Write output to html from php

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Hi,
I'm trying to check a number is even or odd on a php page. I write my code but i have a problem. When i run it, it gets 0 as default number and also after checking the number box become empty. Could you help me?

HTML

<html><head></head>
<body>
<form action="PAGEName.php" method="post">
	<table bgcolor="yellow" border="5" bordercolor="black"> <tr> <td>
	Is   <input type="text" size="6" value="" name="input">  
	<input type="submit" value="even or odd?"> <br>
	Answer: <input type="text" size="8" 
	value=" 
	    $n=$_POST['input'];
	print($n%2==1 ? 'odd' : 'even');
	?>
    " 
	name="result"> <br>	
	</br></br></td> </tr> </table>

</body></form></html>

Posted 21-Oct-12 6:52am

sesenyg

Updated 21-Oct-12 11:31am

Nelek

v2

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Thomas Daniels · Accepted Answer · 2012-10-21T07:00:00

Solution 1

I think you cann't add PHP in the value attribute.
So, you need to add 'even' or 'odd' like this:

PHP

<?
if ((int)$_POST['input'] % 2 == 1)
{
echo "odd";
}
else
{
echo "even";
}
?>

Posted 21-Oct-12 7:00am

Thomas Daniels

Updated 21-Oct-12 7:02am

v2

Comments

sesenyg 21-Oct-12 13:07pm

It still behaves the same, it gets 0 default.

sesenyg 21-Oct-12 13:10pm

or in other words, it gets even in default.

Thomas Daniels 21-Oct-12 13:11pm

Then I think there's something wrong in the POST-form.

sesenyg 21-Oct-12 13:15pm

OK, now i did it truely. I add one more if clause like
<input type="text" size="8"
value="

sesenyg 21-Oct-12 13:17pm

Thanks for your attention:)