A beginner question on passthru() function in PHP
- by Robert
Hi all,
I've used the following code to do an XSLT translation in PHP and save the result SVG file into a sub folder:
$command = $java . $saxon . $target2 . ' ' . $xsl2.' '.$param;
ob_start();
passthru($command, $result);
$content_grabbed=ob_get_contents();
ob_end_clean();
$info = pathinfo($_FILES['file']['name']);
$file_name= basename($_FILES['file']['name'],'.'.$info['extension']);
$myFile = "images/convert/".$file_name."_".$_POST["mydropdown"].".svg";
$fh = fopen($myFile, 'w') or die("can't open file");
fwrite($fh, $content_grabbed);
fclose($fh);
It does a very simple SVG to SVG transformation, and created the resulting SVG file.
the $command is just a simple java -jar saxon.jar input.xml stylesheet java command. The current situation is ,the file gets created perfectly, however, the error gets shown in browser, probably because I inserted the "ob_start" before passthru() command.
I am thinking of making use of an argument that comes with the saxon.jar, that is "-o output_file_name". Because the final purpose is not to display the resulting SVG to the browser, but instead creating an SVG file ,and provide the link to the user for downloading.
Are there any better ways to handle this problem, I am still thinking about how to get rid of the result that gets shown in the browser that is sent by the "passthru()". Thanks in advance for suggestions!
