Question

请帮助我创建一个查询，以确定下表中的最小date_time：

 ID  |  Name  |  Date_Time | Location
---------------------------------------
 001 |  John  | 01/01/2015 |   901
 001 |  john  | 02/01/2015 |   903
 001 |  john  | 05/01/2015 |   905
 001 |  john  | 06/01/2015 |   904
 002 |  Jack  | 01/01/2015 |   903
 002 |  Jack  | 03/01/2015 |   904
 002 |  Jack  | 04/01/2015 |   905
 003 |  Sam   | 01/01/2015 |   904
 003 |  Sam   | 03/01/2015 |   903
 003 |  Sam   | 04/01/2015 |   901
 003 |  Sam   | 06/01/2015 |   903

我尝试了这个查询：

SELECT ID, NAME, MIN(DATE_TIME), LOCATION
  FROM TABLE
 GROUP BY (ID)

但我收到此错误消息：

ORA-00979: not a GROUP BY expression

Answer 1

如果使用聚合功能，则指定应该应用聚合的字段。所以你正在使用group by子句。在这种情况下，您可能需要找到每个id，名称组合的最小date_time。

 select id, name, min(date_time) 
              from my_table group by id, name

Answer 2

当您对某些内容进行分组时，所有其他行将保持群集到该分组键。对于密钥，您只能在SELECT中获取行（实体）之一。

快捷方式是，GROUP BY中的SELECT可以自由地进行AGGREGATE。否则，它们必须包含在001函数中。

按ID分组时，

SELECT键有4行聚集在它上面。想想，当你在MIN(date)中指定非分组列时会发生什么。在使用SELECT ID,MIN(NAME),MIN(LOCATION),MIN(DATE) FROM TABLE GROUP BY ID时，在4个日期中，至少需要一个。

因此，您的查询必须是

SELECT ID,LOCATION,NAME,MIN(DATE)
FROM TABLE
GROUP BY ID,LOCATION,NAME

OR

SELECT ID,LOCATION,DATE,MIN(DATE) OVER(PARTITION BY ID ORDER BY NULL) AS MIN_DATE
FROM TABLE.

OR

分析方法。

SELECT T.ID,T.NAME,T.LOCATION,MIN_DATE
FROM
 (
   SELECT ID,MIN(DATE) AS MIN_DATE
   FROM TABLE T1
   GROUP BY ID
 ) AGG, TABLE T
 WHERE T.ID = AGG.ID
 AND T.DATE = AGG.MIN_DATE

然而，关于如何重写查询，这取决于要求。

编辑：要获取与最短日期相对应的行，我们可以创建一个类似下面的SELF JOIN。

SELECT ID,NAME,LOCATION,MIN_DATE
FROM
( 
  SELECT ID,
         NAME,
         LOCATION,
         MIN(DATE) OVER(PARTITION BY ID ORDER BY NULL) MIN_DATE,
         ROW_NUMBER() OVER(PARTITION BY ID ORDER BY NULL) RNK
  FROM TABLE
)
WHERE RNK = 1;

OR

char   *pri_key[] = "some key"; // ---> some key, that i've got from server

RSA *rsa;
BIO *keybio;
keybio = BIO_new_mem_buf(pri_key, strlen(pri_key));
rsa = PEM_read_bio_RSAPrivateKey(keybio, &rsa, NULL, NULL);

// Decrypt it
// Encoded message is in buff
char *decrypt = new char[BUFF_SIZE];
int decrypt_len = RSA_private_decrypt(BUFF_SIZE, (unsigned char*)buff, (unsigned char*)decrypt,
    rsa, RSA_PKCS1_OAEP_PADDING); // ------> it fails here

Answer 3

尝试对所有其他列进行分组...如果表名为'table'，请尝试将表名更改为架构中的其他名称。

SELECT ID , NAME , MIN(DATE_TIME) , LOCATION FROM TABLE  GROUP BY ID, Name, Location

Answer 4

选择t1.name，t1.id，t1.location，t1.date from（select id，MIN（Date）as min_date from table group by id）t2 inner join TABLE t1 on t1.date = t2.min_date and t1.id = t2.id;

使用聚合函数返回最小值

4 个答案: