Glenskie Posted February 23, 2012 Share Posted February 23, 2012 ok , i made a friends thing and its to hard to work with so i need to change it to where it has it own table ... i had it in an array in a table with other things , and now its hard to get it from the array so i want to make it simpler ... so here is my code and what i want to do , i need help and thank you in advance here is my part 1 of my adding people as friends. <?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(); } $sql_frnd_arry_mem1 = mysql_query("SELECT friend_array FROM myMembers WHERE id='$mem1' LIMIT 1"); while($row=mysql_fetch_array($sql_frnd_arry_mem1)) { $frnd_arry_mem1 = $row["friend_array"]; } $frndArryMem1 = explode(",", $frnd_arry_mem1); if (in_array($mem2, $frndArryMem1)) { echo 'This member is already your Friend'; exit(); } $sql = mysql_query("SELECT id FROM friends_requests WHERE mem1='$mem1' AND mem2='$mem2' Limit 1"); $numRows = mysql_num_rows($sql); if ($numRows > 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(); } $sql = mysql_query("SELECT id FROM friends_requests WHERE mem1='$mem2' AND mem2='$mem1' Limit 1"); $numRows = mysql_num_rows($sql); if ($numRows > 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(); } $sql = mysql_query("INSERT INTO friends_requests (mem1, mem2, timedate) VALUES('$mem1','$mem2',now())") or die (mysql_error("Friend Request Insertion Error")); //$sql = mysql_query("INSERT INTO pms (to, from, time, sub, msg) VALUES('$mem2','XXXXX',now(),'New Friend Request','You have a new Friend Request waiting for approval.<br /><br />Navigate to your profile and check your friend requests. <br /><br />Thank you.')") or die (mysql_error("Friend Request PM Insertion Error")); //$id = mysql_insert_id(); echo '<img src="images/round_success.png" width="20" height="20" alt="Success" /> Friend request sent successfully. This member must approve the request.'; exit(); } ?> Now i need to change my code to where it inserts all this info and stuff not as an array and , in another table which is called "friends" my table is set up like this -- Table structure for table `friends` CREATE TABLE IF NOT EXISTS `friends` ( `id` int(11) NOT NULL AUTO_INCREMENT, `mem_1` int(11) NOT NULL, `mem_2` int(11) NOT NULL, `date` int(11) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; So please if im not clear enough i will post more details Link to comment https://forums.phpfreaks.com/topic/257634-need-to-change-my-code-but-dont-know-where-to-start/ Share on other sites More sharing options...
Glenskie Posted February 23, 2012 Author Share Posted February 23, 2012 Please! Link to comment https://forums.phpfreaks.com/topic/257634-need-to-change-my-code-but-dont-know-where-to-start/#findComment-1320491 Share on other sites More sharing options...
Drummin Posted February 23, 2012 Share Posted February 23, 2012 No promises. Not tested. <?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 $sql2 = "SELECT id FROM friends WHERE mem1='$mem1' AND mem2='$mem2' Limit 1"; $result2 = mysql_query($sql2) or die (mysql_error("sql2 failed")); $2data2 = mysql_fetch_row($result2); if ($2data2[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 $sql3 = "SELECT id FROM friends WHERE mem1='$mem2' AND mem2='$mem1' Limit 1"; $result3 = mysql_query($sql3) or die (mysql_error("sql3 failed")); $2data3 = mysql_fetch_row($result3); if ($2data3[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/257634-need-to-change-my-code-but-dont-know-where-to-start/#findComment-1320514 Share on other sites More sharing options...
Glenskie Posted February 24, 2012 Author Share Posted February 24, 2012 thanks ill try it out Link to comment https://forums.phpfreaks.com/topic/257634-need-to-change-my-code-but-dont-know-where-to-start/#findComment-1320710 Share on other sites More sharing options...
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