Storing related records: Object of class Artists could not be converted to string

Question

Storing related records: Object of class Artists could not be converted to string

kofkof01 May '14

Created May '14	Last Reply Jul '14	Replies 6	Views 2667	Votes 0

kofkof01 8.2k

May '14

Hello,

I'm trying to run the first example in the Storing related records documentation but all I get is an error message reading Object of class Artists could not be converted to string.

In case it could help, the following steps do produce the error:

First create mysql schema and tables:

CREATE SCHEMA TEST;
USE TEST;

CREATE TABLE Artists (
  id INT UNSIGNED NOT NULL AUTO_INCREMENT,
  name VARCHAR(45) NOT NULL,
  country VARCHAR(45) NULL,
  PRIMARY KEY (id)
);

CREATE TABLE Albums (
  id INT UNSIGNED NOT NULL AUTO_INCREMENT,
  name VARCHAR(45) NOT NULL,
  artist INT UNSIGNED NOT NULL,
  year YEAR NULL,
  PRIMARY KEY (id),
  CONSTRAINT fk_Albums_Artists
    FOREIGN KEY (artist)
    REFERENCES Artists (id)
);

Create a brand new project: $ phalcon project test and edit database details in /app/config.php.

Create /app/models/Artists.php:

class Artists extends \Phalcon\Mvc\Model
{
    public $id;
    public $name;
    public $country;

    public function initialize() {
        $this->hasMany('id', 'Albums', 'artist');
    }
}

Create /app/models/Albums.php:

class Albums extends \Phalcon\Mvc\Model
{
    public $id;
    public $name;
    public $artist;
    public $year;

    public function initialize() {
        $this->belongsTo('artist', 'Artists', 'id');
    }
}

Edit /controllers/IndexController.php with the code from the doc:

class IndexController extends ControllerBase
{

    public function indexAction()
    {
        // Create an artist
        $artist = new Artists();
        $artist->name = 'Shinichi Osawa';
        $artist->country = 'Japan';

        // Create an album
        $album = new Albums();
        $album->name = 'The One';
        $album->artist = $artist; //Assign the artist
        $album->year = 2008;

        //Save both records
        $album->save(); // this line triggers the error
    }

}

Run the page and get:

Catchable fatal error: Object of class Artists could not be converted to string

Thanks for any help.

ren
8.1k

Accepted
answer

edited May '14
May '14

You should give your relationship an alias. I guess in your example, since no alias, this should work: $album->artists = $artist

I guess you would want to name that alias 'Artist' but then you have a collision with your column named artist... (case doesn't matter here).

To solve that you can:

rename the artist column (e.g. to artist_id);
use a column map to rename the column;
use another alias for the relationship ('AlbumArtist' perhaps? - there could be more kinds of Artists related to the Album of course..) (but it would still be a good idea to apply option 1 or 2 anyway, to prevent accidental collisioning and to elliminate (some) ambiguity).

Hope this helps (and I am not mistaken)? ;)

Dylan Anderson · Answer 1 · 2014-05-23T10:50:39-07:00

I don't know if this is what's causing this particular error, but your Album/Artist relationship isn't set up properly. In the Artist class, this line:

$this->hasMany('id', 'Albums', 'artist_id');

Is referencing the artist_id property of Albums, but Albums doesn't have that property - it just has an artist property.

kofkof01 · Answer 2 · 2014-05-23T11:05:10-07:00

Thank you for spotting that, quasipickle. You are totally right. However, I still get the same error :-(

I've edited the code in my first post.

kofkof01 · Answer 3 · 2014-05-26T01:03:44-07:00

kofkof01
8.2k

May '14

Nobody seems to have any idea about this. Should I consider it a bug and report it?

Oleg · Answer 4 · 2014-05-26T01:18:27-07:00

$album->artist = $artist; //Assign the artist

your mistake here $artist is object. $album->artist is integer

kofkof01 · Answer 5 · 2014-05-26T01:36:22-07:00

Hi Oleg. I understand that $album->artist is expecting an id, but if I read the documentation correctly, this integer should automagically be extracted from the $artist object when $album gets saved. So is it a Phalcon bug or an error in the doc (or just me missing something obvious)?

Thanks anyway for helping!

Oleg · Answer 6 · 2014-05-26T02:14:29-07:00

I understood your mistake :)

You have defined the wrong relations.


CREATE TABLE Albums (
  id INT UNSIGNED NOT NULL AUTO_INCREMENT,
  name VARCHAR(45) NOT NULL,
  artist INT UNSIGNED NOT NULL,
  year YEAR NULL,
  PRIMARY KEY (id),
  CONSTRAINT fk_Albums_Artists
    FOREIGN KEY (artist)
    REFERENCES Artists (id)
);

artist must have name artist_id, for example - it link to id of artist

Then all code have sense.


class IndexController extends ControllerBase
{

    public function indexAction()
    {
        // Create an artist
        $artist = new Artists();
        $artist->name = 'Shinichi Osawa';
        $artist->country = 'Japan';

        // Create an album
        $album = new Albums();
        $album->name = 'The One';
        $album->artist = $artist; //Assign the artist
        $album->year = 2008;

        //Save both records
        $album->save(); // this line triggers the error
    }

}

You object Albums in this case have not artists property. But have artists_id for relations between Albums and Artists Then you create new (dynamic if you want named) property artists

$album->artist = $artist;

Before your Albums has structure like Albums-> ... any property fron DB. After this operand your Albums have structure Example.

