (filesToUpload.length > 0) {
var formData = new FormData();
for (var i = 0; i < filesToUpload.length; i++) {
var file = filesToUpload[i];
formData.append("file[]", file, file.name);
}
var other_data = $('#form-upload').serializeArray();
$.each(other_data,function(key,input){
formData.append(input.name,input.value);
});
$.ajax({
url: uploadURI,
type: 'post',
data: formData,
processData: false,
contentType: false,
success: function(data) {
console.log(data);
listFilesOnServer();
},
控制器代码：
function upload_project_image() {
if (!empty($_FILES)) {
$this->load->library('upload');
$config['upload_path'] = "./assets/uploads";
$config['allowed_types'] = 'gif|jpg|png|mp4|ogv|zip';
$this->upload->initialize($config);
$this->load->library('upload', $config);
if (!is_dir($config['upload_path'])) {
mkdir($config['upload_path'], 0777, true);
}
$files = $_FILES;
$project_id = $_POST['projectID'];
echo $project_id . "--ID";
$number_of_files = count($_FILES['file']['name']);
$errors = 0;
我认为没有必要单独序列化表单，如果您使用FormData，请尝试以下操作-
// this is what i am using to upload file and forms data with in the form
$(document).on('submit','#form-upload',function(){
var formData=new FormData($('#form-upload')[0]);
$object=$(this);
$.ajax({
url : $($object).attr('action'),
type : 'post',
data : formData,
processData:false,
contentType:false,
success : function(responseText)
{
alert('uploaded successfully');
// do what ever you want with responseText
},
error : function(e)
{
alert('hello from here');
}
});
});