LinuxQuestions.org - safe_mysqld:command not found

Hello all,

Trying to get a simple website up and running from my home but mysql is being a pain in my butt.

It installed fine and all that. When I try running safe_mysqld& as root I get:

Code:

/usr/bin/safe_mysqld: line 162: 5: command not found

Starting mysqld daemon with databases from /usr/var

030124 19:42:44  mysqld ended

The relevant section from the script is:

Code:

NOHUP_NICENESS="nohup"

if test -w /

then

  NOHUP_NICENESS=`nohup nice 2>&1`

  if test $? -eq 0 && test x"$NOHUP_NICENESS" != x0 && nice --1 echo foo > /dev/null 2>&1

  then

    if $NOHUP_NICENESS -gt 0

    then

      $NOHUP_NICENESS="nice --$NOHUP_NICENESS nohup"

    else

      NOHUP_NICENESS="nice -$NOHUP_NICENESS nohup"

    fi

  else

  NOHUP_NICENESS="nohup"

  fi

fi

And line 162:5 is "if"?!?!?
(if $NOHUP_NICENESS -gt 0)
I have tried everything I can think of and can't think of anything else to troubleshoot. I am using Arch Linux on a i686.

TIA