[SOLVED] how to display results in 2 columns

[pixeltrace]

hi,

 

i need help, i have a page wherein the result listing needs to be displayed in 2 columns.

let say i have a total count of 10, the first 5 will be displayed in the first column and the

remaining 5 will be displayed in the next column.

this is my current code and i cant seem to make it work

 

<?			$uSql = "SELECT * FROM jos_classifieds_categories WHERE parent = '$id' AND published=1 LIMIT 0,5";
		$uResult = mysql_query($uSql, $connection);
		if(!$uResult){
		echo 'no data found';
		}else{
		while($uRow = mysql_fetch_array($uResult)){
		$name = $uRow["name"];
		$subid = $uRow["id"];
echo "
<table width='600' border='0' cellspacing='3' cellpadding='0'>
  <tr>
    <td><a href='test.php?id=$id&subid=$subid&view=subview'>$name</a></td>
    <td><a href='test.php?id=$id&subid=$subid&view=subview'>$name</a></td>
  </tr>
</table>

";
}
}
?>

 

hope you could help me with this.

 

thanks!

[pixeltrace]

Hi,

 

Any help on this please?

 

thanks!

[pixeltrace]

hi,

 

i found a code that displays result in 2 columns but i still cant make it work

 

normally what i wanted to happen is if i have 10 results

the display will look like this

 

result1      result6

result2      result7

result3      result8

result4      result9

result5      result10

 

and currently this is the code that i have

 

<?php

$columns = 2;

$sql = "SELECT * FROM jos_classifieds_providers WHERE published=1";     


   
$result = mysql_query($sql) or die("Couldn't execute query."); 

$num_rows = mysql_num_rows($result); 

$rows = ceil($num_rows / $columns); 

while($row = mysql_fetch_array($result)) {  
$data[] = $row['name'];  
$data1[] = $row['address'];  
$data2[] = $row['postalcode']; 

} 

echo "<table border='0'>n"; 

for($i = 0; $i < $rows; $i++) {
echo "<tr>n";        

for($j = 0; $j < $columns; $j++) {    
if(isset($data[$i + ($j * $rows)])) {   
  echo "<td><img src="" . $data[$i + ($j * $rows)] . ""></td>n";        
  echo "<td valign="top">" . $data1[$i + ($j * $rows)] . "<br>" . $data2[$i + ($j * $rows)] . "</td>n";
      }
        } 
  echo "</tr>n";
  }    
  echo "</table>n"; 
  ?> 

 

 

but i am getting an error

Parse error: parse error, unexpected '\"', expecting ',' or ';' in /var/www/html/myoochi/findit/test2.php on line 29

 

and this is my testpage link

http://mango.resonance.com.sg/myoochi/findit/test2.php

 

 

 

hope you could help me with this.

 

thanks!

 

[sasa]

try

<?php

$columns = 2;

$sql = "SELECT * FROM jos_classifieds_providers WHERE published=1";     


   
$result = mysql_query($sql) or die("Couldn't execute query."); 

$num_rows = mysql_num_rows($result); 

$rows = ceil($num_rows / $columns); 

while($row = mysql_fetch_array($result)) {  
$data[] = $row['name'];  
$data1[] = $row['address'];  
$data2[] = $row['postalcode']; 

} 

echo "<table border='0'>\n"; 

for($i = 0; $i < $rows; $i++) {
echo "<tr>\n";        

for($j = 0; $j < $columns; $j++) {    
if(isset($data[$i + ($j * $rows)])) {   
  echo "<td><img src=\"" . $data[$i + ($j * $rows)] . "\"></td>\n";        
  echo "<td valign=\"top\">" . $data1[$i + ($j * $rows)] . "<br>" . $data2[$i + ($j * $rows)] . "</td>\n";
      }
        } 
  echo "</tr>\n";
  }    
  echo "</table>\n"; 
  ?> 

[pixeltrace]

thanks its working now!