我有以下两个联想：
日程日期hasOne图像说明
日程日期hasMany照片
当我执行此查询时：

$dates = $this->AgendaDates
                        ->find()
                        ->order(['AgendaDates.date' => 'ASC'])
                        ->contain(['ImageDescriptive' => function ($q) { // AgendaDates hasOne ImageDescriptive
                            return $q->find('translations');
                        }])
                        ->limit(2);

我有以下错误：

SQLSTATE[42000]：语法错误或访问冲突：1055 ORDER BY子句的表达式#1不在GROUP BY子句中，并且包含非聚集列'brigittepatient.AgendaDates.date'，该列在功能上与GROUP BY子句中的列无关;这与sql_mode=only_full_group_by不兼容

返回错误的SQL查询是：

SELECT 
  FichiersI18n.id AS FichiersI18n__id, 
  FichiersI18n.locale AS FichiersI18n__locale, 
  FichiersI18n.model AS FichiersI18n__model, 
  FichiersI18n.foreign_key AS FichiersI18n__foreign_key, 
  FichiersI18n.field AS FichiersI18n__field, 
  FichiersI18n.content AS FichiersI18n__content 
FROM 
  fichiers_i18n FichiersI18n 
  INNER JOIN (
    SELECT 
      (ImageDescriptive.id) 
    FROM 
      agenda_dates AgendaDates 
      LEFT JOIN fichiers ImageDescriptive ON (
        ImageDescriptive.model = 'AgendaDates' 
        AND ImageDescriptive.field = 'image_descriptive' 
        AND AgendaDates.id = ImageDescriptive.foreign_key
      ) 
    GROUP BY 
      ImageDescriptive.id 
    ORDER BY 
      AgendaDates.date ASC 
    LIMIT 
      2
  ) ImageDescriptive ON FichiersI18n.foreign_key = ImageDescriptive.id 
WHERE 
  FichiersI18n.model = 'Fichiers'

奇怪的是，当我做同样的事情，但与照片（链接到议程日期的hasMany，而不是hasOne）：

$dates = $this->AgendaDates
                        ->find()
                        ->order(['AgendaDates.date' => 'ASC'])
                        ->contain(['Photos' => function ($q) { // AgendaDates hasMany Photos
                            return $q->find('translations');
                        }])
                        ->limit(2);

我没有错误，查询如下所示：

SELECT 
  FichiersI18n.id AS FichiersI18n__id, 
  FichiersI18n.locale AS FichiersI18n__locale, 
  FichiersI18n.model AS FichiersI18n__model, 
  FichiersI18n.foreign_key AS FichiersI18n__foreign_key, 
  FichiersI18n.field AS FichiersI18n__field, 
  FichiersI18n.content AS FichiersI18n__content 
FROM 
  fichiers_i18n FichiersI18n 
  INNER JOIN (
    SELECT 
      (Photos.id) 
    FROM 
      fichiers Photos 
    WHERE 
      (
        Photos.model = 'AgendaDates' 
        AND Photos.field = 'photos' 
        AND Photos.foreign_key in (5, 3)
      ) 
    GROUP BY 
      Photos.id
  ) Photos ON FichiersI18n.foreign_key = Photos.id 
WHERE 
  FichiersI18n.model = 'Fichiers'

我真的不明白为什么我的cakephp查询在hasOne关联的情况下会抛出错误，而在hasMany关联的情况下却不会？