我需要只存储文件的文件名+扩展名时上传和重命名完成我的databse.my问题是无法获得扩展名。PHP:如何获取Codeigniter中上传文件的文件名和扩展名?
$new_file_name=date("mdY")."_".time();
$config['upload_path'] = './assets/images/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = 2048;
$config['max_width'] = 1024;
$config['max_height'] = 768;
$config['file_name'] = $new_file_name;
$this->load->library('upload', $config);
if (! $this->upload->do_upload('uploadFile'))
{
$error = array('error' => $this->upload->display_errors());
var_dump($error);
}
else
{
$data = array('upload_data' => $this->upload->data());
var_dump($data);
echo $config['file_name'] . $config['file_ext'];
}
我的下一个问题加载助手的形式为?或需要加载表单来使用库form_validation?
$this->load->helper(array('form')); for ?
$this->load->library('form_validation'); this for validation rule
的可能的复制(http://stackoverflow.com【如何提取的PHP文件扩展名?]/questions/173868/how-to-extract-a-file-extension-in-php) –
请参阅本用户指南:https://codeigniter.com/user_guide/libraries/file_uploading.html#preferences并应用此答案。 :http://stackoverflow.com/a/5257466/4952944 – Bhavin
您应该看到您从$ this-> upload-> data()创建的var_dump($ data)中的所有File属性。所以你只需要抓住你想要的东西。 – TimBrownlaw