lostjoe Posted April 14, 2013 Share Posted April 14, 2013 Im trying to display the songs for an artist i select on a drop down menu but i just cant get it to work, ive been trying for hours but i have got nowhere. This is my dropbox code, it debugs fine and i can click on my artists and submit fine but when the browser moves onto my process3.php it does'nt show the artist with their data (title, artist, genre, year, price) <html> <body> <form action="process3.php" method="post"> <p>View grocery current quantity in stock? <select name="band"> <option value=""> -- Please select band to view-- </option> <?php $con = mysql_connect("localhost","root"); mysql_select_db("cd_online", $con); $query = "SELECT id, artist FROM songs"; $result=mysql_query($query); while ($row=mysql_fetch_array($result)) { echo "<option value =\"{$row['id']}\"> {$row['artist']} </option>\n"; } mysql_close($con); ?> </select> <br/> <input type="submit" name="submit" value="Submit" /> </form> </html> The sample code i've been given is below i'll give you this version as when i change it it doesn't work (it does'nt debug as it is they give it as an outline but im completly lost) <?php $selectedfruit = $_POST['fruit']; print "$selectedfruit"; $con = mysql_connect("localhost","root"); mysql_select_db("grocery_db", $con); $query = "SELECT `curr_qty` FROM grocery_inventory WHERE item_name =\"$selectedfruit\""; print "$query"; $result = mysql_query($query, $con); while($myrow = mysql_fetch_array($result)) {print $myrow["curr_qty"]; } mysql_close($con); ?> Quote Link to comment Share on other sites More sharing options...
jcbones Posted April 14, 2013 Share Posted April 14, 2013 I will give you a starting point. <?php if($_SERVER['REQUEST_METHOD'] == 'POST') { echo $_POST['band']; } else { echo 'Form not sent!'; } Quote Link to comment Share on other sites More sharing options...
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