Fatal Error: Function name

Question

Im getting an Error:

FATAL ERROR: Function name Must be a string in C:\xampp\localhost\student\viewgrades.php on line 78.

I wanted to make an if Statement if the grade is more than 75 then it should echo 'passed', but if the calculated grade is below 75 i should 'echo failed'.

Sorry for the code structure my knowledge for PHP is still limited

<?php
$resultp = mysql_query("SELECT * FROM prelcent");
$row = mysql_fetch_array($resultp);

$prelcent=$row['percentage'];

$resultpp = mysql_query("SELECT * FROM midcent");

$row = mysql_fetch_array($resultpp);

$midcent=$row['percentage'];

$resultppp = mysql_query("SELECT * FROM precent");

$row = mysql_fetch_array($resultppp);

$precent=$row['percentage'];

$resultpppp = mysql_query("SELECT * FROM fincent");

$row = mysql_fetch_array($resultpppp);

$fincent=$row['percentage'];

$remark = (($pre*$prelcent)+($mid*$midcent)+($prf*$precent)+($fin*$fincent));
    if ($remark() >75)   <<<<<Line 78 {
        $rem1 = "echo 'passed!';";
    } else {
        if ($remark()<75) {
            $rem1 = "echo 'failed';";
        }
    }
    echo "<td>". $rem1 ."</td>"
?>

GautamD31 · Accepted Answer · 2014-02-15 11:14:11Z

4

Can you try with

if ($remark >75)

Because $remark is not a function.Its a variable that to the result of

($pre*$prelcent)+($mid*$midcent)+($prf*$precent)+($fin*$fincent)

answered Feb 15, 2014 at 11:14

GautamD31

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Krish R · Accepted Answer · 2014-02-15 11:39:29Z

2

$remark is variable, not a function, So it should be , also $rem1 = "passed!"; instead of $rem1 = "echo 'passed!';";

if ($remark > 75){
     $rem1 = "passed!";
}

instead of

if ($remark() > 75){
   $rem1 = "echo 'passed!';";
}

Your code should be like,

 $remark = (($pre*$prelcent)+($mid*$midcent)+($prf*$precent)+($fin*$fincent));
 if ($remark > 75) {
    $rem1 = "passed!";
 } else {
    if ($remark < 75) {
        $rem1 = "failed";
    }
 }
 echo "<td>". $rem1 ."</td>";

edited Feb 15, 2014 at 11:39

answered Feb 15, 2014 at 11:14

Krish R

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Comments

AnOldSoul · Accepted Answer · 2014-02-15 11:31:23Z

0

$rem1 = "echo 'passed!';"; makes it look like a function. In order for it to look like a variable it should be $rem1 = "passed!";

answered Feb 15, 2014 at 11:31

AnOldSoul

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