【问题标题】:ERROR in GROUP_BY with query getting from two databases CODEIGNITERGROUP_BY 中的错误,查询来自两个数据库 CODEIGNITER
【发布时间】:2026-01-25 09:00:01
【问题描述】:

我有一个查询,它连接来自两个不同数据库的表。它已经显示了结果,但是我只想显示独特的结果,因为有些结果是多余的。所以我添加了一个 GROUP BY 以仅获取唯一结果但出现错误。

这是我的代码:

public function search_results_accommodations($location,$from_date,$to_date,$bedroom,$guests)
{
    $this->db->select('*, akzapier.bookings.id as BOOKING_ID, akzapier.properties.id as PROPERTY_ID, ci_alexandrohomes.assigned_property.ID as ASSIGNED_PROPERTY_ID, ci_alexandrohomes.listings.ID as LISTING_ID');
    $this->db->from('akzapier.bookings');
    $this->db->join('akzapier.properties', 'akzapier.properties.id=akzapier.bookings.property_id', 'inner');
    $this->db->join('ci_alexandrohomes.assigned_property', 'ci_alexandrohomes.assigned_property.property_id=akzapier.properties.id', 'inner');
    $this->db->join('ci_alexandrohomes.listings', 'ci_alexandrohomes.listings.ID=ci_alexandrohomes.assigned_property.listing_id');
    $this->db->where('akzapier.bookings.check_in !=', $from_date);
    $this->db->where('akzapier.bookings.check_out !=', $to_date);
    $this->db->where('ci_alexandrohomes.listings.city', $location);
    $this->db->where('ci_alexandrohomes.listings.bedrooms', $bedroom);
    $this->db->where('ci_alexandrohomes.listings.guests', $guests);
    $this->db->group_by('akzapier.properties.id', 'ASC')
    $query = $this->db->get();
    return $query->result();
}

错误没有显示在页面中,所以我将其转换为 SQL 以查看实际情况:

SELECT * akzapier.bookings.id as BOOKING_ID, akzapier.properties.id as PROPERTY_ID, ci_alexandrohomes.assigned_property.ID as ASSIGNED_PROPERTY_ID, ci_alexandrohomes.listings.ID as LISTING_ID
FROM akzapier.bookings 
INNER JOIN akzapier.properties ON akzapier.properties.id=akzapier.bookings.property_id
INNER JOIN ci_alexandrohomes.assigned_property ON ci_alexandrohomes.assigned_property.property_id=akzapier.properties.id
INNER JOIN ci_alexandrohomes.listings ON ci_alexandrohomes.listings.ID=ci_alexandrohomes.assigned_property.listing_id
WHERE akzapier.bookings.check_in != '2019-09-21'
AND akzapier.bookings.check_out != '2019-09-30'
AND ci_alexandrohomes.listings.city = ‘1’
AND ci_alexandrohomes.listings.bedrooms = '2'
AND ci_alexandrohomes.listings.guests = '4'
GROUP BY akzapier.bookings.property_id ASC

错误提示:

1055 - SELECT 列表的表达式 #1 不在 GROUP BY 子句中,并且包含在功能上不依赖于 GROUP BY 子句中的列的非聚合列 'akzapier.bookings.id';这与 sql_mode=only_full_group_by 不兼容

【问题讨论】:

标签: php mysql codeigniter activerecord


【解决方案1】:

一般来说,当您使用 GROUP BY 时,您的 SELECT 语句必须包含聚合(例如 MAX(...)、COUNT(...) 等)或列必须出现在 GROUP BY 中。您已选择所有带星号 * 的非聚合字段。在这种情况下,它抱怨的是 akzapier.bookings.id 字段,该字段既不是聚合的,也不是在您的 GROUP BY 中。

如果您确实想要唯一值,请尝试 SELECT DISTINCT,这将从结果中删除重复的行。

【讨论】:

  • 你能告诉我它是怎么做的吗?我有点不知所措
  • 只需将 SELECT 替换为 SELECT DISTINCT 并删除 GROUP BY。如果这不符合您的要求,请更新您的问题以更具体(即输出有什么问题以及您希望它如何不同。)
  • 我不能只使用 distinct,因为我需要很多其他列