[SOLVED] need help spotting the error

[webguync]

Hi,

 

I have an error somewhere in my code, which produces a blank white page. I don't have access to the .ini file to change the error reporting, so was hoping someone could spot it. It might be obvious and I am just not seeing it.

 

<?php
$con = mysql_connect("localhost","username","pw");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("DBName", $con);


$result = mysql_query("SELECT * 
		FROM tablename ORDER BY ORDER BY employee_id, date_created ASC");
$count = 0;
$CallNumber = 1;

while($row = mysql_fetch_array($result))
  {

$currentID = $row['employee_id'];

    if($count == 0 || $currentID != $previousID)
        $CallNumber = 1;
    else
        $CallNumber++;

  echo "<tr>";
  echo "<td>" . $row['score_id'] . "</td>";
  echo "<td>" . $row['employee_id'] . "</td>";
  echo "<td>" . $row['employee_name'] . "</td>";
  echo "<td>" . $row['score1'] . "</td>";
  echo "<td>" . $row['score2'] . "</td>";
  echo "<td>" . $row['score3'] . "</td>";
  echo "<td>" . $row['score4'] . "</td>";
  echo "<td>" . $row['score5'] . "</td>";
  echo "<td>" . $row['score6'] . "</td>";
  echo "<td>" . $row['assessor_name'] . "</td>";
  echo "<td>" . $row['assessor_id'] . "</td>";
  echo "<td>" . $CallNumber . "</td>";
  echo "<td>" . $row['date_created'] . "</td>";
  echo "<td>" . $row['date_uploaded'] . "</td>";
  echo "</tr>";

$previousID = $currentID;
    
    $count++;

  }

mysql_close($con);

?>

[taquitosensei]

I think you forgot brackets here

    if($count == 0 || $currentID != $previousID)
        $CallNumber = 1;
    else
        $CallNumber++;

should be

    if($count == 0 || $currentID != $previousID) {
        $CallNumber = 1;
}
    else {

}

plus you can change error reporting by adding this

ini_set("display_errors","1");
ERROR_REPORTING(E_ALL);

to the top of your page

 

        $CallNumber++;

[webguync]

thanks, I added the brackets, and still get an error. With the error reporting, I now know what the error is:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/etsi-dataservices.com/httpdocs/NNPrinceton/Jan09/report/Individual_Scores.php on line 27

 

my current code is:

 

<?php
ini_set("display_errors","1");
ERROR_REPORTING(E_ALL);
$con = mysql_connect("localhost","username","pw");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("DBName", $con);


$result = mysql_query("SELECT * 
		FROM TableName ORDER BY ORDER BY employee_id, date_created ASC");
$count = 0;
$CallNumber = 1;

while($row = mysql_fetch_array($result))
  {

$currentID = $row['employee_id'];

    if($count == 0 || $currentID != $previousID) {
        $CallNumber = 1; }
    else {
        $CallNumber++;
	}

  echo "<tr>";
  echo "<td>" . $row['score_id'] . "</td>";
  echo "<td>" . $row['employee_id'] . "</td>";
  echo "<td>" . $row['employee_name'] . "</td>";
  echo "<td>" . $row['score1'] . "</td>";
  echo "<td>" . $row['score2'] . "</td>";
  echo "<td>" . $row['score3'] . "</td>";
  echo "<td>" . $row['score4'] . "</td>";
  echo "<td>" . $row['score5'] . "</td>";
  echo "<td>" . $row['score6'] . "</td>";
  echo "<td>" . $row['assessor_name'] . "</td>";
  echo "<td>" . $row['assessor_id'] . "</td>";
  echo "<td>" . $CallNumber . "</td>";
  echo "<td>" . $row['date_created'] . "</td>";
  echo "<td>" . $row['date_uploaded'] . "</td>";
  echo "</tr>";

$previousID = $currentID;
    
    $count++;

  }

mysql_close($con);

?>

 

 

[premiso]

You have two ORDER BY's in your sql query.

 

Alternatively, using mysql_query would have pointed this out.

 

$result = mysql_query("SELECT * FROM TableName ORDER BY employee_id, date_created ASC") or die(mysql_error());

[webguync]

doh! thanks, I should have seen that.