Call to undefined function

tobimichigan · April 20, 2010

Fatal error: Call to undefined function int() in C:\xampp\htdocs\details.php on line 135

<?php
						$grade = array("A","B","C","D","E","F");                     

						$adno = $_GET['adno'];
						//$grade[0]=A;
						//$grade[1]=B;
						//$grade[2]=C;
						//$grade[3]=D;
						//$grade[4]=E;
						//$grade[5]=F;
						$adno=$_GET['adno'];
$select = mysql_query("select * from acadinfo where adno='$adno'") or die('<p>Error Retrieving<br/>'.'Error: ' .mysql_error() . '</p>');					echo "<table width='548' border='1'>
	<tr>
		<th width='26' scope='col'>ADMISSION NO.</th>
		<th width='26' scope='col'>MATHEMATICS</th>        
		<th width='26' scope='col'>ENGLISH</th>        
		<th width='26' scope='col'>PHYSICS</th>        
		<th width='26' scope='col'>CHEMISTRY</th>        
		<th width='365' scope='col'>BIOLOGY</th>    
	</tr>
<tr><th scope='row'>$adno</th>";    
$datalist = mysql_fetch_array($select);
//$datalist = htmlspecialchars($datalist);

foreach($datalist as $data){    
if (int($data) >= 80) {
 echo "<td>$grade[0]</td>";
} else {        
	if (int($data) >= 70  && int($data) < 80) { 
	   echo "<td>$grade[1]</td>";
	    } else {
		  if (int($data) >= 60  && int($data) < 70) {
		   echo "<td>$grade[2]</td>";
		   }else {
				if (int($data) >= 50  && int($data) < 60) {
				    echo "<td>$grade[3]</td>";
					  } else {
						if (int($data) >= 40  && int($data) < 50) {
						  echo "<td>$grade[4]</td>";
						     } else {
								if (int($data) <= 39  && int($data) < 40) {
								    echo "<td>$grade[5]</td>";                        
									}                    
								}                
							}            
						}        
					}    
				}    
			}
echo "</tr></table>"; 
?>

Please can I stand corrected? I m working on student grades...

Line 135-if (int($data) >= 80) {

Gekk0 · April 20, 2010

try intval() instead of int()

Psycho · April 20, 2010

Correct usage

(int) $someVar;

Example:

$foo = 1.23;
$bar = (int) $foo;
echo $bar; //Output 1

tobimichigan · April 20, 2010

Hello Gekko, please could you tell what exactly is making the code output F,F,F,FF, instead of the real grades?

Here's my code:

<?php
						$grade = array("A","B","C","D","E","F");                     

						$adno = $_GET['adno'];
						$grade[0]=A;
						$grade[1]=B;
						$grade[2]=C;
						$grade[3]=D;
						$grade[4]=E;
						$grade[5]=F;
						$adno=$_GET['adno'];
$select = mysql_query("select * from acadinfo where adno='$adno'") or die('<p>Error Retrieving<br/>'.'Error: ' .mysql_error() . '</p>');					echo "<table width='548' border='1'>
	<tr>
		<th width='26' scope='col'>ADMISSION NO.</th>
		<th width='26' scope='col'>MATHEMATICS</th>        
		<th width='26' scope='col'>ENGLISH</th>        
		<th width='26' scope='col'>PHYSICS</th>        
		<th width='26' scope='col'>CHEMISTRY</th>        
		<th width='365' scope='col'>BIOLOGY</th>    
	</tr>
<tr><th scope='row'>$adno</th>";    
$datalist = mysql_fetch_array($select);
$num=mysql_num_rows($select);
//$datalist = htmlspecialchars($datalist);

foreach($datalist as $data){    
if (intval($data) >= 80) {
 echo "<td>$grade[0]</td>";
} else {        
	if (intval($data) >= 70  && intval($data) < 80) { 
	   echo "<td>$grade[1]</td>";
	    } else {
		  if (intval($data) >= 60  && intval($data) < 70) {
		   echo "<td>$grade[2]</td>";
		   }else {
				if (intval($data) >= 50  && intval($data) < 60) {
				    echo "<td>$grade[3]</td>";
					  } else {
						if (intval($data) >= 40  && intval($data) < 50) {
						  echo "<td>$grade[4]</td>";
						     } else {
								if (intval($data) <= 39  && intval($data) < 40) {
								    echo "<td>$grade[5]</td>";                        
									}                    
								}                
							}            
						}        
					}    
				}    
			}
echo "</tr></table>"; 
?>

