Glenskie Posted May 16, 2012 Share Posted May 16, 2012 ok , this is to request someone to be your friend and i keep getting an error saying "mysql_error() expects parameter 1 to be resource" i dont know what is wrong , it worked before ... please help. <?php if ($_POST["request"] == "requestFriendship") { $mem1 = preg_replace('#[^0-9]#i', '', $_POST['mem1']); $mem2 = preg_replace('#[^0-9]#i', '', $_POST['mem2']); // if (!$mem1 || !$mem2 || !$thisWipit) { echo 'Error: Missing data'; exit(); } // if ($mem1 == $mem2) { echo 'Error: You cannot add yourself as a friend'; exit(); } //See if already friends $sql1 = "SELECT id FROM friends WHERE mem2='$mem1' AND mem1='$mem2' LIMIT 1"; $result1 = mysql_query($sql1) or die (mysql_error("sql1 failed")); $data1 = mysql_fetch_row($result1); if ($data1[0]) { echo 'This member is already your Friend'; exit(); } //Check if user already requested to be friends $sql2 = "SELECT id FROM friends_requests WHERE mem1='$mem1' AND mem2='$mem2' Limit 1"; $result2 = mysql_query($sql2) or die (mysql_error("sql2 failed")); $data2 = mysql_fetch_row($result2); if ($data2[0]) { echo '<img src="images/round_error.png" width="20" height="20" alt="Error" /> You have a Friend request pending for this member. Please be patient.'; exit(); } //Check if friend already requested to be friends $sql3 = "SELECT id FROM friends_requests WHERE mem1='$mem2' AND mem2='$mem1' Limit 1"; $result3 = mysql_query($sql3) or die (mysql_error("sql3 failed")); $data3 = mysql_fetch_row($result3); if ($data3[0]) { echo '<img src="images/round_error.png" width="20" height="20" alt="Error" /> This user has requested you as a Friend already! Check your Requests on your profile.'; exit(); } //If still here make insert $mem1=mysql_real_escape_string($mem1); $mem2=mysql_real_escape_string($mem2); $sql = mysql_query("INSERT INTO friends (mem1, mem2, timedate) VALUES('$mem1','$mem2',now())") or die (mysql_error("Friend Request Insertion Error")); echo '<img src="images/round_success.png" width="20" height="20" alt="Success" /> Friend request sent successfully. This member must approve the request.'; exit(); } ?> Quote Link to comment https://forums.phpfreaks.com/topic/262589-minor-error/ Share on other sites More sharing options...
requinix Posted May 16, 2012 Share Posted May 16, 2012 mysql_error("sql1 failed") That's not how you use it. die("sql1 failed: " . mysql_error()) Quote Link to comment https://forums.phpfreaks.com/topic/262589-minor-error/#findComment-1345855 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.