Re: Ok next php problem
| From: | Jo�o C�ndido de Souza Neto | Date: | Mon, 19 Jun 2006 01:38:54 +0000 |
| Subject: | Re: Ok next php problem | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-238250@lists.php.net to get a copy of this message | ||
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@fiberuplink.com> escreveu na mensagem
news:00aa01c69336$4c12dee0$f31f10ac@rob...
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