子查询：尝试添加1列

发布于 2025-01-22 08:10:23 字数 582 浏览 1 评论 0原文

我是SQL的新手。实际上，我正在尝试为此添加一个新列：

SELECT ville_departement, COUNT(*) FROM villes_france_free 
GROUP BY ville_departement
HAVING ville_departement IN
(SELECT ville_departement FROM departement
INNER JOIN villes_france_free 
ON departement.departement_code = villes_france_free.ville_departement);

我有每个部门的城市数量。但是，除了Deptionlement_Code外，我还想在最终结果

问题上显示Depertament_Name：显然，我无法与1列面对1列。Dection_Code

和Dectionlement_name之间的链接定义为：有一个Deptifement_Code用于多个Deptionlement_name。

感谢您的帮助！！ :)

Léandre

原文

I'm novice with SQL. Actually, I'm trying to add a new column to this :

SELECT ville_departement, COUNT(*) FROM villes_france_free 
GROUP BY ville_departement
HAVING ville_departement IN
(SELECT ville_departement FROM departement
INNER JOIN villes_france_free 
ON departement.departement_code = villes_france_free.ville_departement);

I have the number of cities from each departements. But in addition of departement_code, I want to show departement_name on the final result

Problem : Obviously, I can't confront 1 column with 2.

The link between departement_code and departement_name is defined as : there is one departement_code for several departement_name.

Thanks for your help !! :)

Léandre

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吃不饱 2025-01-29 08:10:23

将我的awser更新到另一篇文章

SELECT 
      departement_code, 
      departement_nom,
      COUNT(*) as NombreVilles
FROM 
      villes_france_free as ville inner join
      departement on departement.departement_code = ville.ville_departement
GROUP BY 
      departement_code, 
      departement_nom
HAVING 
      ville_departement IN ( Your filter ) --- No need to put a subquery select in here

To update my awser to the other post

SELECT 
      departement_code, 
      departement_nom,
      COUNT(*) as NombreVilles
FROM 
      villes_france_free as ville inner join
      departement on departement.departement_code = ville.ville_departement
GROUP BY 
      departement_code, 
      departement_nom
HAVING 
      ville_departement IN ( Your filter ) --- No need to put a subquery select in here

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