[Resolved] table columns layout

[beginphp04]

hello
i am trying to format a table with four columns and as many rows as there are records. but i get a warning "mysql_num_rows(): supplied argument is not a valid MySQL result resource on line 21"

here is my code so far:

<?php
require_once ('mysql_connect.php');

if (isset($GET['name_id'])) {
$name_id=$GET['name_id'];

$sql = mysql_query("SELECT employee_photo, employee_name FROM employees WHERE name_id = '$name_id'") or die (mysql_error());

$result = mysql_query($sql);
$rows = mysql_num_rows($result);
if($rows > 0) {

$break = 4;
$nowbreak = 4;
$i = 0;

echo "<table>\n";
echo "<tr>\n";

while($row=mysql_fetch_array($results)) {
$i++;
echo "<td><img src=\"images/".$row['employee_photo']."\"></td><br />".$employee_name."\n";

  if($i == $nowbreak) { 
     if($nowbreak < $rows) { 
        $nowbreak += $break; 
        echo "</tr><tr>\n"; 
     } 
  } 

}
echo "</tr></table>";
}
}
?>

many thanks

[ClevaTreva]

Hi

$result = mysql_query($sql); 

should be

$result = mysql_fetch_array($sql); 

😉

Trevor

[beginphp04]

oops!

thank you clevatreva. incorporating the mistake you pointed out i also managed to massively simplify the code too (in case anyone is interested):

$i=0;
$break=4;

while($row=mysql_fetch_array($result)) {
echo "<td><img src=\"images/".$row['employee_photo']."\"><br />".$row['employee_name']."</td>\n";

$i++;
if($i == $break) {
echo "</tr><tr>";
$i=0;
}
}

[leatherback]

$sql = mysql_query("SELECT employee_photo, employee_name FROM employees WHERE name_id = '$name_id'") or die (mysql_error()); 

$result = mysql_query($sql); 
$rows = mysql_num_rows($result); 
if($rows > 0) { 

$break = 4; 
$nowbreak = 4; 
$i = 0; 

echo "<table>\n"; 
echo "<tr>\n"; 

while($row=mysql_fetch_array($results)) { 
$i++; 
echo "<td><img src=\"images/".$row['employee_photo']."\"></td><br />".$employee_name."\n"; 

if($i == $nowbreak) { 
if($nowbreak < $rows) { 
$nowbreak += $break; 
echo "</tr><tr>\n"; 
} 
} 
} 
echo "</tr></table>"; 
} 
} 

reworked to:

$sql = "SELECT employee_photo, employee_name FROM employees WHERE name_id = '$name_id'";

$result = mysql_query($sql)  or die (mysql_error());  

$rows = mysql_num_rows($result); 
if($rows > 0) { 

$nowbreak = 4; 
$i = 0; 

echo "<table>\n"; 
echo "<tr>\n"; 

while($row=mysql_fetch_array($result)) { 
$i++; 
echo "<td><img src=\"images/".$row['employee_photo']."\"><br />".$employee_name."</td>\n"; 

if($i == $nowbreak) { 
echo "</tr><tr>\n"; 
$i=0;
} 
} 
echo "</tr></table>"; 
} 
} 

Hope it helps.

J.

[Weedpacket]

Reworked further to:

$sql = "SELECT employee_photo, employee_name FROM employees WHERE name_id = '$name_id'";

$result = mysql_query($sql)  or die (mysql_error());  

$rows = mysql_num_rows($result); 
if($rows > 0) { 

$nowbreak = 4; 

echo "<table>\n"; 
echo "<tr>\n"; 

for($i=1; $i<=$rows; $i++) {
	$row=mysql_fetch_array($result);
	echo "<td><img src=\"images/" . $row['employee_photo'] . "\"><br />" . $row['employee_name'] . "</td>\n"; 

	if($i%$nowbreak==0) { 
		echo "</tr><tr>\n"; 
	} 
} 
echo "</tr></table>"; 
} 

Saves some reduplication of effort.