Re: Ok next php problem

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If i understood right, you want to list a sequence of numbers and follow 
some data like:

1 - data of number 1
2 - data of number 2
...

If it's right, you can do it.

$query="SELECT * FROM db ORDER BY numbers DESC limit 1";

$maxnum=$numbers;

for ($num=1;$num<=$maxnum; $num++) {
    $query="SELECT * FROM db where numbers=".$num;
    echo $num." - ".$data."<br>";
}



""Rob W."" <rob@xxxxxxxxxxxxxxx> escreveu na mensagem 
news:00aa01c69336$4c12dee0$f31f10ac@xxxxxx
I got the previous question answered, Now here's my next problem.

With the numbers displaying correctly again I got:

1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
18
19
20
21
22
25

listed in that order in the database

Now i'm trying to figure out how to write a syntax saying that if like 
number 8 isnt listed, display it.

I've tried doing a

if ($count != $data) {
  echo "$data";
}
$count++;

But when I get to like id number 9 it dont work right because the next 
entrie is displayed as 10 in the db. So that's my problem is to try and 
display only them numbers that are not in there. I have also tried putting 
the numbers in to an array and matching from there but it still come's up as 
the same as above.

- Rob

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