harry gaillac wrote:
hello,
use database apache serweb logs to find problem --- Chris Mason lists@masonc.com a écrit :
Harry: I found the problem as shown below. The applicication is reuqesting a column that does not exist in the database. I checked the contents of scripts/ser_mysql.sh from ser-0.9.3_src.tar.gz ftp://ftp.berlios.de/pub/ser/0.9.3/src/ser-0.9.3_src.tar.gz and find that column does not exist in the subscriber table. There's no way I can get it to work. Is the ser distribution broken or the serweb?
Chris
Query as executed by user ser
####################################################################### mysql> select phplib_id from subscriber where username='admin' and password='heslo' and perms='admin' and domain='mason.home' -> ; ERROR 1054 (42S22): Unknown column 'perms' in 'where clause'
####################################################################### create table statement from scripts/ser_mysql.sh #######################################################################
# # Table structure for table 'subscriber' -- user database #
CREATE TABLE subscriber ( phplib_id varchar(32) NOT NULL default '', $USERCOL varchar(64) NOT NULL default '', domain varchar(128) NOT NULL default '', password varchar(25) NOT NULL default '', first_name varchar(25) NOT NULL default '', last_name varchar(45) NOT NULL default '', phone varchar(15) NOT NULL default '', email_address varchar(50) NOT NULL default '', datetime_created datetime NOT NULL default '0000-00-00 00:00:00', datetime_modified datetime NOT NULL default '0000-00-00 00:00:00', confirmation varchar(64) NOT NULL default '', flag char(1) NOT NULL default 'o', sendnotification varchar(50) NOT NULL default '', greeting varchar(50) NOT NULL default '', ha1 varchar(128) NOT NULL default '', ha1b varchar(128) NOT NULL default '', allow_find char(1) NOT NULL default '0', timezone varchar(128) default NULL, rpid varchar(128) default NULL, domn int(10) default NULL, uuid varchar(64) default NULL, UNIQUE KEY phplib_id (phplib_id), PRIMARY KEY ($USERCOL, domain), KEY user_2 ($USERCOL) ) $TABLE_TYPE;