Error trying to output result of calculation based on user input

Question

I recently started PHP and I am not able to figure out how to echo the result of this. Any help would be appreciated.

<html>
    <head>
    <style>
    input[type=submit]{
        background-color: #4CAF50;
        border: none;
        color: white;
        padding: 16px 32px;
        text-decoration: none;
        margin: 4px 2px;
        cursor: pointer;
    }
    input[type=text]{
        border: 3px solid black;
    }
    </style>
    </head>

    <body>
    <form action="fitness1.php" method="post">
    <b>weight:</b>  <input type="text" name="weight"><br>
    <br>
    <b>height:<b>   <input type="text" name="height"><br>
    <br>
    <input type="submit">
    </body>
</html>

This is the PHP code:

<html>
<body>
<?php

$WEight = $_Post['weight'];
$HEight = $_post['height'];

if (is_numeric($WEight)&& is_numeric($HEight)){

    $bmi= $WEight / $HEight * $HEight;
  }     

echo ("$bmi");


else {
    echo "please enter your weight and height";
}
?>
</body>
</html>

Parse error: syntax error, unexpected '}', expecting ',' or ';' in G:\wamp\www\test\fitness1.php on line 19 — aye.boss
– aye.boss, Commented Jul 3, 2016 at 7:44
Move the curly brace before the echo BMI line to be after that line. — Tieson T.
– Tieson T., Commented Jul 3, 2016 at 7:45

Will · Accepted Answer · 2016-07-03 09:58:45Z

2

The case of variables matters in PHP. You're looking for this:

$WEight = $_POST['weight'];
$HEight = $_POST['height'];

Then, this is the logic we would use:

if (is_numeric($WEight)&& is_numeric($HEight)) {
    $bmi = $WEight / $HEight * $HEight;
    echo $bmi;
} else {
    echo "please enter your weight and height";
}

Your echo needs to be inside the if (...) { ... } block.

Edit:

As requested, here's how I would improve it:

HTML:

<html>
    <head>
        <style>
        input[type=submit]{
            background-color: #4CAF50;
            border: none;
            color: white;
            padding: 16px 32px;
            text-decoration: none;
            margin: 4px 2px;
            cursor: pointer;
        }
        input[type=text]{
            border: 3px solid black;
        }
        </style>
    </head>

    <body>
        <form action="fitness1.php" method="post" />
        <b>Weight:</b> <input type="text" name="weight" /><br />
        <br />
        <b>Height:<b>  <input type="text" name="height" /><br />
        <br />
        <input type="submit" />
    </body>
</html>

PHP:

<html>
    <body>
<?php

$weightInKg = (int) $_POST['weight'];
$heightInCm = (int) $_POST['height'];

if ($heightInCm && $weightInKg) {
    $bmi = $weightInKg / pow($heightInCm, 2);
    echo "Your BMI is ${bmi}.";
} else {
    echo "Please enter your height and weight as a number greater than 0.";
}
?>
    </body>
</html>

edited Jul 3, 2016 at 9:58

answered Jul 3, 2016 at 7:45

Will

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11 Comments

Will Over a year ago

No problem, glad to help. Comment if you still can't get it working.

aye.boss Over a year ago

Parse error: syntax error, unexpected 'else' (T_ELSE) in G:\wamp\www\test\fitness1.php on line 11

aye.boss Over a year ago

hey it was still giving error on line 8 and 9 i changed it to $weightInkg and $heightincm it echo bmi but with an error on line 8 of undefined variable

aye.boss Over a year ago

when i changed variable on line 8 to weightinkg and heightincm it echo's the else statement not the bmi

Will Over a year ago

Oops my bad as well. Corrected :) You have to use the proper case; PHP variables are case sensitive.

|

Perdeep Singh · Accepted Answer · 2016-07-03 07:48:14Z

0

Main issue of error message is :

echo("$bmi"); // this is wrong you don't need brackets.

replace this with

echo $bmi;

That's the reason of error but if would suggest to move echo $bmi inside if condition or initialise $bmi else you can get undefined variable error if it doesn't go through if condition.

answered Jul 3, 2016 at 7:48

Perdeep Singh

4924 silver badges7 bronze badges

5 Comments

aye.boss Over a year ago

still doesent work now its giving errors on line 5 and 6 and 10 can you write the php code for me it would be more understandable for me

aye.boss Over a year ago

undefined variable on line 5 6

Perdeep Singh Over a year ago

fair enough you need to fix typo mistake replace line 4 and 5 ( $WEight = $_Post['weight']; $HEight = $_post['height'];) by $WEight = $_POST['weight']; $HEight = $_POST['height'];

aye.boss Over a year ago

yeah but now its not doing the calculation right like 100/2*2 =25 but its echo's 100

Perdeep Singh Over a year ago

ok did you initialised it with 100? Can you please show your exact code what it looks like now?

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