Mysql Clojure HoneySQL-如何将连接后的字符串值聚合到一行中?
我正在执行以下查询,该查询跨3个表联接,以提取训练和与之相关的标记Mysql Clojure HoneySQL-如何将连接后的字符串值聚合到一行中?,mysql,clojure,honeysql,Mysql,Clojure,Honeysql,我正在执行以下查询,该查询跨3个表联接,以提取训练和与之相关的标记 (db/query {:select [:workouts.id :workouts.name :tag.tag_name] :from [:workouts] :left-join [[:workout_tags :workout_tag] [:= :workout_tag.workout_id :workouts.id]
(db/query {:select [:workouts.id :workouts.name :tag.tag_name]
:from [:workouts]
:left-join [[:workout_tags :workout_tag] [:= :workout_tag.workout_id :workouts.id]
[:tags :tag] [:= :tag.tag_id :workout_tag.tag_id]]
:where [:= :workouts.id 1]}))
这将返回以下内容:
({:id 1, :name "Short", :tag_name "cardio"}
{:id 1, :name "Short", :tag_name "No weights"})
理想情况下,我希望向最终用户返回一个结果,tag\u name
合并到一个字段中。比如:
{:id 1, :name "Short", :tag_name ["cardio" "No weights"]}
它会发现我可以很容易地完成这项工作,但我想看看是否有一个内置的MySQL函数可以完成我希望完成的任务。似乎
GROUP\u CONACT
可能会实现我想要的功能,但在HoneySQL环境下,我似乎无法让它正常工作。我会在事后再做:
(let [data [{:id 1, :name "Short", :tag_name "cardio"}
{:id 1, :name "Short", :tag_name "No weights"}
{:id 2, :name "Long", :tag_name "Tall"}
{:id 2, :name "Long", :tag_name "Hills"}]
grouped (group-by :id data)
id-tags (vec (for [[id data-maps] grouped]
(let [tags (mapv :tag_name data-maps)]
{:id id :tags tags})))]
(is= id-tags
[{:id 1, :tags ["cardio" "No weights"]}
{:id 2, :tags ["Tall" "Hills"]}]))
中间结果分组
如下所示
grouped =>
{1
[{:id 1, :name "Short", :tag_name "cardio"}
{:id 1, :name "Short", :tag_name "No weights"}],
2
[{:id 2, :name "Long", :tag_name "Tall"}
{:id 2, :name "Long", :tag_name "Hills"}]}
有关完整配置的详细信息,请参阅。以下内容应接近您的需要:
(require '[honeysql.core :as hc])
(hc/format {:select [:workouts.id :workouts.name [(hc/call :group_concat :tag.tag_name) :tag_name]]
:from [:workouts]
:left-join [[:workout_tags :workout_tag] [:= :workout_tag.workout_id :workouts.id]
[:tags :tag] [:= :tag.tag_id :workout_tag.tag_id]]
:where [:= :workouts.id 1]
:group-by [:tag.tag_name]})
这将生成以下SQL:
SELECT workouts.id, workouts.name, group_concat(tag.tag_name) AS tag_name
FROM workouts
LEFT JOIN workout_tags workout_tag ON workout_tag.workout_id = workouts.id
LEFT JOIN tags tag ON tag.tag_id = workout_tag.tag_id
WHERE workouts.id = ?
GROUP BY tag.tag_name
你没有展示你尝试过什么或失败了什么,这肯定会有助于我们确定我们应该建议什么作为答案