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do not put --user into command line if the value comes from [mysqld] or [server] sections

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Bug #2163
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unknown
2003-12-19 23:08:20 +01:00
parent fb36199281
commit cb12a8b827
1 changed files with 9 additions and 3 deletions
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12
scripts/mysqld_safe.sh
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@ -46,8 +46,9 @@ parse_arguments() {
--user=*) --user=*)
if test $SET_USER -eq 0 if test $SET_USER -eq 0
then then
user=`echo "$arg" | sed -e "s;--[^=]*=;;"` ; SET_USER=1 user=`echo "$arg" | sed -e "s;--[^=]*=;;"`
fi fi
SET_USER=1
;; ;;
# these two might have been set in a [mysqld_safe] section of my.cnf # these two might have been set in a [mysqld_safe] section of my.cnf
@ -129,7 +130,6 @@ fi
# these rely on $DATADIR by default, so we'll set them later on # these rely on $DATADIR by default, so we'll set them later on
pid_file= pid_file=
err_log= err_log=
SET_USER=0
# Get first arguments from the my.cnf file, groups [mysqld] and [mysqld_safe] # Get first arguments from the my.cnf file, groups [mysqld] and [mysqld_safe]
# and then merge with the command line arguments # and then merge with the command line arguments
@ -147,7 +147,13 @@ else
fi fi
args= args=
parse_arguments `$print_defaults --loose-verbose $defaults mysqld server mysqld_safe safe_mysqld` SET_USER=2
parse_arguments `$print_defaults --loose-verbose $defaults mysqld server`
if test $SET_USER -eq 2
then
SET_USER=0
fi
parse_arguments `$print_defaults --loose-verbose $defaults mysqld_safe safe_mysqld`
parse_arguments PICK-ARGS-FROM-ARGV "$@" parse_arguments PICK-ARGS-FROM-ARGV "$@"
if test ! -x $ledir/$MYSQLD if test ! -x $ledir/$MYSQLD