Question

我在这里尝试了最好的答案：Select count / duplicates 并没有得到准确的结果，也许是因为这个例子适用于一个领域？

我原本试过这个，但似乎也没有用：

   SELECT T.STATE, T.CITY, COUNT(T2.CITY)
     FROM (SELECT DISTINCT T.STATE, T.CITY 
             FROM TABLE ) T
LEFT JOIN TABLE T2 ON T2.STATE = T.STATE 
                  AND T2.CITY = T.CITY
 GROUP BY T.STATE, T.CITY

这也会产生不准确的结果。我正在尝试获取表中2个字段的重复值的计数。

示例数据：

 Los Angeles, CA, Bob
 New York, NY, Bill
 Los Angeles, CA, Jim
 Los Angeles, CA, Joe
 New York, NY, Gary
 Los Angeles, CA, Roger

应该回归：

 Los Angeles, CA, 4
 New York, NY, 2

建议？

Answer 1

DECLARE @TABLE TABLE(
    CITY VARCHAR(100),
    [STATE] CHAR(2),
    NAME VARCHAR(50)
)

INSERT INTO @TABLE VALUES
 ('Los Angeles', 'CA', 'Bob'),
 ('New York', 'NY', 'Bill'),
 ('Los Angeles', 'CA', 'Jim'),
 ('Los Angeles', 'CA', 'Joe'),
 ('New York', 'NY', 'Gary'),
 ('Los Angeles', 'CA', 'Roger')

 SELECT CITY,[STATE],COUNT(*) [COUNT]
 FROM @TABLE
 GROUP BY CITY,[STATE]
 HAVING COUNT(*) > 1


 --if there are erroneous white spaces and inconsistant case, and collation is case sensitive use this
 DELETE FROM @TABLE
 INSERT INTO @TABLE VALUES
 ('  Los Angeles', 'CA', 'Bob'),
 ('New York', 'NY', 'Bill'),
 ('Los ANgeles   ', 'CA', 'Jim'),
 ('LOS Angeles', 'CA', 'Joe'),
 ('New YorK  ', 'NY', 'Gary'),
 ('  Los Angeles ', 'CA', 'Roger')

 SELECT RTRIM(LTRIM(UPPER(CITY))) [CITY] ,RTRIM(LTRIM(UPPER([STATE]))) [STATE],COUNT(*) [COUNT]
 FROM @TABLE
 GROUP BY RTRIM(LTRIM(UPPER(CITY))),RTRIM(LTRIM(UPPER([STATE])))
 HAVING COUNT(*) > 1

Answer 2

根据我的理解，您希望获得所有具有计数的记录＆gt; 1。

select T.State, T.City, COUNT(1)
FROM TABLE T
GROUP BY T.State, T.City
HAVING COUNT(1) > 1

Answer 3

SELECT T.STATE, T.CITY, COUNT(*) as cnt
FROM Table X T
GROUP BY T.STATE, T.CITY
HAVING COUNT(*)>1

如果输入不准确，您可以尝试

SELECT RTRIM(LTRIM(T.STATE)), LTRIM(RTRIM(T.CITY)), COUNT(*) as cnt
FROM Table X T
GROUP BY RTRIM(LTRIM(T.STATE)), LTRIM(RTRIM(T.CITY))
HAVING COUNT(*)>1

计算具有2个重复/匹配列的行

3 个答案: