Nomally when i run it. it will return the page after task2 is completed.
task1();
print "task1 done.\n";
task2();
print "task2 done.\n";
I wonder if is there any why to print " task1 done" right before run task2()
so even task2 takes time running, but i still get the task 1 result in early time.
Many Thanks and Merry Xmas to all