Don't allow multiple --user options to mysqld

(First one is used)

scripts/safe_mysqld.sh

@@ -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;--[^=]*=;;"` ;;

sql/mysqld.cc

@@ -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':