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Home/ Questions/Q 8779865
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T19:54:15+00:00 2026-06-13T19:54:15+00:00

What I’m trying to do is set up a query that calls from a

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What I’m trying to do is set up a query that calls from a Database (the Chinook database in particular – http://chinookdatabase.codeplex.com/)

Now, in order to display the query output correctly, I need to join multiple tables – each that share similar column names. With the Chinook database, there are multiple tables (Artist, Album, Genre, Media Type, Track, etc.). Let’s say I’m trying to display the contents of an Artist’s Name with:

public function fetchArtistName(){
    return $row['Name'];
}

This is very ambiguous since this row would go and find the Name field for Media Type or Album Name before it finds the artist name.

My SQL statement is as follows:

SELECT *  
FROM track AS t  
LEFT JOIN album al 
ON t.AlbumId = al.AlbumId  
LEFT JOIN artist ar ON al.ArtistId = ar.ArtistId  
LEFT JOIN mediatype m ON t.MediaTypeId = m.MediaTypeId 
LIMIT 0,15; 

Is there a way I could go about changing the table that is outputted to be like
| t.Name | al.Name | ar.Name | m.name |

so that I could essentially call these much easier with

public function fetchArtistName(){
    return $row['ar.Name'];
}

It has been so long since I’ve played with SQL and I’m really stuck. Any help would be greatly appreciated, I’ve Googled and search here already.

Thank you.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T19:54:16+00:00Added an answer on June 13, 2026 at 7:54 pm

    I’m not sure that you can do this with SELECT *.... You’d need to do something like:

    SELECT t.Name as track_name,
      al.Name as album_name, ar.Name as artist_name, m.Name as media_name 
    FROM track AS t  
    LEFT JOIN album al 
    ON t.AlbumId = al.AlbumId  
    LEFT JOIN artist ar ON al.ArtistId = ar.ArtistId  
    LEFT JOIN mediatype m ON t.MediaTypeId = m.MediaTypeId 
    LIMIT 0,15; 
    

    …such that your PHP function is defined as

    public function fetchArtistName(){
        return $row['artist_name'];
    }
    
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