Easy if....question

[Eze]

Hi!!!!

I need to create a table of 4 columns x n rows. The number of row will
depend on the number of images for that specific page.

I think that I have to use de mod function saying that if (number of
image mod 4 equal 0 then start another row)

Well....first thing...I find the function fmod in the help that I
think that will be useful.

I wrote this small code, but it does not work

while($row = mysql_fetch_arr ay($respuesta))
{
echo('<td width=15% align="center"> <img src=" ' .$row["url"]. '
"></td>');
$i++;
if (fmod ($i,4)==0)
{
<tr>
}
}

Anyone can help me?

Thanks in advance

Ezequiel

[Jan Pieter Kunst]

Re: Easy if....question

In article <e11a3cdf.04041 70835.4da88b88@ posting.google. com>,
esapoznik@hotma il.com (Eze) wrote:
[color=blue]
> I think that I have to use de mod function saying that if (number of
> image mod 4 equal 0 then start another row)[/color]

It seems simpler to me to just check if the number is equal to 4, and if
it is, reset it to zero again.

$i = 0;
while($row = mysql_fetch_arr ay($respuesta)) {
if ($i < 4) {
do_something();
$i++;
} elseif ($i == 4) {
do_something_el se();
$i = 0;
}
}

JP

--
Sorry, <devnull@cauce. org> is een "spam trap".
E-mail adres is <jpk"at"akamail .com>, waarbij "at" = @.

[Tom Thackrey]

Re: Easy if....question


On 17-Apr-2004, esapoznik@hotma il.com (Eze) wrote:
[color=blue]
> I need to create a table of 4 columns x n rows. The number of row will
> depend on the number of images for that specific page.
>
> I think that I have to use de mod function saying that if (number of
> image mod 4 equal 0 then start another row)
>
> Well....first thing...I find the function fmod in the help that I
> think that will be useful.
>
> I wrote this small code, but it does not work
>
> while($row = mysql_fetch_arr ay($respuesta))
> {
> echo('<td width=15% align="center"> <img src=" ' .$row["url"]. '
> "></td>');
> $i++;
> if (fmod ($i,4)==0)
> {
> <tr>
> }
> }
>
> Anyone can help me?[/color]

You have several problems. The fmod() or % will work, I find counting
easier. Your HTML is wrong (spaces in the src= parm for example). The <tr>
in the if needs an echo or something.

This should be close (untested):

echo '<table>';
$i = 0;
while($row = mysql_fetch_arr ay($respuesta))
{
if ($i==4)
{
echo '</tr><tr>';
$i=0;
}
echo '<td width="15%" align="center"> <img src="' .$row["url"]. '"></td>';
$i++;
}
while($i<4)
{
echo '<td> </td>';
$i++;
}
echo '</tr></table>';


--
Tom Thackrey

https://www.creative-light.com

tom (at) creative (dash) light (dot) com
do NOT send email to jamesbutler@wil lglen.net (it's reserved for spammers)

[Five Cats]

Re: Easy if....question

In message <7gigc.23633$gO 1.8706@newssvr2 7.news.prodigy. com>, Tom
Thackrey <use.signature@ nospam.com> writes[color=blue]
>You have several problems. The fmod() or % will work, I find counting
>easier. Your HTML is wrong (spaces in the src= parm for example). The
><tr> in the if needs an echo or something.
>
>This should be close (untested):
>
>echo '<table>';[/color]

echo '<tr>';
[color=blue]
>$i = 0;
>while($row = mysql_fetch_arr ay($respuesta))
>{
> if ($i==4)
> {
> echo '</tr><tr>';
> $i=0;
> }
> echo '<td width="15%" align="center"> <img src="' .$row["url"].
>'"></td>';
> $i++;
>}
>while($i<4)
>{
> echo '<td> </td>';
> $i++;
>}
>echo '</tr></table>';[/color]

--
Five Cats
Email to: cats_spam at uk2 dot net