well someone is confused ;) , PHP_OS is a PHP constant , simply <?echo PHP_OS;?> Simon Wheeler firepages "Luis Moreira" <luis.moreira@xxxxxxxxxxxxxxx> wrote in message news:407509D5.7010808@xxxxxxxxxxxxxxxxxx > David Scott wrote: > > > In the PHP on Windows chapter of Programming PHP, there is mention of > > a constant, PHP_OS, that can be used to determine the OS running the > > server that PHP is on. > > > > The example code is as follows, > > > > <?php > > if (PHP_OS == "WIN32" || PHP_OS == "WINNT") { > > define("INCLUDE_DIR","c:\\myapps"); > > } else { > > // some other platform > > define("INCLUDE_DIR", "/include"); > > } > > ?> > > > > However, rather than checking this value, I'd like to see if flat out. > > I tried this, > > > > <?php > > echo "PHP_OS"; > > ?> > > > > But it didn't work. How can I see the value contained in this constant? > > > There is a lot of confusion on your head. > That, and you didn't read the whole thing... > The example MUST include the setting of a variable first, named PHP_OS > (or better $PHP_OS), and then proceed to the part you show. > > If you write > echo "PHP_OS"; > you are echoing a string, not the contents of a variable. > > If you do > if (PHP_OS == "WIN32" || PHP_OS == "WINNT") { > you are doing nothing, since PHP_OS is not a variable. It must be > preceded by a dollar sign. > > > -- PHP Windows Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
