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Displaying MySql Results table side by side


richiejones24
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need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. 

 


require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");


print "<table width=\"293\" height=\"111\" border=\"0\">
  <tr>\n";

while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['star'] . "</td>\n";
print "  </tr>\n";
print "  <tr>\n";
print "<td>" . $row['username_search'] . "</td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['phone_search'] . "</td>\n";
print "  </tr>   \n";

  }
print "</table>";

mysql_close($con);
?>


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Try this:

<?php
require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");

print "<table width=\"293\" height=\"111\" border=\"0\">
	<tr>";
$right = false;
while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /><br />";
	. $row['star'] . "<br />"
	. $row['username_search'] . "<br />";
	. $row['phone_search'] . "</td>";
if($right)
{
	print "</tr><tr>";
}
$right=!$right;
  }
print "</tr></table>";

mysql_close($con);
?>

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