我想以相同的形式上传音频文件和图像,并将其名称存储在数据库中......所发生的事情只是上传了一个文件(行中的第一个) 另一个给出错误的不允许类型
这是我的观点:
<?php echo form_open_multipart('index.php/ena_dyo_edit');?>
Title:<input type='text' name="title" size="40" id="title" />
<p>Title Photo:<input type="file" name="title_photo" size="20" id="title_photo" /></p>
<p>Body:<textarea name="body" rows = "10" cols="60" id="body"></textarea><p>
<p>Soundfile:<input type="file" name="soundfile" size="20" id="soundfile" /></p>
<p>Author:<input type="text" name="author" size="40" id="author"/></p>
<p><input type="submit" value="Submit New Post"/></p>
<?php echo form_close(); ?>
这是我的模型功能:
function soundfile_upload($userfile = 'userfile'){
//uploads a sound file
$config = array(
'upload_path' => './uploads/soundfiles/',
'allowed_types' => 'mp3|wav|aif|aiff|ogg',
'max_size' => '256000' //250 MB max size
);
$this->load->library('upload', $config);
if (!$this->upload->do_upload($userfile)) {
$error = array('error' => $this->upload->display_errors());
$this->load->view('upload_error', $error);
}
else{
$soundfile_data = $this->upload->data();
$entry_title = $this->input->post('title'); //using the name of the post to relate the audio file
$entry_id = $this->db->insert_id(); //the id of our last entry!
//create array to load to database
$insert_data = array(
'soundfile' => $soundfile_data['file_name']
);
//$this->db->where('id', $entry_id);
$this->db->update('entries', $insert_data,array('id' => $entry_id)); //load array to database
}
}
function title_photo_upload($userfile = 'userfile'){
//uploads a photo for a post title
$config = array(
'upload_path' => './uploads/title_photos/',
'allowed_types' => 'jpg|jpeg|gif|png',
'max_size' => '256000' //250 MB max size
);
$this->load->library('upload', $config);
if (!$this->upload->do_upload($userfile)) {
$error = array('error' => $this->upload->display_errors());
$this->load->view('upload_error', $error);
}
else{
$soundfile_data = $this->upload->data();
$entry_id = $this->db->insert_id(); //the id of our last entry!
//create array to load to database
$insert_data = array(
'title_photo' => $soundfile_data['file_name']
);
$this->db->update('entries', $insert_data,array('id' => $entry_id)); //load picture where id = entry_id
}
}
答案 0 :(得分:2)
根据CI文档,配置的'allowed_types'部分是mime类型的管道分隔列表,而不是文件扩展名。它看起来像你正在使用文件扩展名,如果扩展名与mime类型不同,可能会产生不允许类型的错误。例如,我的mp3文件往往具有mime类型'audio / mpeg'。所以你的allowed_types配置应该如下所示:
'allowed_types' => 'audo/mpeg|audio/x-wav|audio/x-aiff|application/ogg'