flemingmike Posted September 27, 2011 Share Posted September 27, 2011 hello, im only getting the second half of my results showing in my dropdown box. here is my code. if(isset($_GET['id'])) { $uid = $_GET['id']; } $forward = mysql_query("SELECT * FROM players WHERE status = '1' AND position = 'forward' ORDER BY name"); $forward1=""; while($row1 = mysql_fetch_array($forward)) { $forwardid=$row1["id"]; $forwardname=$row1["name"]; $forwardteam=$row1["team"]; $forward1.="<OPTION TITLE=\"$forwardteam\" VALUE=\"$forwardid\">".$forwardname.'</option>'; } $result = mysql_query("SELECT * FROM picks WHERE uid = '$uid' AND pickid = 'forward1'"); while($row = mysql_fetch_array($result)) { $playerid=$row['playerid']; } if(isset($forwardpick1)) { $result1 = mysql_query("SELECT * FROM players WHERE id = '$playerid'"); while($row1 = mysql_fetch_array($result1)) { $forwardpick1=$row1['name']; } $forwardpick1="<br /><br /><font color ='#000000' size = '5'><b>$forwardpick1"; } else { $forwardpick1=" <form method='POST' action='draftpick.php' name='forward1'> <p><font color='#000000'><select size='1' name='player'> <?php echo '$forward1'; ?> </select><br><br><input type='submit' value='Submit' name='submit'></font></p> </form> "; } any direction is appreciated. Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted September 27, 2011 Share Posted September 27, 2011 Variables aren't interpolated within single quotes. Moreover, you don't need any quotes at all to simply echo a variable. <?php echo '$forward1'; ?> Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.