我使用了排序功能

sub sort_users { 
  my $self = shift; 
  return $self->search(undef, { order_by => [{ -desc => 'data_begin' }, 
    { -asc => 'username' }]}); 
} 

根据开始日期过滤用户，如果相等则按用户名排序，数据库如下所示:

user1 2019-09-26 00:00:00 
user2 2019-09-26 00:00:00 
user3 2019-09-26 00:00:00 
user4 2019-09-26 00:00:00 

以下排序有时会以以下方式显示用户

user2, user3, user1, user4 

其他时间

user1, user2, user3, user4. 

编辑

我使用 DBIC_TRACE = 1 运行并得到:

SELECT COUNT( * ) FROM user_access_rights me WHERE ( ( access_rights_id = ? AND date_end >= ? ) ): 'level1', '2019-10-28 07:43:30+0000' 
SELECT access_rights_recipient.username, access_rights_recipient.last_login, me.access_rights_id FROM user_access_rights me  JOIN users access_rights_recipient ON access_rights_recipient.username = me.username WHERE ( ( access_rights_id = ? AND date_end >= ? ) ) ORDER BY date_begin DESC, username ASC LIMIT ?: 'level1', '2019-10-28 07:43:30+0000', '2' 
 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level1', '2019-10-28T07:43:29', '2019-10-29T07:43:29', 'user1' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level1', '2019-10-28T07:43:29', '2019-10-29T07:43:29', 'user2' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level1', '2019-10-28T07:43:29', '2019-10-29T07:43:29', 'user3' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level1', '2019-10-28T07:43:29', '2019-10-29T07:43:29', 'user4' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level1', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user5' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user6' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user7' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user8' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user9' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user10' 
INSERT INTO user_access_rights ( access_rights_id, date_begin, date_end, username) VALUES ( ?, ?, ?, ? ) RETURNING id: 'level2', '2019-10-28T07:43:30', '2019-10-29T07:43:30', 'user11' 

要求显示 user1 和 user2，而不是显示 user5 和 user1

请您参考如下方法:

这是因为与具有级别 1 的用户相比，user5 在数据库中的添加时间晚了一秒钟，因此他们按该顺序显示，因为他们按 data_begin 的功能降序排列。