我想要一个多文件上传代码。
例如:
Koala.jpg
Penguins.jpg
Jellyfish.jpg
有输入文本,用户可以在其中设置图像的新名称。
用户现在将上传图像,输入的新图像名称为#34;动物"
现在,我想要的是当上传输出时应该是Animals1.jpg,Animals2.jpg,Animals3.jpg。
问题是当我尝试上传所有这些图片时,只上传了一张图片。
我尝试进行研究并在我的程序中应用了一些代码,但仍然无效。
控制器
public function do_upload() {
$config = array(
'image_library' => 'gd2',
'file_name' => $this->input->post('finame'),
'upload_path' => './public/img/uploads',
'upload_url' => base_url().'public/img/uploads',
'allowed_types' => 'gif|jpg|jpeg',
'max_size' => '1024KB',
'max_width' => '1024',
'max_height' => '768',
'maintain_ratio'=> TRUE,
'overwrite' => false,
);
$this->load->library('upload', $config);
if (!$this->upload->do_upload()) {
$error_msg = "<div class='alert alert-error'>".$this->upload->display_errors()."</div>";
$error = array('error' => $error_msg);
}
else {
$upload_data = $this->upload->data();
$data['thumbnail_name'] = $upload_data['raw_name']. '_thumb' .$upload_data['file_ext'];
$file_array = array(
'image' => $data['thumbnail_name'],
'image_name' => $upload_data['file_name'],
//'description' => "",
'date_created' => date('Y-m-d H:i:s', now()),
'date_modified' => date('Y-m-d H:i:s', now()),
'author' => $this->session->userdata('username'),
'size' => $upload_data['file_size'],
'type' => $upload_data['image_type'],
'width' => $upload_data['image_width'],
'height' => $upload_data['image_height'],
//'document_name' => $field,
//'department' => $field2,
//'notes' => "",
);
$this->session->set_userdata('image_print', $file_array);
$this->load->database();
$this->db->insert('tbl_image', $file_array);
$data = array('upload_data' => $this->upload->data());
$user_level['records']=$this->user_model->get_records();
$this->load->view('v_dashboard/page/header_view', $user_level);
$this->load->view('v_document/upload/upload_result_view', $data);
$this->load->view('v_dashboard/page/footer_view');
}
}
我在HTML上有这个
<label for="file"><strong>Select File To Upload:</strong></label>
<input type="file" name="userfile[]" multiple class="btn transcolor btn-file"/>
<br/><br/>
顺便说一下,我在我的数据库上使用了BLOB。
我试图引用这个链接
答案 0 :(得分:0)
您必须运行for循环,直到上传的图像文件的计数。
像这样:
for ($i=0; $i < count($_FILES['userfile']['name']); $i++)
{
// function to add the image name one by one into database.
}