a calculator in php

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I need a PHP code from a calculator, which adds 4 notes and divides by 4 and if the final average is greater than or equal to 6 that it shows approved, if not disapproved. SCRR

<?php

       $nt1 = $_GET[$nt1];
       $nt2 = $_GET[$nt2];
       $nt3 = $_GET[$nt3];
       $nt4 = $_GET[$nt4];
       $mf = ( $nt1 + $nt2 + $nt3 + $nt4)/4;
       echo " $nt1, $nt2, $nt3, $nt4,  $mf ";

      if ( $mf >= 6  ) {
       echo "aprovado" ;

      }
      else{
       echo "Reprovado";

        } 
    ?>

HTML code

<!DOCTYPE html>
<html>
 <head>
  <meta charset="utf-8">

  <title> calculadora </title>
 </head>
  <body>
  <form action="calculadora.php" method = "get">
  <table border="1" cellpadding="1" cellspacing="1" style="width: 500px;">
    <tbody>
        <tr>
            <td>Primeiro bimestre</td>
            <td><input name="$nt1" type="text" /></td>
        </tr>
        <tr>
            <td>Segundo bimestre</td>
            <td><input name="$nt2" type="text" /></td>
        </tr>
        <tr>
            <td>terceiro bimestre</td>
            <td><input name="$nt3" type="text" /></td>
        </tr><tr>
            <td>quarto bimestre</td>
            <td><input name="$nt4" type="text" /></td>
        </tr>
        <tr>
            <td></td>
            <td><input name="bt_validar" type="submit" value="Calcular" /></td>
        </tr>
    </tbody>
</table>


  </body> 
</html>
php html
  • 1

    You could post the HTML code of your form here?

  • can yes, more aq in the comments does not fit

  • I’ll edit the question and post there may be??

  • Elaborate on your question Cristina. Stackoverflow doesn’t work like this: "I want code X for yesterday because I have a deadline." Which part of the code did you not understand? What is your question exactly? Where is the corresponding HTML code?

  • can’t you do the calculation? I think it’s best to change the names of inputs, take out that cipher.

  • It is pq in the code in PHP, it should show the result of the sums qnd do the division and if it was greater or equal to 6 it was to give approved and if it was smaller it was to fail, only goes to the condition of Else, regardless of the value that gives. and I don’t find the error

  • HTML code put in question

  • rray, input numbers are the variables I used in PHP

  • If it is falling only on Else its variables must be as zero or null look: $nt1 = $_GET[$nt1]; so you say for php, want the Intel $_GET only he doesn’t exist ...

  • More want the user to give the value of the variable understand? what can I put in place? Can help me?

  • In all variable php starts with $ just take it out of the name of inputs and $nt1 = $_GET[$nt1];, let alone $nt1 = $_GET['nt1'];

  • Got it, it was missing simple quotes in the variable. Obg rray

Show 7 more comments

2 answers

2


You are picking up your form data in a wrong way.

Instead of $_GET[$nt1]; should be $_GET['nt1']; and so in others.

See the working code here.

Follow the full example code based on your code:

HTML:

<!-- não esqueça de abrir com tag <html>, inserir o <head> e fechar ele e tmabém de abrir <body> -->

<form id="formulario" name="formulario" method="get" action="calcular.php">
Nota 1 <input id="nt1" name="nt1" type="text" /><br />
Nota 2 <input id="nt2" name="nt2" type="text" /><br />
Nota 3 <input id="nt3" name="nt3" type="text" /><br />
Nota 4 <input id="nt4" name="nt4" type="text" /><br />
<input id="btnenviar" name="btnenviar" type="submit" value="Calcular" />
</form>

<!-- feche o <body> e feche o <html> -->

PHP file calculate.php: 

<?php

    $nt1 = $_GET['nt1'];
    $nt2 = $_GET['nt2'];
    $nt3 = $_GET['nt3'];
    $nt4 = $_GET['nt4'];
    $mf = ( $nt1 + $nt2 + $nt3 + $nt4)/4;
    //echo " $nt1, $nt2, $nt3, $nt4,  $mf "; Esta linha vai imprimir as notas e e também a média

    if ( $mf >= 6  ) {
        echo "aprovado" ;
    }
    else{
        echo "Reprovado";
    } 
?>

  • It was the lack of even simple quotes. Muuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuo Obliaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

  • 1

    Yeah, use the name in his html without the $. And similarly call him with the quotes and without the $ inside $_GET[];

2

When you put $nt4 = $_GET[$nt4];

He understands the $nt4 as a variable of PHP, since to declare them you use $, even if you put between "" not right, and only replace by "nt4", does not forget to change in html too.

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