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[SOLVED] Needs help Urgent Please!

studgate

By studgate
in PHP Coding Help

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Barand Prolific Member

Barand

Posted
Posted

If they are on the same page I would use a form with radio buttons for the user to select the report required rather than links.

 

<html>
<head>
<meta name="generator" content="PhpED Version 4.5 (Build 4513)">
<title>Sample reports code</title>
<meta name="author" content="Barand">

</head>
<body>
<form method="get">
<h3>Choose report type </h3>
<input type="radio" name="name" value="Artist"> by Artist<br>
<input type="radio" name="name" value="Genre"> by Genre<br>
<input type="radio" name="name" value="Year"> by Year<br>
<input type="radio" name="name" value="Type"> by Type<br> <br>
<input type="submit" name="action" value="Show report">
</form>
<?php
include 'db.php'; // database connection code

if (isset ($_GET['name']))   {
        switch ($_GET['name']) {
            case 'Artist' :
                $sql = "SELECT a.artist, COUNT(*) as num
                        FROM music m INNER JOIN artist a
                        ON m.artist_id = a.id
                        GROUP BY a.artist
                        ORDER BY num DESC";
                break;
                
            case 'Genre' :
                $sql = "SELECT g.genre, COUNT(*) as num
                        FROM music m INNER JOIN genre g
                        ON m.genre_id = g.id
                        GROUP BY g.genre
                        ORDER BY num DESC";
                break;
                
            case 'Type' :
                $sql = "SELECT t.type, COUNT(*) as num
                        FROM music m INNER JOIN musictype t
                        ON m.type_id = t.id
                        GROUP BY t.type
                        ORDER BY num DESC";
                break;
                
            case 'Year' :
                $sql = "SELECT m.year, COUNT(*) as num
                        FROM music m 
                        GROUP BY m.year
                        ORDER BY num DESC";
                break;
        }

        $res = mysql_query ($sql);
        echo '<table border="1" cellpadding="4">';
        while (list($descrip, $num) = mysql_fetch_row($res)) {
             echo "<tr><td>$descrip</td><td>$num</td></tr>";
        }
        echo '</table>';
}
?>
</body>
</html>

studgate Member

studgate

Posted
  • Author
Posted

 

CREATE TABLE music (
    id int not null auto_increment primary key,
    song_title varchar(100),
    artist_id int,
    genre_id int,
    year int,
    type_id int
)

Hi Barand, artist, genre, & type cannot be integers, I just realized that.

& also my form add the data to the music database & then insert the

artist, genre, year, & type in their respectives database, should I consider

re-writing the form.

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