Reply to Re: Problem with table...

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Posted by shagnast on 10/13/27 11:36

Barbara de Zoete wrote:
> On Fri, 06 Jan 2006 16:37:02 +0100, Beauregard T. Shagnasty
> <a.nony.mous@example.invalid> wrote:
>
>> Is this what you mean? <tr> starts a new row; you do not want to
>> put each "column" in its own row.
>
> <prediction> Next question: How do I display the data cells in a
> table underneith one an other, like it was a column (instead next to
> eachother in a row)? </prediction>

You are probably correct. :-0


echo "<table align=\"center\" width=\"100%\">";
// data display
$wynik2=mysql_query($hotel_query2);
while($wiersz2=mysql_fetch_array($wynik2,MYSQL_NUM))
{
echo"<tr><td>Room:</td><td>.return_data($wiersz2[0])."</td></tr>";
echo"<tr><td>Price:</td><td >".return_data($wiersz2[1])."</td></tr>";
echo"<tr><td>How many
rooms?:</td><td>".return_data($wiersz2[2])."</td></tr>";
}
echo "</table>";


This of course means Leszek only ever expects three items to be returned
by the query. The above is not the normal way to display a table of data.

--
-bts
-Warning: I brake for lawn deer

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