variables in fopen filename

[MrDoug]

I have an html file that takes user input and sends it to a php file which looks somthig like this...

 

<?php

$name = $_POST['name'];
$input = $_POST['input'];
$fp = fopen('$name' . '.html', 'a');

fwrite($fp, '$input');

fclose($fp);

?>

 

so basicly i want it to open the html file that the user wants, and there write the user input to it.

however, instead it just creates a new html file called "$name.html".  My question

how do I make it read "$name" as a php variable instead of just a file name?  can it be done?  is there a better way?

Thanks in advance

 

 

[ron8000]

Have you tried to take it out of quotes? or try ->

 

$fp = fopen($name . '.html', 'a');

 

 

[bluejay002]

you do not need to put quotes intentionally to variables unless it has other literal characters together with it.

 

<?php

 

$name = $_POST['name'];

$input = $_POST['input'];

$fp = fopen($name . '.html', 'a');

 

fwrite($fp, $input);

 

fclose($fp);

 

?>