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perl - 按多列对数据库中的现有用户进行排序

转载 作者:行者123 更新时间:2023-12-02 04:22:29 26 4
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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

最佳答案

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。

关于perl - 按多列对数据库中的现有用户进行排序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58568732/

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