RE: [PHP-GENERAL] evaluating PHP source to a file.

Hi, >> --> $content=eval($template); > What's a sample template, and what's coming out on the browser > and what's being stored in $content. Because I would expect > this to "work" if what you describe is what you really want. no, it's not working that way. eval() always writes it's output to the current document and there seems to be no way of catching it into the $content. I tried playing around by opening special filehandlers (like doing $fd=open("php://stdout") and trying to see if I can catch the output through that, but nothing helped. then I decided the only way to go was to write down the template to a file on the disk and do exec("php -q template.php > output.html"). but it didn't work either - just as the template or content went a little more complex (db connect / couple of queries) the output.html started containing just the headers (even -q means NO HEADERS). the same thing executed from the shell worked just fine. I even tried writing down a shell script and then executing it by exec() - still the same. the same script when run from the shell command line works like a charm. I currently had to build the system especially ugly way - I write down the shell script and then a flag file from the PHP script, and then have a every-2-minutes cron job what checks the presence of the flag file and if it exists, executes the shell script.. PS. I'm still not subscribed to the lists, so please CC: an answer to my e-mail address. Thanks. Rgds, Tfr --==< tfr@cafe.ee >==< http://tfr.cafe.ee/ >==< +1-504-4467425 >==--