By the way the int problem was solved by implementing your correction...thanks

Psycho · April 20, 2010

$datalist = mysql_fetch_array($select);
$num=mysql_num_rows($select);
//$datalist = htmlspecialchars($datalist);

foreach($datalist as $data){    
if (intval($data) >= 80) {
 echo "<td>$grade[0]</td>";

$data is an ARRAY. You can't get the integer of an arryay. Although there is a lot more wrong with that code.

Psycho · April 20, 2010

What does your DB table look like?

Gekk0 · April 20, 2010

Okidoki, a few things....

                     $grade = array("A","B","C","D","E","F");                     
                     
                     $adno = $_GET['adno'];
                     $grade[0]=A;
                     $grade[1]=B;
                     $grade[2]=C;
                     $grade[3]=D;
                     $grade[4]=E;
                     $grade[5]=F;

this is kind of redundant isn't it?

you can remove all of this:

                     $grade[0]=A;
                     $grade[1]=B;
                     $grade[2]=C;
                     $grade[3]=D;
                     $grade[4]=E;
                     $grade[5]=F;

you've already defined your array of possible grades, there's no need to do it again!

for this section of code:

foreach($datalist as $data){    
   if (intval($data) >= 80) {
    echo "<td>$grade[0]</td>";
   } else {        
      if (intval($data) >= 70  && intval($data) < 80) { 
         echo "<td>$grade[1]</td>";
          } else {
           if (intval($data) >= 60  && intval($data) < 70) {
            echo "<td>$grade[2]</td>";
            }else {
               if (intval($data) >= 50  && intval($data) < 60) {
                   echo "<td>$grade[3]</td>";
                    } else {
                     if (intval($data) >= 40  && intval($data) < 50) {
                       echo "<td>$grade[4]</td>";
                          } else {
                           if (intval($data) <= 39  && intval($data) < 40) {
                               echo "<td>$grade[5]</td>";                        
                              }                    
                           }                
                        }            
                     }        
                  }    
               }    
            }

I'd look at investing in some elseif() statements, seriously

so something like this:

foreach($datalist as $data)
{
if(intval($data) >= 80)
	{

	 echo "<td>$grade[0]</td>";

	} elseif(intval($data) >= 70  && intval($data) < 80) {

	echo "<td>$grade[1]</td>";

	} elseif(intval($data) >= 60  && intval($data) < 70) {

	echo "<td>$grade[2]</td>";

	} elseif(intval($data) >= 50  && intval($data) < 60) {

	echo "<td>$grade[3]</td>";

	} elseif(intval($data) >= 40  && intval($data) < 50) {

	echo "<td>$grade[4]</td>";

	} elseif(intval($data) <= 39  && intval($data) < 40) {

	echo "<td>$grade[5]</td>";

	}
}

thats what the elseif is for!

but, check that $data variable, i'm guessing that what you actually want there is something like $data['mark'] instead just $data, because your looking at the element of that array $data

jus editing this again .. yeah so basically you downloaded your database into that variable $datalist (with entries of each row you've retrieved from the database), which is an array, and each element in that array will be another array - this will be an array for example $dataList[0]['name'], $dataList[0]['mark'], $dataList[0]['birthday'] - or whatever, just depends on how your tables are structured

so when you launch into that foreach statment your going through each $dataList element and giving it the name $data, but $data is still an array - because each element of $dataList was an array..

am I making sense?

