Thanks for your help. As i said im quite new to php so all your help in
simplifying my code has been appeciated.
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> Is there a mysql or array variable for the current row of a query
array?
> I want to alternate background colours for each row of the query
output so
> i
> need to know the current row number.
There is a way with SQL variables, but you're better off just letting
PHP keep track of it.
---John H
[mailto:[EMAIL PROTECTED]]
> Sent: Thursday, September 26, 2002 11:42 AM
> To: 'Patrick Lebon'; [EMAIL PROTECTED]
> Subject: RE: [PHP-DB] Current row of query
>
> > -Original Message-
> > From: Patrick Lebon [mailto:[EMAIL PROTECTED]]
> > Sent: 26 Septembe
> -Original Message-
> From: Patrick Lebon [mailto:[EMAIL PROTECTED]]
> Sent: 26 September 2002 16:20
>
> This is how im currently doing it...
>
> $rowNum = 0;
> while ( $row = mysql_fetch_array($result) )
> {
> $rowNum++;
> if ($rowNum % 2) { $bgCol = "#EADBC6"; } else { $bgCol = "
Last I checked there was no way to retrieve the current index into the result
array, you can set the index manually with mysql_seek_data(), but I don't think
that is what you want, so the best solution IMHO, is to do what you are doing
and manually keep track of the index.
-Brad
Patrick Lebon wro
I want it to.
>
> > -Original Message-
> > From: Brad Bonkoski [mailto:[EMAIL PROTECTED]]
> > Sent: Thursday, September 26, 2002 11:10 AM
> > To: Patrick Lebon; [EMAIL PROTECTED]
> > Subject: Re: [PHP-DB] Current row of query
> >
> >
> >
Thanks, im currently doing something similar to this but I was wondering if
there was an already defined variable that i could use rather then have to
create a value and increment it myself. I guess am being picky, but was just
curious as im new to php.
This is how im currently doing it...
$row
rying to
figure out why the code works, but not the way I want it to.
> -Original Message-
> From: Brad Bonkoski [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, September 26, 2002 11:10 AM
> To: Patrick Lebon; [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Current row of query
>
&g
why not
if (!($1%2))
{
echo "even!";
}
else
{
echo "odd!";
}
Adam
On Thu, 26 Sep 2002, Brad Bonkoski wrote:
> oops...
>
> I meant:
> (if $i is odd)
> ...
>
> because $i would be the current index in the array.
>
>
> Brad Bonkoski wrote:
>
> > why not this?
> >
> > $result = mys
oops...
I meant:
(if $i is odd)
...
because $i would be the current index in the array.
Brad Bonkoski wrote:
> why not this?
>
> $result = mysql_query("Select * from table');
> $num_rows = mysql_num_rows($rows);
>
> for ($i=0; $i<$num_rows, $i++)
> {
> $row = mysql_fetch_array($result);
>
why not this?
$result = mysql_query("Select * from table');
$num_rows = mysql_num_rows($rows);
for ($i=0; $i<$num_rows, $i++)
{
$row = mysql_fetch_array($result);
if ($num_rows is odd)
{
print $row with BG color of red;
}
else
{
printf $row with BG color o
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