Perl to PHP

iceraider

I just came out of perl, and having a very hard time translating this line.

my @ref = ();
$username = $mem_info[2];
my @names = $username;
my $base_sql=q{SELECT username,status FROM members WHERE referral };

while (scalar(@names)) {
 my $sql=$base_sql.($#names?'in ('.join(',',map('?',@names)).')':'=?');
 my $sth=$dbh->prepare($sql);
 my $rv=$sth->execute(@names);
 my ($res,$status) = $sth->fetchall_arrayref();
 @names=map($res->[$_][0],0..$#$res);
 push @ref, join '|',@names if (scalar(@names));
 last if $#ref == 2;
}

If you dont know perl, Ill tell you what I'm trying to do here.

In my database I have so that rows are like this

Username|Referral

penti|apple

So thats in my database, apple was referred by no one, scott, pete, penti was referred by apple. Tim was referred by pete. Im trying to get my results like this.

Level 1: apple
Level 2: scot|pete|penti
Level 3: tim|billyjoe

To get that in array I need it saved like this

$array = array('apple','scot|pete|penti','tim|billyjoe');

Is that a little better?

iceraider

this is a hard ?! :eek:

iceraider

You know a site with perl and php in it, that I would get more results from?

diego25

I don't get your example...

tim and billyjoe were referred by different people, so why are they in the same level? And what does each level mean?

Diego

iceraider

Its like a tree

 
Level 1:                 apple
                       /   |  \
Level 2:           scot|pete|penti
                   /       |
Level 3:       bilyjoe   tim

That bettter?

diego25

You can do that recursively:

get all root nodes (i.e. ppl not referred by anyone)
get all child nodes (i.e. ppl referred by root node
get nodes referred by child nodes
keep doing that recursively.

And to get it into the array, you will do something like
$array[] = implode("|", $nodes)

Diego

iceraider

In my admin panel, I want to able to pick when I want it to stop. Like 4 levels. I need to do it in a loop somehow. There has to be a way. In my member section, they click on referrals, and it shows them under there username. So i would assign them to the root.

diego25

Then make sure the recursion doesn't pass the 4th recursion call...Maybe pass an argument of the total level (4), and the current level...

Diego

iceraider

Hmm, im going to have troble doing this!

🙁

Ill give you $5.00 PayPal or E-Gold if you do it. LOL :p

diego25

Which part are you having trouble with?

Diego

iceraider

I know perl real well just not PHP, I barely know were to start.

diego25

Well, since you know Perl, and PHP being similar to Perl, I assume you already know for loops, while loops and how to assign arrays.

So, the function you need will be the MySQL functions:
http://www.php.net/manual/en/ref.mysql.php
More specifically, mysql_connect, mysql_select_db, and mysql_query.

Then you're going to need implode(), which is PHP's version of join().

Also, to assign to the end of an array, you can just do:
$array[] = "something";
i.e. if you leave out the array index, PHP will add "something" as the last element of the array (aka push into the array in Perl term)

Diego

iceraider

Thank you, for that. I know all the MySQL don't worry about that 🙂. Ill see what I can do.

See but won't impload make new array elements when I only need like 1 for each level.

$ar[0] = //Level 1
$ar[1] = //Level 2
$ar[2] = //Level 3

diego25

Yeah, but for each level, you're doing

push @ref, join '|',@names if (scalar(@names));

So in PHP, the join part, would be

implode('|', $names);

Diego

iceraider

So if I had

array('a','b','c','d');

would implode do this?

array('a','|','b','|','c','|','d');

diego25

No, it'd do "a|b|c|d".

Diego

iceraider

SWEET!

And join just does the exact same thing

diego25

Yeah, that's why I said

Then you're going to need implode(), which is PHP's version of join().

a few posts ago...

Diego

iceraider

  $username = $mem_info[2];
  $result = mysql_query("SELECT username,status FROM members WHERE referral='$username'");
  $strong = array();
  $strong2 = array();
  $strong3 = array();

  while ($a = mysql_fetch_row($result)) {
        $result2 = mysql_query("SELECT username,status FROM members WHERE referral='$a[0]'");
        array_push($strong, $a[0]);

    while ($b = mysql_fetch_row($result2)) {
 		$result3 = mysql_query("SELECT username,status FROM members WHERE referral='$b[0]'");
            array_push($strong2, $b[0]);

		while ($c = mysql_fetch_row($result3)) {
		array_push($strong3, $c[0]);
 		}
    }
  }

  $ref[] = implode('|', $strong);
  $ref[] = implode('|', $strong2);
  $ref[] = implode('|', $strong3);

I did that, but I need a way to do it so I can say $level = 3; Then it will loop through a loop three times or something

diego25

You should be able to combine a for loop with that while loop. There's only need for 1 while loop if you add the for loop:

for ($i = 0; $i<3; $i++) {
  while ($row = mysql_fetch_array($result)) {
    // build array with this level's referrals
  }
   // implode array and push to another array
   // then build SQL for next level
}

You'll have to change your SQL statement to the one similar in Perl.

Diego