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multiple-to-one relationship mysql, submissions

开发者 https://www.devze.com 2023-02-10 16:54 出处:网络
I have the following problem. Basically I have a form with an option to submit up to 3 images. Right now, after each submission it creates 3 records for album table and 3 records for images. I need it

I have the following problem. Basically I have a form with an option to submit up to 3 images. Right now, after each submission it creates 3 records for album table and 3 records for images. I need it to be one record for album and 3 for images, plus to link images to the album. I hope it all makes sense...

Here is my structure.

TABLE `albums` (
  `id` int(11) NOT NULL auto_increment,
  `title` varchar(50) NOT NULL,
  `fullname` varchar(40) NOT NULL,
  `email` varchar(100) NOT NULL,
  `created_at` datetime NOT NULL,
  `theme_id` int(11) NOT NULL,
  `description` int(11) NOT NULL,
  `vote_cache` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;

TABLE `images` (
  `id` int(11) NOT NULL auto_increment,
  `album_id` int(11) NOT NULL,
  `name` varchar(30) NOT NULL,

and my code

function create_album($params)
    {
       db_connect();

         $query = sprintf("INSERT INTO albums set
                                         albums.title = '%s',
                                                                     albums.email = '%s',
                                                                     albums.discuss_url = '%s',
                                                                     albums.theme_id = '%s',
                                                                     albums.fullname = '%s',
                                                                     albums.description = '%s',
                                                                     created_at = NOW()",
                                                                     mysql_real_escape_string($params['title']),
                                                                     mysql_real_escape_string($params['email']),
                                                                                                                 mysql_real_escape_string($params['theme_i开发者_如何学编程d']),
                                                                     mysql_real_escape_string($params['fullname']),
                                                                     mysql_real_escape_string($params['description'])
                                                                     );

         $result = mysql_query($query);
         if(!$result)
         {
              return false;
         }

         $album_id = mysql_insert_id();

         return $album_id;
    }

    if(!is_uploaded_file($_FILES['userfile']['tmp_name'][$i])) 
{ 
$warning = 'No file uploaded'; 
} 
elseif is_valid_file_size($_FILES['userfile']['size'][$i])) { 
$_POST['album']['theme_id'] = $theme['id']; 
create_album($_POST['album']); mysql_query("INSERT INTO images(name) VALUES('$newName')"); 
copy($_FILES['userfile']['tmp_name'][$i], './photos/'.$original_dir.'/' .$newName.'.jpg');


I must be missing something. Your query to insert images doesn't insert an album id, while that field is mandatory in images. So that query should fail.

I'm guessing that this is not all of your code and you actually execute a different query to insert the image. If so, that statement probably looks something like this:

$query = "
    INSERT INTO images (name, album_id) 
    VALUES ('albumname', " . create_album($params);

If so, you're creating an album for each image you save. You should save the resulting album id in a variable and use that variable in the three statements that create the image records.

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