mysql_fetch_assoc help

i got this function that create's a dynamic HTML table the outputs SOL query result the code works if i manually enter a SQL query but if i add assign a variable to the query it brings up an error.

Warning: mysql_fetch_assoc(): on line 61

Notice: Undefined variable: array on line 64

PHP Code:

---

<?PHP
include("dbandy.php");

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to mysql');

$dbname = 'db_andrewb';
mysql_select_db($dbname)or die('error finding DB');

$qstring = $_SERVER['QUERY_STRING'];

$qstring = str_replace('queries=', '',$qstring);

echo urldecode($qstring);

$query = $qstring; //if i use this it dosen't work

//$query = "Select * from students"; if i use this it works


$result = mysql_query($query);


?>

<?PHP
function array2table($arr,$width)
   {
   $count = count($arr);
   if($count > 0){
       reset($arr);
       $num = count(current($arr));
       echo "<table align=\"center\" border=\"1\"cellpadding=\"5\" cellspacing=\"0\" width=\"$width\">\n";
       echo "<tr>\n";
       foreach(current($arr) as $key => $value){
           echo "<th>";
           echo $key."&nbsp;";
           echo "</th>\n";    
           }    
       echo "</tr>\n";
       while ($curr_row = current($arr)) {
           echo "<tr>\n";
           $col = 1;
           while ($curr_field = current($curr_row)) {
               echo "<td>";
               echo $curr_field."&nbsp;";
               echo "</td>\n";
               next($curr_row);
               $col++;
               }
           while($col <= $num){
               echo "<td>&nbsp;</td>\n";
               $col++;        
           }
           echo "</tr>\n";
           next($arr);
           }
       echo "</table>\n";
       }
   }

?>
<?php

while($row = mysql_fetch_assoc( $result)){ 
  $array[] = $row; }
       
array2table($array,600); // Will output a table of 600px width

?>

---

can anyone please help?

try adding $array = array(); before the while loop. this might fix the undefined variable

PHP Code:

---

<?php
$array = array();
while($row = mysql_fetch_assoc( $result)){
  $array[] = $row; }
       
array2table($array,600); // Will output a table of 600px width

?>

---

also, since this is all PHP code, you can get rid of those unnecessary <?PHP ?> TAG.

PHP Code:

---

<?PHP
include("dbandy.php");

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to mysql');

$dbname = 'db_andrewb';
mysql_select_db($dbname)or die('error finding DB');

$qstring = $_SERVER['QUERY_STRING'];

$qstring = str_replace('queries=', '',$qstring);

echo urldecode($qstring);

$query = $qstring; //if i use this it dosen't work

//$query = "Select * from students"; if i use this it works


$result = mysql_query($query);

function array2table($arr,$width)
   {
   $count = count($arr);
   if($count > 0){
       reset($arr);
       $num = count(current($arr));
       echo "<table align=\"center\" border=\"1\"cellpadding=\"5\" cellspacing=\"0\" width=\"$width\">\n";
       echo "<tr>\n";
       foreach(current($arr) as $key => $value){
           echo "<th>";
           echo $key."&nbsp;";
           echo "</th>\n";    
           }    
       echo "</tr>\n";
       while ($curr_row = current($arr)) {
           echo "<tr>\n";
           $col = 1;
           while ($curr_field = current($curr_row)) {
               echo "<td>";
               echo $curr_field."&nbsp;";
               echo "</td>\n";
               next($curr_row);
               $col++;
               }
           while($col <= $num){
               echo "<td>&nbsp;</td>\n";
               $col++;        
           }
           echo "</tr>\n";
           next($arr);
           }
       echo "</table>\n";
       }
   }

$array = array();

while($row = mysql_fetch_assoc($result)){
  $array[] = $row; 
}
       
array2table($array,600); // Will output a table of 600px width

?>

---

it won't fix anything, but just clean the code, hope that helped