2 rows in one cell going down in 2 columns

nem

hey,
i just got this code, altered it as well so it could go well with my site.

What i want is a cell with 2 bits of information which means i will need 2 rows.
Now i want 2 columns that retirieves these 2 bits of information to each cell referring to its ID.

here is my code.

<? include("header.php"); ?>
<?
$columns = 2;

include('connect.php');
$query = "SELECT * FROM members ORDER BY id DESC";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
echo "<TABLE BORDER=\"0\">\n";
for($i = 0; $i < $num_rows; $i++) {
    $row = mysql_fetch_array($result);
    if($i % $columns == 1) {
	   echo "<TR>\n";
    }

echo '<tr><td><img src="' . $row[image]  . '" /></td></tr><tr><td>' . $row['username'] . '</td></tr>';
if(($i % $columns) == ($columns - 1) || ($i + 1) == $num_rows) {
    //if there is a remainder of 1, end the row
    //or if there is nothing left in our result set, end the row
    echo "</TR>\n";
}
}
echo "</TABLE>\n";
?> 
<? include("footer.php");?>

Why is it that it only shows in one column?? I want all my information to show in 2 columns vertically and not 1 column going down in rows

sidney

any use of </tr> will start a new row

two options creat a table within table cell or use <br/> within cell

also $i%2=1
should be

$i%2=0

  echo "<TABLE BORDER=\"0\">\n"; 
for($i = 0; $i < $num_rows; $i++) { 
    $row = mysql_fetch_assoc($result); 
    if($i % $columns == 0) { 
       echo "<TR>\n"; 
    } 

echo '<td><table><tr><td><img src="' . $row['image']  . '" /></td></tr><tr><td>' . $row['username'] . '</td></tr></table></td>'; 
if(($i % $columns) == ($columns - 1) || ($i + 1) == $num_rows) { 
    //if there is a remainder of 1, end the row 
    //or if there is nothing left in our result set, end the row 
    echo "</TR>\n"; 
} 
} 
echo "</TABLE>\n";

nem

im getting a parse error on:

echo echo "<TABLE BORDER=\"0\">\n";

it says expecting '" or '"

tekky

Originally posted by nem
im getting a parse error on:
echo echo "<TABLE BORDER=\"0\">\n";  
it says expecting '" or '" [/B]

its because of the echo (literally speaking)

remove one of the "echo"'s

sidney

Originally posted by nem
im getting a parse error on:
echo echo "<TABLE BORDER=\"0\">\n";  
it says expecting '" or '" [/B]

i see 2 echos in a row remove one

nem

lol i realised that as soon as i sent the post, i was like D'OH.

thanks guys for replying so quickly... but i know now im stuck lol.

Ok i added an "array" i think thats what its called, and thats because i wanted to add a link.

  [code=php]$member = $row['username'] 
echo "<td><table><tr><td><img src=/" . $row['image']  . " /></td></tr><tr><td><a href=/"profile.php?id=$member"/></a></td></tr></table></td>";  [/code]

now, i try to make it work but it aint! 🙁

mrhappiness

echo '<td><table><tr><td><img src="'.$row['image'] .'"  /></td></tr>
<tr><td><a href="profile.php?id='.$member.'">title of link</a></td></tr></table></td>';

sidney

missing a semi colon on first line and no text in <a href...>text</a>

  $member = $row['username'];  

    echo "<td><table><tr><td><img src=\"" . $row['image']  .
	"\" /></td></tr><tr><td><a href=\"profile.php?id=$member\">".
	$member."</a></td></tr></table></td>";

mrhappiness

won't work since you did not escape the " ...