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Armstrong Number In Php


prathameshkakade

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Hello everyone.

I tried to find out wether a number is armstrong or not.My logic was right but few php errors are creeping.I am new to PHP.So please help me out.

Here is the code

armstrong.php

<html>
<head>
<title>
Armstrong number
</title>
</head>
<body>
<?php
$arm=0;
$i;
$k;
$i=$_POST["i"];
$j=$i;
while($j!=0)
{
$k=$j%10;
$arm=$arm+($k*$k*$k);
$j=$j/10;
}
if($i==$arm)
{echo "Yes it is an Armstrong number";}
else
{echo "No,it is not an Armstrong Number";}
?>
<form method="POST" action="armstrong.php">
<input type="text" name="i" />
<br>
<br>
<input type="submit" name="submit" />
<input type="reset" name="reset" />
</form>
</body>
</html>

 

The error is it shows "Yes it is an Armstrong number" by default when we open armstrong.php in browser.And when we enter a number and click submit,every thing goes fine.

Here is a screenshot of it.

8965345ac30891d41da6a57f6732a47a.PNG

<hr>

 

 

 

 

ad3dcaeda45662d4cedd6bd5711e3b40.PNG

And also some php errors are visible(undefined variable.....).

Please help.

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You need to make sure that your code only runs if you have submitted a number. You can do that by wrapping the whole block of code in a if() statement that checks whether $_POST['i'] exists or not

 

if (isset($_POST['i'])){
....
}

 

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You need to make sure that your code only runs if you have submitted a number. You can do that by wrapping the whole block of code in a if() statement that checks whether $_POST['i'] exists or not

 

if (isset($_POST['i'])){
....
}

 

Thanks man.It worked out perfectly.But can you tell me why we added that lines to code?I mean the function of (isset($_POST['i'])).

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