How to return multiple results in a php function

Question

I have the following function that aims to fetch all credits from an artist for a particular song using the ID from the url ($id), and a foreach statement that displays the information on the Web page. At the moment it's displaying the artist names fine, but the IDs aren't being displayed. How would I go about returning the ID information so it's displayed as well?

function getArtistsBySongId($id)
{
$query = "SELECT * FROM `Credit_To_Artist` AS c2a 
INNER JOIN `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN `Artist` AS a ON a.artist_id = c2a.artist_id
LEFT OUTER JOIN `Song` AS s ON s.song_id = c2a.song_id
LEFT OUTER JOIN `Project` AS p ON p.project_id = s.project_id
WHERE c2a.song_id = $id";

$res = mysql_query($query);

$artists = Array();
$artisttoid = Array();
$songtoid = Array();

while( $row = mysql_fetch_array($res) ) {
    $artist = $row[artist_name];
    $credit = $row[credit_name];
    $songcr = $row[song_id];

    if(!array_key_exists($artist, $artists) ) {
        $artists[$artist] = Array();
        $artisttoid[$artist] = $row[artist_id];
        $songtoid[$songcr] = $row[song_id];
    }

    $artists[$artist][] = $credit;

}

return $artists;
return $songtoid;
return $artisttoid;

}

I've used include's in the code because I'm still green to PHP and find it easier to understand.

<table border="0" cellspacing="5" cellpadding="5" class="cdinfo" width="100%;">
    <tr>
        <?php
        if (getArtistsBySongId($id) == NULL) {
            echo "<th style='font-size: 13px'>Credits:</th>";
            echo "<td style='font-size: 13px'>There are currently no artists linked to this song.</td>";
        } else {
            include 'songs/getsongcredits.php';
        }
        ?>
    </tr>
</table>

songs/getsongcredits.php

<?php foreach (getArtistsBySongId($id) as $artist => $creditarr) {
        $credits = implode( ", ", $creditarr );
        echo "<a href='star.php?id={$artisttoid[$artist]}'>{$artist}</a> ({$credits})<br />";
} ?>

If you can, you should stop using mysql_* functions. They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and consider using PDO, it's really not hard. — Jay Blanchard
– Jay Blanchard, Commented Jun 4, 2015 at 20:31
You should put quotes around the string indexes of your arrays, e.g. $row['artist_name']. Don't depend on PHP automatically quoting unrecognized constants. — Barmar
– Barmar, Commented Jun 4, 2015 at 20:35

Machavity · Accepted Answer · 2015-06-04 20:32:25Z

4

Objects or arrays are the way to do it

return array('artists' => $artists, 'songtoid' => $songtoid, 'artisttoid' => $artisttoid);

answered Jun 4, 2015 at 20:32

Machavity♦

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7 Comments

Mitch Over a year ago

Thanks for your reply @machavity. I tried this solution in my code but the output has gone all weird. For example, a song (song_id 455) with artist (artist_id 18) returns the result: 0 (Array) 1 (455) 2 (18) Do you know what's gone wrong?

Machavity Over a year ago

Try my code above and do a var_dump() of the returned value and see if that works

Machavity Over a year ago

All the array does is hold your data for the return. You still have to take it and output it correctly. Like I said, var_dump() it and then output it

Mitch Over a year ago

Is it possible to ask for an example of using the var_dump() and outputting it? I had a look on PHP.net but am unsure where exactly to place it. Sorry to be a pain. :/

Machavity Over a year ago

var_dump(getArtistsBySongId($id)); It does its own output

|

Francisco Pacheco · Accepted Answer · 2015-06-05 13:24:03Z

1

I recomend you to return an object with each item you want in attributes. To return an object, firstly you need a class:

class MyClass {
    private $artists;
    private $songtoid;
    private $artisttoid;

    public function __construct($arg1, $arg2, $arg3){
        $this->artists = $arg1;
        $this->songtoid = $arg2;
        $this->artisttoid = $arg3;
    }

    public function getArtists(){return $this->artists;}
    public function getSongtoid(){return $this->songtoid;}
    public function getArtisttoid(){return $this->artisttoid;}
}

In your function

function getArtistsBySongId(){
    ...
    return new MyClass($artists, $songtoid, $artisttoid);
}

Also you can return an associative array like this

return array(
    "artists"=>$artists,
    "songtoid"=>$songtoid,
    "artisttoid"=>$artisttoid
);

Or, if you want, you can return an array (as Machavity answered) and read the result using list()

function getArtistsBySongId(){
    ...
    return array($artists, $songtoid, $artisttoid);
}

list($artists, $songtoid, $artisttoid) = getArtistsBySongId();

edited Jun 5, 2015 at 13:24

answered Jun 4, 2015 at 20:38

Francisco Pacheco

266 bronze badges

3 Comments

Mitch Over a year ago

Thanks for you reply @Francisco Pacheco. However, the above array return example doesn't seem to work in my code, would you be able to give an example of how to return an object?

Francisco Pacheco Over a year ago

To return an object, firstly you need a class: class MyClass { public $artists; public $songtoid; public $artisttoid; public function __construct($arg1, $arg2, $arg3){ $this->artists = $arg1; $this->songtoid = $arg2; $this->artisttoid = $arg3; } After this, you can return the object: return new MyClass($artists, $songtoid, $artisttoid);

Francisco Pacheco Over a year ago

I editted the answer, now it has the code to return an object

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