Temperature Conversion using HTML form and PHP functions - php

Question

I am trying to convert deg -> c and c -> deg using php. The objective is to use functions containing the formulas and then having a form with radio buttons and a text box. The user should be able to enter the degree in the text box, click the radio button of their choice (F or C) and then submit the form to receive their conversion. I have seen similar posts but the methodology is not specific to the issue I am having.
I have updated my code but am now getting the "White page of death"
Can anyone see an error that I cannot see? Thank you!
HTML
<h1>Temperature Conversion</h1>
<form action='lab_exercise_6.php' method ='POST'>
<p>Enter a temp to be converted and then choose the conversion type below.</p>
<input type ='text' maxlength='3' name ='calculate'/>
<p>Farenheit
<input type="radio" name='convertTo' value="f" />
</p>
<p>Celsius
<input type="radio" name='convertTo' value="c" />
</p>
<input type='submit' value='Convert Temperature' name='convertTo'/>
PHP
<?php
//function 1
function FtoC($deg_f) {
return ($deg_f - 32) * 5 / 9;
}
//function 2
function CtoF($deg_c){
return($deg_c + 32) * 9/5;
}
if( isset($_POST['convertTo']) && $_POST['convertTo'] ==="c" ){
$farenheit = FtoC($deg_f);
print('This temperature in FARENHEIT is equal to ' . $celsius . ' degrees celsius! </br>');
}else if(isset($_POST['convertTo'])&& $_POST['convertTo']==='f'){
$celsius = CtoF($deg_c);
print('This temperature in CELSIUS is equal to ' . $farenheit . ' degrees farenheit! </br>');
}
?>

Answer 1

it's this part that makes it behave unexpected:
$farenheit = FtoC($deg_f);
// And then a few lines lower:
if(isset($farenheit)){ /* ... */}
You set it to a value right there, so it will always try to calculate an outcome.
With a small tweak you will use the radiobutton more as it is intended:
<input type="radio" name="convertTo" value="f" />
<input type="radio" name="convertTo" value="c" />
The value in PHP will now always have the same name, you can now just check it's value:
if( isset($_POST['convertTo']) && $_POST['convertTo']==="c" ){
$farenheit = FtoC($deg_f);
print('This temperature in FARENHEIT is equal to ' . $celsius . ' degrees celsius! </br>');
}

Answer 2

You haven't added values to the radio boxes.
Then change your radio name same as the other so for example
<p>Farenheit
<input type = 'radio' name ='temp_type' value='f'/>
</p>
<p>Celsius
<input type = 'radio' name ='temp_type' value='c'/>
</p>
Now you can access those with PHP.
if($_POST['temp_type'] && ($_POST['calculate']){
$temp2calculate = $_POST['calculate'];
$temp_type = $_POST['temp_type'];
}

Answer 3

You have to give same name for the radio buttons
<p>Farenheit
<input type = 'radio' value ='farenheit' name='units'/>
</p>
<p>Celsius
<input type = 'radio' value ='celsius' name='units'/>
</p>
Check whether which radio button is selected:
<?php
$units = $_POST['units'];
if($units == "fahrenheit"){
//call function to convert to fahrenheit
}
else{
//call function to convert to celsius
}
?>