Trying to read the file name of a file that is uploaded as part of a form. The print_r command that I'm using to test just shows a blank screen. I have read the manual (near the bottom here) pertaining to this and don't understand what I'm doing wrong.
Controller:
public function do_upload() {
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png|txt|pdf|xlsx|csv|xls|bmp';
$config['max_size'] = 1000;
$this->load->library('upload', $config);
$file_name = $this->upload->data('file_name');
print_r($file_name);
View:
<?php echo form_open_multipart('Corpmuns/do_upload', array('method' => 'post'));?>
... // some drop-down menus and text fields here
<INPUT TYPE="file" NAME="userfile" id="userfile" >
</form>
答案 0 :(得分:1)
您没有执行上传操作,因此该文件尚未上传。这就是为什么你无法获得上传文件的名称。因为它不存在。
代码$this->upload->do_upload('userfile')
并确保在获取文件名之前成功上传文件。
答案 1 :(得分:0)
从上传库调用do_upload
功能。
更新控制器:
public function do_upload() {
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png|txt|pdf|xlsx|csv|xls|bmp';
$config['max_size'] = 1000;
$this->load->library('upload', $config);
if ($this->upload->do_upload('userfile')) { //use this function
$data['error'] = false;
$upload_data = $this->upload->data();
$data['data'] = $upload_data;
$data['msg'] = 'Image Successfully Uploaded.';
} else {
$data['msg'] = $this->upload->display_errors('', '<br>');
}
print_r($data)
}
答案 2 :(得分:0)
if ( ! $this->upload->do_upload('input_name'))
{
echo $this->upload->display_errors();
}
else
{
$file=$this->upload->data();
echo $image=$file['file_name'];//Set file name to varilable
}
}
答案 3 :(得分:-1)
<form action="" enctype="multipart/form-data" method="post"
name="uploadfile">
you should add enctype="multipart/form-data"