Question

很抱歉，如果我没有正确解释这一点，我对SQL相对较新。

在oracle中有一个描述属性的表（城市，房产类型，每月租金，其他信息）

我的问题是：假设3种独特的房产类型（酒店，房屋，空地），我如何显示哪些城市没有所有3种类型的房产？

Answer 1

GROUP BY解决方案，确保每个城市返回的属性类型少于3种：

select city
from tablename
where property_type in ('hotel', 'house', 'empty lot')
group by city
having count(distinct property_type) < 3

Answer 2

您的SQL查询应该是

SELECT City 来自YourTable 酒店＆lt;＆gt; 'hotelname'和house＆lt;＆gt; 'housename'和emptylot＆lt;＆gt; '名称'

假设酒店，House，Emptylot是您数据库中的列名。

Answer 3

（已修改）尝试此查询：

  select t.city  from table_name t 
  where t.city NOT IN 
  (select city from table_name 
   where ( property_type ='hotel' or 
           property_type ='house' or 
           property_type ='Empty lot')
  );

（查询返回不存在所有三种属性的城市）：

 select t.city from table t inner join 
  (select city from table 
   where property_type not in ('House','Hotel','Empty lot')) x 
 on t.city=x.city 
 group by t.city 
 having count(*)<3 ;

Answer 4

可能是这样的：

SELECT City
FROM YourTable
WHERE [property type] != 'hotel' OR
      [property type] != 'empty lot' OR
      [property type] != 'house'

Answer 5

有两种方法，

GROUP BY
分析COUNT() OVER()

例如，假设我有3个城市的样本数据，其中城市1的所有属性类型都满意，其他城市则没有所有必需的属性类型。

使用GROUP BY

SQL> -- sample table data
SQL> WITH DATA AS(
  2  SELECT 1 city, 'hotel' property FROM dual UNION ALL
  3  SELECT 1 city, 'house' property FROM dual UNION ALL
  4  SELECT 1 city, 'empty' property FROM dual UNION ALL
  5  SELECT 2 city, 'hotel' property FROM dual UNION ALL
  6  SELECT 2 city, 'house' property FROM dual UNION ALL
  7  SELECT 2 city, 'scrap' property FROM dual UNION ALL
  8  SELECT 3 city, 'empty' property FROM dual UNION ALL
  9  select 3 city, 'house' property from dual
 10  )
 11  -- query
 12  SELECT city
 13     FROM data
 14   WHERE property IN ('hotel', 'house', 'empty')
 15  GROUP BY city
 16  HAVING COUNT(property) < 3
 17  /

      CITY
----------
         2
         3

SQL>

使用分析COUNT（）OVER（）

SQL> -- sample table data
SQL> WITH DATA AS(
  2  SELECT 1 city, 'hotel' property FROM dual UNION ALL
  3  SELECT 1 city, 'house' property FROM dual UNION ALL
  4  SELECT 1 city, 'empty' property FROM dual UNION ALL
  5  SELECT 2 city, 'hotel' property FROM dual UNION ALL
  6  SELECT 2 city, 'house' property FROM dual UNION ALL
  7  SELECT 2 city, 'scrap' property FROM dual UNION ALL
  8  SELECT 3 city, 'empty' property FROM dual UNION ALL
  9  select 3 city, 'house' property from dual
 10  )
 11  -- query
 12  SELECT DISTINCT city
 13  FROM
 14    (SELECT t.* ,
 15      COUNT(property) OVER(PARTITION BY city ORDER BY city) rn
 16    FROM DATA t
 17    WHERE property IN ('hotel', 'house', 'empty')
 18    )
 19  WHERE rn < 3
 20  /

      CITY
----------
         2
         3

SQL>

Oracle SQL显示来自多个记录的数据

5 个答案: