Not sure if this is relevant anymore, but...
> i.e. "1223123" into ['1','2,','2','3','1,'2','3'] ?
$num = "1223123";
$nums = array_filter(preg_split('//', $nums));
Or you can use this function.
It's probably better since the array_filter will probably get rid of any 0's
in the string.
function string_to_array($str) {
$arr = preg_split('//', $str);
array_shift($arr); // removing leading null from split
array_pop($arr); // remove training null from split
return $arr;
}
Or, if you just want the numbers from your student id (and you wanted only
numbers):
$num = "A123456789";
$nums = string_to_array(preg_replace("/[^0-9]+/", "", $nums));
- Jon L.
On Wed, Mar 26, 2008 at 1:35 PM, Richard Dunne <richarddunne1971@xxxxx>
wrote:
> Using this extract from
> http://ie.php.net/manual/en/control-structures.foreach.php
> Amaroq
> 09-Mar-2008 06:40
> Even if an array has only one value, it is still an array and foreach will
> run it.
>
> <?php
> $arr[] = "I'm an array.";
>
> if(is_array($arr))
> {
> foreach($arr as $val)
> {
> echo $val;
> }
> }
> ?>
>
> The above code outputs:
> I'm an array.
> -------------------------------------
> So if I use:
>
> $query = "Select answer from answers where studentID='A123456789'";
> $result = mysql_query($query,$connection) or die(mysql_error());
> while($row = mysql_fetch_assoc($result))
> {
> foreach($row as $answer)
> {
> echo $answer . "\n";
> }
> }
> I thought I would be able to print out each array element, but I am not
> getting any output. Has anyone got a better idea?
>
>
> ---------- Forwarded message ----------
> From: Evert Lammerts <evert.lammerts@xxxxxxxxx>
> To: Richard Dunne <richarddunne1971@xxxxx>
> Date: Wed, 26 Mar 2008 18:22:53 +0100
> Subject: Re: numeric string to single digit array
>
> > Tried that as well and got the same result.
> > I tried "Select count(answer) as total from answers where
> studentID='A123456789'";
> > from the CLI and got total = 2 as a result.
> >
>
> So, we got rid of the Invalid Resource error and we know that the
> student id you use occurs in both rows in your table and that your query
> works fine.
>
> Did you get rid of the semicolon @ line 15 "while($row =
> mysql_fetch_assoc($result));", as Jason suggested? Also, an:
> error_reporting(E_ALL);
> at the top of your code might help in backtracing.
>
>
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