CentOS remove press any key to enter the menu 2017-02-18 23:45

After fresh installing CentOS and starting the system there is a sentence on the screen "Press any key to enter the menu, Booting boot in __seconds". Editing the /etc/grub.conf to remove the sentence and start CentOS quickly. The content is like following.

# grub.conf generated by anaconda 
# Note that you do not have to rerun grub after making changes to this file 
# NOTICE:  You have a /boot partition.  This means that 
#          all kernel and initrd paths are relative to /boot/, eg. 
#          root (hd0,0) 
#          kernel /vmlinuz-version ro root=/dev/mapper/VolGroup-lv_root 
#          initrd /initrd-[generic-]version.img 
title CentOS 6 (2.6.32-642.el6.i686) 
        root (hd0,0) 
        kernel /vmlinuz-2.6.32-642.el6.i686 ro root=/dev/mapper/VolGroup-lv_root rd_NO_LUKS LANG=en_US.UTF-8 rd_NO_MD rd_LVM_LV=VolGroup/lv_swap SYSFONT=latarcyrheb-sun16 crashkernel=auto rd_LVM_LV=VolGroup/lv_root  KEYBOARDTYPE=pc KEYTABLE=us rd_NO_DM rhgb quiet 
        initrd /initramfs-2.6.32-642.el6.i686.img

Change the value of timeout to 0 and save the file. Next time you start the system "Press any key to enter the menu, Booting boot in __seconds" will be disappeared.

If you want know the detail click here -> The GRUB Boot Loader