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help with creating average between data in fields from mysql DB

webguync

By webguync
in PHP Coding Help

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webguync Advanced Member

webguync

Posted
Posted

I have a database with 8 fields all containing numbers from 1-7 and I have another field called 'average' where I am seeking to average the data. I am getting an error with the code I have which is "Warning: Cannot use a scalar value as an array in /var/www/vhosts/etsi-dataservices.com/httpdocs/NNPrinceton/Phase1A_May27_2010/report/Individual_Scores.php  on line 59

 

Warning: Cannot use a scalar value as an array in /var/www/vhosts/etsi-dataservices.com/httpdocs/NNPrinceton/Phase1A_May27_2010/report/Individual_Scores.php on line 62"

 

not sure what this means. Anyway what I have is not working! Here is the entire code.

 

<html>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Phase1A Award Results - May 2010</title>
<link href="report.css" rel="stylesheet" type="text/css" />
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
</head>
<body>
<html>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Phase1A Award Results - May 2010</title>
<link href="css/report.css" rel="stylesheet" type="text/css" />
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
</head>
<body>
<?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 employee_id,employee_name, roster_Phase1A_May27_2010.title,assessor_id,assessor_name,score1,score2,score3,score4,score5,score6,score7,score8,average,date_created,date_uploaded FROM rpc1A_May27_2010 LEFT JOIN  roster_Phase1A_May27_2010 USING (employee_id) ORDER BY employee_id, date_uploaded ASC") or die(mysql_error());
echo "<table>
<tr>

<th>Employee ID</th>
<th>Employee Name</th>
<th>Employee Title</th>
<th>score 1</th>
<th>score 2</th>
<th>score 3</th>
<th>score 4</th>
<th>score 5</th>
<th>score 6</th>
<th>score 7</th>
<th>score 8</th>
<th>Average Score (1-</th>
<th>Assessor Name</th>
<th>Assessor ID</th>
<th>Call Number (1-4)</th>
<th>Date Created</th>
<th>Date Uploaded</th>
</tr>";
//take sum
$row = 0;
$sum = 0;
for ($i = 1; $i <= 8; $i++){
$sum += $row['score'.$i];
}
$row['average'] = $sum/8;

//round $average
$row['average'] = round($row['average'], 1);

//count Call Numbers and list by date entered
$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>" . $row['employee_id'] . "</td>";
  echo "<td>" . ucwords($row['employee_name']) . "</td>";

  echo "<td>" . $row['title'] . "</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['score7'] . "</td>";
  echo "<td>" . $row['score8'] . "</td>";

  echo "<td>" . $row['average'] . "</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++;

  }
echo "</table>";
mysql_close($con);

?>
</body>
</html>




</body>
</html>

 

 

the specific code I am trying to do the averaging is...


//take sum
$row = 0;
$sum = 0;
for ($i = 1; $i <= 8; $i++){
$sum += $row['score'.$i];
}
$row['average'] = $sum/8;

//round $average
$row['average'] = round($row['average'], 1)

 

 

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kenrbnsn Regular Member

kenrbnsn

Posted
Posted

You can't use a variable as both a scalar (single value) and an array (multiple values) at the same time:

<?php
$row = 0;
$sum = 0;
for ($i = 1; $i <= 8; $i++){
$sum += $row['score'.$i];
}
$row['average'] = $sum/8;

//round $average
$row['average'] = round($row['average'], 1);
?>

 

Instead of using $row['average'] just use a normal variable:

<?php
$average = 0;
$sum = 0;
for ($i = 1; $i <= 8; $i++){
$sum += $row['score'.$i];
}
$average = round($sum/8);
?>

 

But this code isn't going to work where you have it. It has to be inside the while loop where you're fetching the rows.

 

Ken

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webguync Advanced Member

webguync

Posted
  • Author
Posted

the way I have it now, the only row being displayed is the average score, but it shows up as all zeros. Here is my while loop.

 

while($row = mysql_fetch_array($result))
  {
  //take sum
$row = 0;
$sum = 0;
for ($i = 1; $i <= 8; $i++){
$sum += $row['score'.$i];
}
$average = round($sum/8);

//round $average
$average = round($row['average'], 1);
//count Call Numbers and list by date entered
$count = 0;
$CallNumber = 1;

$currentID = $row['employee_id'];

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

  echo "<tr>";

  echo "<td>" . $row['employee_id'] . "</td>";
  echo "<td>" . ucwords($row['employee_name']) . "</td>";

  echo "<td>" . $row['title'] . "</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['score7'] . "</td>";
  echo "<td>" . $row['score8'] . "</td>";

  echo "<td>" . $average . "</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++;

  }
echo "</table>";
mysql_close($con);

?>

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