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Don't allow multiple --user options to mysqld

(First one is used)


scripts/safe_mysqld.sh:
  Don't allow multiple --user options
sql/mysqld.cc:
  Don't allow multiple --user options
This commit is contained in:
unknown
2003-03-10 12:24:22 +02:00
parent 281ad254dc
commit 73c2d4ad49
2 changed files with 10 additions and 3 deletions

View File

@ -33,8 +33,12 @@ parse_arguments() {
--basedir=*) MY_BASEDIR_VERSION=`echo "$arg" | sed -e "s;--[^=]*=;;"` ;;
--datadir=*) DATADIR=`echo "$arg" | sed -e "s;--[^=]*=;;"` ;;
--pid-file=*) pid_file=`echo "$arg" | sed -e "s;--[^=]*=;;"` ;;
--user=*) user=`echo "$arg" | sed -e "s;--[^=]*=;;"` ; SET_USER=1 ;;
--user=*)
if [ $SET_USER == 0 ]
then
user=`echo "$arg" | sed -e "s;--[^=]*=;;"` ; SET_USER=1
fi
;;
# these two might have been set in a [safe_mysqld] section of my.cnf
# they get passed via environment variables to safe_mysqld
--socket=*) MYSQL_UNIX_PORT=`echo "$arg" | sed -e "s;--[^=]*=;;"` ;;

View File

@ -3639,7 +3639,10 @@ static void get_options(int argc,char **argv)
use_temp_pool=1;
break;
case 'u':
mysqld_user=optarg;
if (!mysqld_user)
mysqld_user=optarg;
else
fprintf(stderr, "Warning: Ignoring user change to '%s' becasue the user is set to '%s' earlier on the command line\n", optarg, mysqld_user);
break;
case 'v':
case 'V':