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(); } ?> 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()) Link to comment https://forums.phpfreaks.com/topic/262589-minor-error/#findComment-1345855 Share on other sites More sharing options...
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