Question

我有一个包含两列的表格：

市
国家

我想要一个包含所有City值的结果集。对于我想要包含国家/地区列的相同城市值的行。我想要的结果集的一个例子如下：

City
----------------
Chicago
New York
Toronto
London, Canada
London, England
Los Angeles

注意

兴趣点是“伦敦”的重复城市名称，包括逗号，空格和国家/地区名称以区分它们。

如何使用MySQL语句实现此目的？我使用group by和having条款尝试了一些事情，但我无法得到我想要的结果。

Answer 1

使用：

  SELECT a.city AS city
    FROM YOUR_TABLE a
GROUP BY a.city
  HAVING COUNT(a.country) = 1
UNION ALL
 SELECT CONCAT(b.city, ', ', b.country) AS city
   FROM YOUR_TABLE b
  WHERE EXISTS (SELECT NULL
                  FROM YOUR_TABLE c
                 WHERE c.city = b.city
              GROUP BY c.city
                HAVING COUNT(c.country) > 1)
ORDER BY city

Answer 2

您可以使用以下SQL语句，假设cityTable中的表名称

    SELECT cityTable.City  , cityTable.Country  FROM cityTable INNER JOIN ( SELECT City FROM cityTable GROUP BY City HAVING COUNT(City) > 1 ) as tempTable ON cityTable.City = tempTable.City

    SELECT City FROM Table GROUP BY City HAVING  COUNT(City) = 1

第一个查询将为所有重复城市提供国家/地区名称，第二个查询将为所有不重复的城市提供仅城市名称。

Answer 3

accepted answer完全可行。这是解决问题的另一种方法：

   SELECT CASE
          WHEN repeated IS NULL THEN city
          ELSE                       city || ', ' || country
          END AS city
LEFT JOIN (SELECT city AS repeated
             FROM cities
         GROUP BY 1
           HAVING COUNT(1) > 1) d
          ON cities.city = d.repeated

Answer 4

这在MySQL中不起作用，它还不支持窗口化（AFAIK！），但是对于记录，在做的RDBMS中，你可以做到

SELECT CASE WHEN (count(*) > 1) THEN (city || ', ' || country) ELSE city END
OVER (PARTITION BY city)
FROM city_table;

Answer 5

DELETE FROM table WHERE city IN (
SELECT city, 
 COUNT(city) AS NumOccurrences
FROM table
GROUP BY city
HAVING ( COUNT(city) > 1 )

如何为第一列上具有重复项的行选择第二列？

注意

5 个答案: