Posted by mypetprogrammer on 06/05/07 09:01

On Jun 4, 5:53 pm, Denis Gerina <denisREMOVET...@cced.ba> wrote:
> vinnie wrote:
> > it only displays the name, but the last name and the address seems not
> > to exist! why?
>
> > <?php
> > @ $db=mysql_pconnect('host', 'UserID', 'PWD');
> > if (!$db)
> > {
> > echo 'conneciton eror';
> > exit;
> > }
> > else
> > {
> > echo 'connection on!';
> > }
> > mysql_select_db('list');
> > $query="select * from index";
> > $result=mysql_query($query);
> > $num_results=mysql_num_rows($result);
> > if ($num_results == 0)
> > {
> > echo'<br><br>nothing to dispaly';
> > }
> > else
> > {
> > echo'<br><br>here are the results: '.$num_results;
> > }
> > for ($i<0; $i<$num_results; $i++)
> > {
> > $row=mysql_fetch_array($result);
> > echo '<br>name: ';
> > echo htmlspecialchars(stripslashes($row['name']));
> > echo '<br>last name: ';
> > echo htmlspecialchars(stripslashes($row['last']));
> > }
> > ?>
>
> You don't seem to be printing address anywhere. As for last name, check
> the name of the column in your table (index).
>
> And, to avoid notices, the code
>
> for ($i<0; $i<$num_results; $i++)
>
> should be
>
> for ($i=0; $i<$num_results; $i++)
>
> Although I would probably write something like
>
> while ($row = mysql_fetch_array($result))
> {
> ...
>
> }

See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_array($result)) != null)
{

So as not to be operating on a null on the last iteration of the loop.
Agree with the while, though.

~A!

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