You are confused :)
What you did is:
echo expr1 . expr2;
meaning echo the expr1 concatenated with expr2. Both expressions are
evaluated before the concatenation occurs. so print_r(posix_uname()) is
evaluated first and prints out the posix uname info. The return value is 1
for success so what you get is echo expr1 . 1;
You should do:
echo "posix_uname:";
print_r(posix_uname());
echo "\n\n";
Andi
At 03:55 PM 4/29/2001 -0400, Scott Wolf wrote:
>Here is the script I was running to get some info.
>
><?
>echo "<html><head><title>System Info</title></head><body>";
>echo "<pre>";
>echo "posix_getpid: " . posix_getpid() . "\n\n";
>echo "posix_getppid: " . posix_getppid() . "\n\n";
>echo "posix_uname: " . print_r(posix_uname()) . "\n\n";
>echo "posix_times: " . print_r(posix_times()) . "\n\n";
>echo "</pre>";
>echo "</body></html>";
>?>
>
>Here is the output.
>
>
>posix_getpid: 1741
>
>posix_getppid: 1735
>
>Array
>(
> [sysname] => Linux
> [nodename] => red
> [release] => 2.4.1
> [version] => #1 Tue Feb 20 11:09:36 EST 2001
> [machine] => i686
>)
>posix_uname: 1
>
>Array
>(
> [ticks] => 345752069
> [utime] => 4
> [stime] => 0
> [cutime] => 0
> [cstime] => 0
>)
>posix_times: 1
>
>
>
>
>
>Either I am really confused or the output is being made in the wrong order.
>
>This is the latest cvs as of 3:30 EST with Apache 1.3 on linux.
>
>
>
>
>
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