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PHP populate variable with function

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# 1  
Old 11-20-2010
PHP populate variable with function
Hi,
I've a php scirpt that calls in an external class. The external class, myClass, has several public and private functions.
There is one public function which derives a value that I want be able to bring it to the main php scirpt. Here is the code.
PHP Code:
....snip....
include 'myClass.class.php';
$createFile = new myClassFunction();

                echo '<div id="preview" style="display:block"><p>Accessing  your link. Please wait.</p>';
                // still working on this part
                flush();
// Main Program

            if ($createFile->DownloadPDF(trim($_POST['userPDFurl'])))
            {
               //User link is good
               echo $createFile->GetFileName();  //This is the file user should be able to download
               echo ($createFile->GeneratePDF('$_POST[quality]')) ? '<p>Success!</p>' : '<p>Error generating PDF file!</p>';
            }
            else
            {
               // Could not reach user link
               echo '<p>Error downloading PDF!</p>';
            }
 ....snip.... 
If you notice in above code, I can echo the newly created file name and path 'echo $createFile->GetFileName();'
What I want to do is to put the value from the function to some variable and create a link. Something like this:
PHP Code:
echo'
Here is your file <a href="http://mysite.cc/"$createFile->GetFileName()></a>. Click here 
It doesn't really work, and I link appears messed up.Smilie Can I put $createFile->GetFileName() in to yet another variable and echo that? Smilie

Thank you,
Nitin
# 2  
Old 11-21-2010
Hi,
I figured that my syntax was not entirely correct. Now I can display the entire link as text.
PHP Code:
echo "http://mystie.cc/",$createFile->GetFileName(),"<br />\n"; 
The comma before and after the function was needed. Smilie
Next step is to make this thing text as a click-able link. But that's another story.

Nitin
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