Barand

If the user chooses the "All" option then you can leave that column and the condition value out of the query

You can't bind column names, only values $pp1 = 'Beginner'; $pp2 = 'Intermediate'; $ll = 7; $room_no = 4; $query = "SELECT md.Member_reg_id, md.fname, md.lname, md.email, md.cell, ms.level, ms.diff, ms.score, r.ID_Status FROM register as r JOIN member_detail as md ON r.ID = md.Member_reg_id JOIN memstatus as ms On r.ID = ms.ID WHERE r.CENTERCODE = ? AND r.ID_Status ='A' AND ms.level IN (?,?) AND ms.diff <= ? ORDER by level, diff, score DESC"; $stmt=$fcon->prepare($query); $stmt->bind_param('issi',$room_no,$pp1,$pp2,$ll);

The string that you are echoing finishes at the ' after location.href= You need to escape the single quotes within the string echo'<input name="Enabled" type="button" value="Enabled" onclick="location.href=\'../edit.php\'"/>';

One connection will suffice, and when you decide on MySQLi or PDO, use a prepared query. And you do not specify which column the value $killforumid should be written to

Of the two I think confusion is the more likely option. You brilliant coding merely takes an array that looks like this $row = array( array('CID'=>100,'Country'=>'France'), array('CID'=>200,'Country'=>'Germany'), array('CID'=>300,'Country'=>'Holland') ); and converts it to an array that looks like this $countries = array( array('0'=>100,'1'=>'France'), array('0'=>200,'1'=>'Germany'), array('0'=>300,'1'=>'Holland') ); when what is really required is an array that looks like this $countries = array( '100'=>'France', '200'=>'Germany', '300'=>'Holland' ); CroNiX already gave the solution in #17, so why are you muddying the water?

I'd test for localhost otherwise remote.

You need to to connect to the server before you can select a database. You also need to send the POST data from the form. Check that data has been posted before trying to use the posted data ( isset() )

All you are doing in the code in insert.php is defining a string. You have to execute the query defined in that string.

echo

Those COUNTs need to be SUMs SELECT SUM(IF(level='P1',1, NULL)) 'P1', SUM(IF(level='P2',1, NULL)) 'P2', SUM(IF(level='P3',1, NULL)) 'P3' FROM faults

Strange, then, that the examples in the manual have the colons http://php.net/manual/en/pdostatement.execute.php (examples #1, #2)

So. There a 3 level 2 users (Tom, Tommy, Kim) so they each receive £4.50/3 (ie £1.50 each) There is 1 level 3 user (Henry) so he receives £13.50.

So given this situation frank --- tommy --- tom jimmy --- kim --- --- henry Frank (2 refers) gets 1.05 Jimmy (1 refer) gets 1.00 Kim (1 refer) gets 1.00 Is that right?

We live and learn. I haven't used that one before.

$a[2014][123456]['EMPS']=114350; $b[2014][60]['EMPS']=1470; $c = $a; foreach ($b as $y => $ydata) { foreach ($ydata as $k => $v) { $c[$y][$k] = $v; } } echo '<pre>',print_r($c, true),'</pre>'; /* RESULT Array ( [2014] => Array ( [123456] => Array ( [EMPS] => 114350 ) [60] => Array ( [EMPS] => 1470 ) ) ) */

Are you sure those are the correct columns for your join?

Yes, I know. But there is no way that I can I help with that one line of code quoted out of context. The code was tested before posting. BTW, the recursive code I just posted is not the same as that posted earlier in this thread.

