Perl 按多列对数据库中的现有用户进行排序

我使用了排序函数

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

要根据开始日期筛选用户，如果按用户名排序相等，则数据库如下所示：

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

以下排序有时以以下模式显示用户

user2, user3, user1, user4

其他时间

user1, user2, user3, user4.

---

编辑

我使用

DBIC_TRACE=1运行，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这是因为与具有级别1的用户相比，user5在数据库中添加的时间要晚一秒，因此它们按该顺序显示，因为它们是按数据按函数顺序降序排列的。\u begin.
请向我们展示输出有序用户名的代码，因为您的函数没有问题。这应该可以。。。您可以通过设置环境变量
DBIC_TRACE
（
export DBIC_TRACE=1
）来实现吗？然后可以在控制台或日志中检查SQL查询的文本。
orderby
子句是否始终相同？谢谢跟踪。您可以看到查询中的
orderby
子句与您的代码匹配：
orderby date\u begin DESC，username ASC
。您当时在代码中获得了正确顺序的结果吗？有时我得到了正确的答案，有时我没有，这是不可靠的。它必须显示user1和user2，但显示为user5和user1。此外，我还看到顺序by正常工作。