contra10 Posted January 2, 2009 Share Posted January 2, 2009 maybe im attempting this the wrong way but i want to put a form in the brackets as im attempting to have an add friend, reject friend. If i put the form seperate then it will only say add friend and reject friend once instead for each person who adds the user my code is <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("registration") or die(mysql_error()); $requestint = 1; $data = ("SELECT * FROM friend_request WHERE requestname = '$username' AND value = '$requestint' ORDER BY frid DESC") or die(mysql_error()); $postrequest = mysql_query($data); while($info = mysql_fetch_array($postrequest)) { $userrequesting= "{$info['username']}"; echo "$userrequesting wants to be your friend <input type='submit' name='add' value='Add Friend'> <input type='submit' name='reject' value='Reject Friend'></a><br>"; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <table border="0" width="500" align="center" height="100"> <tr> <td align="right"><input type="submit" name="add" value="Add Friend"></td> <td align="right"><input type="submit" name="reject" value="Reject Friend"></td></tr> </table> </form> the form in the php brackets or the buttons of the form are the way that i want my request to look like, but i need to do an action and i can't put $_SERVER['PHP_SELF']; tell me if im making any sense Quote Link to comment Share on other sites More sharing options...
GingerRobot Posted January 2, 2009 Share Posted January 2, 2009 1.) You need to place that loop between the <form> tags - otherwise those submit buttons you're echoing wont be part of the form. 2.) You're going to need some way of identifying which friend requests are being accepted/rejected. Perhaps make the submit button's name an array with the id of the friend/friend request as the key. Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 2, 2009 Author Share Posted January 2, 2009 something like this <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <table border="0" width="500" align="center" height="100"> <tr> <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("registration") or die(mysql_error()); $requestint = 1; $data = ("SELECT * FROM friend_request WHERE requestname = '$username' AND value = '$requestint' ORDER BY frid DESC") or die(mysql_error()); $postrequest = mysql_query($data); while($info = mysql_fetch_array($postrequest)) { $userrequesting= "{$info['username']}"; echo "$userrequesting wants to be your friend <input type='submit' name='add' value='Add Friend'> <input type='submit' name='reject' value='Reject Friend'></a><br>"; } ?> </table> </form> ? Quote Link to comment Share on other sites More sharing options...
contra10 Posted January 2, 2009 Author Share Posted January 2, 2009 where exaclty should i set up the array? and what should i input? Quote Link to comment Share on other sites More sharing options...
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