Gekk0 · April 20, 2010

I dunno if this is necessary, maybe some other forum members could correct me, but this is the way i'd get your $datalist array from the database


		$query = mysql_query($select);

		//Build the array
		$i = 0;

		while ($row = mysql_fetch_assoc($query))
			{

			$keys = array_keys($row);

			foreach($keys as $key)
				{

				$datalist[$i] = $row[$key];

				}

			$i++;

			}

instead of


$datalist = mysql_fetch_array($select);
$num=mysql_num_rows($select);

Psycho · April 20, 2010

Nah, if you want to go that route, I would set the index of the array as the grade percentage and the value as the letter grade. Then you can get the letter grade by a simple foreach loop

$grades = array(80=>'A', 70=>'B', 60=>'C', 50=>'D', 40=>'E', 0=>'F');

$studentPercent = 75; //Some number from the DB query

foreach($grades as $gradePercent => $letterGrade)
{
    if($studentPercent>$gradePercent) { break; }
}

echo $letterGrade

However, I would take a whole different approach. After looking over your code some more, it seems that you should be getting multiple grades from the DB query (one for each column header). I assume the grades for each subject are in a different field name in the database. Here is a rewrite of what you had. Replace the field names as appropriate:

<?php

function getLetterGrade($percent)
{
    if(!is_numeric($percent)) { return '--'; }
    if($percent >= 80) { return 'A'; }
    if($percent >= 70) { return 'B'; }
    if($percent >= 60) { return 'C'; }
    if($percent >= 50) { return 'D'; }
    if($percent >= 40) { return 'E'; }
    return 'F';
}

$adno = (int) $_GET['adno'];
$query = "SELECT * FROM acadinfo WHERE adno='{$adno}'";
$result  = mysql_query($query) or die('<p>Error Retrieving<br/>'.'Error: ' .mysql_error() . '</p>');

if(!mysql_num_rows($result))
{
    echo "No records for admission number {$adno}";
}
else
{
    $studentGrades = mysql_fetch_assocay($result);
    echo "<table width=\"548\" border=\"1\">\n";
    echo "  <tr>\n";
    echo "    <th width=\"26\" scope=\"col\">ADMISSION NO.</th>\n";
    echo "    <th width=\"26\" scope=\"col\">MATHEMATICS</th>\n";
    echo "    <th width=\"26\" scope=\"col\">ENGLISH</th>\n";    
    echo "    <th width=\"26\" scope=\"col\">PHYSICS</th>\n";  
    echo "    <th width=\"26\" scope=\"col\">CHEMISTRY</th>\n";
    echo "    <th width=\"365\" scope=\"col\">BIOLOGY</th>\n";  
    echo "  </tr>\n";
    echo "  <tr>\n";
    echo "    <td>{$adno}</td>\n";
    echo "    <td>" . getLetterGrade($studentGrades['MATHEMATICS']) . "</td>\n";
    echo "    <td>" . getLetterGrade($studentGrades['ENGLISH']) . "</td>\n";
    echo "    <td>" . getLetterGrade($studentGrades['PHYSICS']) . "</td>\n";
    echo "    <td>" . getLetterGrade($studentGrades['CHEMISTRY']) . "</td>\n";
    echo "    <td>" . getLetterGrade($studentGrades['BIOLOGY']) . "</td>\n";
    echo "  </tr>\n";
    echo "</table>\n";
}
?>

jcbones · April 20, 2010

My take on this script;

<?php
                    
function get_grade((int)$i) {
if($i > 79) return 'A';
elseif($i > 69 && $i < 80) return 'B';
elseif($i > 59 && $i < 70) return 'C';
elseif($i > 49 && $i < 60) return 'D';
elseif($i > 39 && $i < 50) return 'E';
else return 'F';
}

$adno = $_GET['adno'];
                     
$select = mysql_query("select * from acadinfo where adno='$adno'") or die('<p>Error Retrieving<br/>'.'Error: ' .mysql_error() . '</p>');              
echo "<table width='548' border='1'>
      <tr>
         <th width='26' scope='col'>ADMISSION NO.</th>
         <th width='26' scope='col'>MATHEMATICS</th>        
         <th width='26' scope='col'>ENGLISH</th>        
         <th width='26' scope='col'>PHYSICS</th>        
         <th width='26' scope='col'>CHEMISTRY</th>        
         <th width='365' scope='col'>BIOLOGY</th>    
      </tr>
<tr><td scope='row'>$adno</td>";    

$datalist = mysql_fetch_assoc($select);

foreach($datalist as $data){	

	echo '<td>' . get_grade($data) . '</td>';

}

   echo '</tr></table>';  
   
   
?>

tobimichigan · April 21, 2010

Wow, way 2 go guz..I'm gonna implement ur corrections and get back 2 u soon...thanks a mil..

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