我无法通过ajax上传多个文件。这是我的代码。
HTML代码: -
<input type="file" id="txtBusinessImage" class="form-control" name="txtBusinessImageName[]" multiple >
<input type="hidden" id="selectBusinessHiddenID" name="selectBusinessHiddenID" value="<?php echo $viewCompanyResult->company_id; ?>">
<input type="button" id="uploadBusinessImg" value="Upload" >
Ajax代码: -
$("#uploadBusinessImg").on("click",function(e)
{
var fd = new FormData();
var file_data = $("#txtBusinessImage")[0].files; // for multiple files
for(var i = 0;i<file_data.length;i++){
fd.append("file"+[i], file_data[i]);
}
var other_data = $("#selectBusinessHiddenID").serializeArray();
$.each(other_data,function(key,input){
fd.append(input.name,input.value);
});
$.ajax({
url: '<?php echo site_url('Main_ctrl/upload_business_photo_do'); ?>',
data: fd,
enctype: 'multipart/form-data',
contentType: false,
processData: false,
type: 'POST', async : true,
success: function(data){
alert(data);
}
});
});
当我通过Ajax调用upload_business_photo_do()函数时,它无法重新获取图片名称$ _FILES [&#39; file&#39;] [&#39; name&#39;]
upload_business_photo_do()
{
$business_hidden_id=$this->input->post('selectBusinessHiddenID');
/*code for image*/
$config['upload_path']='./upload_101/';
$config['allowed_types']= 'jpg|png|jpeg';
$config['max_width'] = '6000';
$config['max_height'] = '4500';
$this->load->library('upload',$config);
for($i=0; $i<count($_FILES['file']['name']); $i++)
{
$_FILES['userfile']['name']= $_FILES['file']['name'][$i];
$_FILES['userfile']['type']= $_FILES['file']['type'][$i];
$_FILES['userfile']['tmp_name']= $_FILES['file']['tmp_name'][$i];
$_FILES['userfile']['error']= $_FILES['file']['error'][$i];
$_FILES['userfile']['size']= $_FILES['file']['size'][$i];
if(! $this->upload->do_upload())
{
/*----set flash message*/
echo "error";
}
else
{
echo "done";
}
}
}
答案 0 :(得分:2)
尝试使用这样,简单易用
$("#uploadBusinessImg").on("click",function(e)
{
var formData = new FormData($("#form_name")[0]);
$.ajax({
url: '<?php echo site_url('Main_ctrl/upload_business_photo_do'); ?>',
processData: false,
contentType: false,
data: formData,
type: 'POST', async : true,
success: function(data){
alert(data);
}
});
});
并在控制器中使用像这样
if($_FILES['txtBusinessImageName'])
{
$file_ary = $this->reArrayFiles($_FILES['txtBusinessImageName']);
foreach ($file_ary as $file)
{
print 'File Name: ' . $file['name'];
print 'File Type: ' . $file['type'];
print 'File Size: ' . $file['size'];
}
}
并且还使用此功能将文件数据转换为多个图像数据的数组
function reArrayFiles(&$file_post) {
$file_ary = array();
$file_count = count($file_post['name']);
$file_keys = array_keys($file_post);
for ($i=0; $i<$file_count; $i++) {
foreach ($file_keys as $key) {
$file_ary[$i][$key] = $file_post[$key][$i];
}
}
return $file_ary;
}
它的工作完美,只是尝试使用它。你不需要用ajax添加额外的文件代码。
答案 1 :(得分:0)
使用表单标记和提交按钮进行文件上传。
<form method="post" enctype="multipart/form-data">
<input type="file" id="txtBusinessImage" class="form-control" name="txtBusinessImageName[]" multiple >
<input type="hidden" id="selectBusinessHiddenID" name="selectBusinessHiddenID" value="<?php echo $viewCompanyResult->company_id; ?>">
<input type="submit" id="uploadBusinessImg" value="Upload">
</form>
并从ajax调用中删除enctype:'multipart / form-data'并尝试。
更改以下内容以获取文件:
var file_data = $('#txtBusinessImage').prop('files')[0];
var fd = new FormData();
fd.append('file', file_data);