= 0; $j < $RowsPerCol; $j++)
{
$Row = mysql_fetch_array($Results);
if (isset($Row))
print $Row[0] . "";
}
print "";
}
print "";
-Original Message-
From: Jason Soza [mailto:[EMAIL
mysql_fetch_array($Results);
if (isset($Row))
print $Row[0] . "";
}
print "";
}
print "";
-Original Message-
From: Jason Soza [mailto:[EMAIL PROTECTED]]
Sent: Thursday, July 11, 2002 3:18 PM
To: Lazor, Ed
Cc: [EMA
Kevin
- Original Message -
From: "Jason Soza" <[EMAIL PROTECTED]>
To: "Michael Davey" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Thursday, July 11, 2002 3:43 PM
Subject: Re: [PHP] Re: Table Making
> I really wish I was at home right now so I could test
ific date.
I'm hoping the code I posted will create something like:
1941
1942
1943
...
Make sense?
Jason Soza
- Original Message -
From: "Lazor, Ed" <[EMAIL PROTECTED]>
Date: Thursday, July 11, 2002 2:08 pm
Subject: RE: [PHP] Re
Will this work for you?
$numFields = mysql_num_fields($Results);
print "";
while ($Row = mysql_fetch_array($Results))
{
print "";
for ($i = 0; $ < $numFields; $i++)
print "".$Row[$i]."";
echo ""\n; //close secondary table at
10 cells
}
echo "\n"; //close main row/cell
}
echo "\n"; //close main table
Jason Soza
- Original Message -
From: "Michael Davey" <[EMAIL PROTECTED]>
Date: Thursday, July 11,
How about working out the length of the column (by dividing the number of
rows by the number of cols you want), dump your results into an array and
using the col length as an offset to pick through the resulting table?
// $data is an array of results
$rows = count ($data)
$row_len = round ($rows
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