$artist = new Artists();
        $artist->name = 'Shinichi Osawa';
        $artist->country = 'Japan';

        // Create an album
        $album = new Albums();
var_dump($album);
        $album->name = 'The One';
        $album->artist = $artist; //Assign the artist
        $album->year = 2008;

var_dump($album);

Before :


object(Albums)#2 (4) {
  ["id"]=>
  NULL
  ["name"]=>
  NULL
  ["artist_id"]=>
  NULL
  ["year"]=>
  NULL
}

After :


object(Albums)#2 (5) {
  ["id"]=>
  NULL
  ["name"]=>
  string(7) "The One"
  ["artist_id"]=>
  NULL
  ["year"]=>
  int(2008)
  ["artist"]=>
  object(Artists)#1 (3) {
    ["id"]=>
    NULL
    ["name"]=>
    string(14) "Shinichi Osawa"
    ["country"]=>
    string(5) "Japan"
  }
}


class Artists 
{
    public $id;
    public $name;
    public $country;

}

class Albums 
{
    public $id;
    public $name;
    public $artist_id;
    public $year;

}

kofkof01 · Answer 7 · 2014-05-26T03:19:24-07:00

Oleg, what I understand you're telling me is that the code example in the doc is wrong and should be changed this way:

// Create an artist
$artist = new Artists();
$artist->name = 'Shinichi Osawa';
$artist->country = 'Japan';
$artist->save(); // Added so that we get an id for the artist

// Create an album
$album = new Albums();
$album->name = 'The One';
// $album->artist = $artist; // Wrong
$album->artist = $artist->id; // Right
$album->year = 2008;

//Save both records
$album->save();

Please note that I haven't renamed the artist property/field "artist_id", I'm perfectly conscious it is an id (integer) but for the sake of simplicity I'll stick to the naming used in the doc. If the naming is important and I have to use the form <table>_id, please tell me.

Of course this would most probably work, but:

the code in the doc is radically different, it only does one save() operation to save both records, and it uses a whole object to link records (not just an id property)
I don't see any magic here (see "Magic properties can be used to store a records and its related properties")

I'm not sure what to think...

Oleg · Answer 8 · 2014-05-26T03:37:15-07:00

Code in the docs are right.

Authors of docs are suppose that user know database and relations.

Usualy you can create relations as Albums.artists_id -> Artists.id , where Albums.artists_id is INTEGER and Artists.id is INTEGER. When you need, you can tie in Albums and Artists as complex Object.

An exemplary sequence of operations :

you request album,
then request artist with id albums.artists_id equal
you insert object Artists to Album object as a new property artist ( $album->artist = $artist )

Then you get object Album, which have property artists as object. And you can get, for example :

album->artist->name

P.S. Magic property - is a big name, of course. In fact - this is a common operation in PHP.

Your main mystake is wrong relations between tables in dabase. Artist must have relation to Album via id, not object.

kofkof01 · Answer 9 · 2014-05-26T04:37:21-07:00

Oleg, sorry but I am not convinced at all. Don't take it personally though, you've dedicated a good deal of your time answering my doubts and I thank you so much for that. But I still don't see why the doc authors would have created this section about something which is not in any way a Phalcon feature ; I can't think of a reason why they would mention a "magic" operation where there is nothing more than plain, dead common, variable assignments ; or why they speak about "Saving the album and the artist at the same time" if in reality you simply can't save them at the same time ; or why they would provide two code examples that just don't work...

If a phalcon dev stops by, I'd appreciate a confirmation of all of this, but meanwhile I'll go back to developping without taking this (hypothetic) feature into account. Thank you again for your time, Oleg!

kofkof01 · Answer 10 · 2014-05-28T02:24:15-07:00

Renskii, you are completely right!

I understand now that the Album model does NOT have any "artist" property and so $album->artist is a magic property, meaning that through PHP's magic methods, it gets to either a relationship with an "Artist" model or one aliased with this name.

Thanks, case closed!

amin-ajami · Answer 11 · 2014-07-16T12:55:49-07:00

you change artist column to artist_id like this :

CREATE TABLE Albums (
  id INT UNSIGNED NOT NULL AUTO_INCREMENT,
  name VARCHAR(45) NOT NULL,
  artist INT UNSIGNED NOT NULL,
  year YEAR NULL,
  PRIMARY KEY (id),
  CONSTRAINT fk_Albums_Artists
    FOREIGN KEY (artist_id)
    REFERENCES Artists (id)
);