Why when I type the new the new exam does it continue to do the sum of the new exam with the average of the old one?

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Why when I type the new the new exam does it continue to do the sum of the new exam with the average of the old one?

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#1 by (0 votes)

-1

The question is this in php: Read four values for four student grades and print a message stating that the student has passed if the average grade is higher than or equal to 7. If the average grade is less than 7, request the exam grade , add up to the average value and get a new mean. If the new average is greater than or equal to 5, present a message stating that the student has passed the exam. If the student has not been approved, indicate a message stating this condition. Present together with the messages the value of the student average for any condition.

this is the main form:

<!DOCTYPE html>
<html>
<title>Questões</title>
<body>
<form action="q6php.php" method="POST">
 Digite a primeira nota: <input type="number" name="n1"><br>
 Digite a segunda nota: <input type="number" name="n2"><br>
 Digite a terceira nota: <input type="number" name="n3"><br>
 Digite a quarta nota: <input type="number" name="n4"><br>
 <input type="submit" value="Salvar" name="menu">
</form>
</body>
</html>

This is php:

<?php
$n1 = $_POST['n1'];
$n2 = $_POST['n2'];
$n3 = $_POST['n3'];
$n4 = $_POST['n4'];
$med = (($n1+$n2+$n3+$n4)/4);
if($med>=7){
      echo("aprovado com ". $med);
    }else if($med<7){
    header('location: q6formdir.php');
    $k= $_POST['n5'];
    if(isset($_POST['salvar'])){
       $mednova = (($med + $k)/2);
      if($mednova>=5){
      echo("aprovado em exame com ".$mednova);

}else{
      echo("reprovado");
}
}

    }

?>

and this other is the form that if the mean is less than 7 it will ask for another test:

<!DOCTYPE html>
<html>
<title>Questões</title>
<body>
<form action="q6php.php" method="POST">
  Digite a nota do novo exame: <input type="number" name="n5">
  <input type="submit" value="salvar" name="salvar">
  <input type="hidden" value="n1" name="n1">
  <input type="hidden" value="n2" name="n2">
  <input type="hidden" value="n3" name="n3">
  <input type="hidden" value="n4" name="n4">
</form>
</body>
</html>

php

asked by anonymous 03.07.2016 / 03:32

1 answer

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score 0 · Answer 1

I do not see why using two forms for the same function, just return using header('Location'); .

   $med = (($n1+$n2+$n3+$n4)/4);

     if($med>=7){
        echo("aprovado com ". $med);
     }else if($med<7){
        header('location: q6formdir.php');

     $k= $_POST['n5'];

     if(isset($_POST['salvar'])){
        $mednova = (($med + $k)/2);
     if($mednova>=5){
     echo("aprovado em exame com ".$mednova);
     }else{
    echo("Reprovado, Faça outro teste!");
    header('Location: exame.php');
   //retorna ao formulario;
    }
   }
 }

?>