It's just the same problem in reverse. If you use the query method, reverse the fields in the joins $sql = "SELECT r.id , r.acct_name , r.ref_id , r1.acct_name as first , r2.acct_name as second , r3.acct_name as third , r4.acct_name as fourth FROM test_referral r LEFT JOIN test_referral r1 ON r1.refer_id = r.ref_id LEFT JOIN test_referral r2 ON r2.refer_id = r1.ref_id LEFT JOIN test_referral r3 ON r3.refer_id = r2.ref_id LEFT JOIN test_referral r4 ON r4.refer_id = r3.ref_id WHERE r.acct_name = 'john' ORDER by id"; +----+-----------+--------+-------+---------+-------+--------+ | id | acct_name | ref_id | first | second | third | fourth | +----+-----------+--------+-------+---------+-------+--------+ | 1 | john | J1234 | frank | tommy | NULL | NULL | | 1 | john | J1234 | doe | girly | NULL | NULL | | 1 | john | J1234 | bull | ronaldo | NULL | NULL | | 1 | john | J1234 | jimmy | kim | henry | fred | | 1 | john | J1234 | frank | tom | NULL | NULL | +----+-----------+--------+-------+---------+-------+--------+ Or if you want to go to any depth, use the recursion method $sql = "SELECT id , acct_name , ref_id , refer_id FROM test_referral"; $data = array(); $res = $mysqli->query($sql); while ($row = $res->fetch_assoc()) { $names[$row['ref_id']] = $row['acct_name']; $data[$row['refer_id']][] = $row['ref_id']; } $id = 'J1234'; getRefs($data, $names, $id); // // recursive function to get list of referrees // function getRefs(&$data, &$names, $id, $level=0) { $indent = str_repeat('--- ', $level); echo "$indent {$names[$id]}<br>"; if (isset($data[$id])) { foreach ($data[$id] as $refid) { getRefs($data, $names, $refid, $level+1); } } } /// OUTPUTS ////////////////////// john --- bull --- --- ronaldo --- doe --- --- girly --- frank --- --- tommy --- --- tom --- jimmy --- --- kim --- --- --- henry --- --- --- --- fred

use exit instead of break

These are my results mysql> select * from test_referral; +----+-----------+--------+----------+ | id | acct_name | ref_id | refer_id | +----+-----------+--------+----------+ | 1 | john | J1234 | 0 | | 2 | bull | B3456 | J1234 | | 3 | doe | D5567 | J1234 | | 4 | frank | F7788 | J1234 | | 5 | jimmy | J9990 | J1234 | | 6 | tommy | T6784 | F7788 | | 7 | tom | T9988 | F7788 | | 8 | girly | G8866 | D5567 | | 9 | fred | F0099 | H7654 | | 10 | ronaldo | R7722 | B3456 | | 11 | henry | H7654 | K1234 | | 12 | kim | K1234 | J9990 | +----+-----------+--------+----------+ SELECT r.id , r.acct_name , r.ref_id , r1.acct_name as first , r2.acct_name as second , r3.acct_name as third , r4.acct_name as fourth FROM test_referral r LEFT JOIN test_referral r1 ON r.refer_id = r1.ref_id LEFT JOIN test_referral r2 ON r1.refer_id = r2.ref_id LEFT JOIN test_referral r3 ON r2.refer_id = r3.ref_id LEFT JOIN test_referral r4 ON r3.refer_id = r4.ref_id WHERE r.acct_name = 'fred' ORDER by id; +----+-----------+--------+-------+--------+-------+--------+ | id | acct_name | ref_id | first | second | third | fourth | +----+-----------+--------+-------+--------+-------+--------+ | 9 | fred | F0099 | henry | kim | jimmy | john | +----+-----------+--------+-------+--------+-------+--------+

Add a TIMESTAMP type column and ORDER BY that.

I have run your query against the test_referral table that you provided (after adding adding a couple of extra referrals to give a chain of 4 names) and it works fine. I can only assume the problem is when you run it against your ca_categories table, which I cannot help with.

If you are after the id of the record just inserted into movimiento, use mysql_insert_id() http://php.net/manual/en/function.mysql-insert-id.php

To get the values automatically you close your eyes and make a wish. However, as this seldom works, you have to write the code. Why are you looping 3 times through the same array? $maxshipping1=$maxshipping2=$maxqty=0; foreach ($_SESSION['products'] as $prod) { $maxshipping1 = max($maxshipping1, $prod['shipping1']); $maxshipping2 = max($maxshipping2, $prod['shipping2']); $maxqty = max($maxqty, $prod['qty']); if ($prod['qty'] >= 2) { echo $prod['product'] . '<br>'; echo "Ship1: " . $prod["shipping1"] . '<br>'; echo "Ship2: " . $prod["shipping2"] . '<br><br>'; } }

With a left join, cust_fname and cust_number will be NULL where there is no customer record matching the site record