Question

我正在尝试使用两个日期之间的uniq序列号获取所有记录。但我无法写出正确的sql。

这是我当前的SQL查询和结果。共有3条记录，带有两个XYZABC1230序列号。

select * from t_recorded_test
where  office_id = '1710011001123'
and    record_date > '2017-11-01'
and    record_date < '2017-12-08'
and    test_result = 'true';

MODEM_MODEL MODEM_SERIAL_NUMBER OFFICE_ID       RECORD_DATE             RECORD_ID                           TEST_RESULT
Type2ModelB XYZABC1230          1710011001123   2017-11-01 19:35:54.0   ccf57f20d585424abc9bce781ada9dcc    TRUE
Type2ModelB XYZABC1230          1710011001123   2017-11-01 19:33:54.0   168ce13ed9644f128f7769432ad6ba2f    TRUE
Type2ModelB XYZABC12312         1710011001123   2017-12-03 19:33:54.0   ab727f836c354f159703565b9eed3331    TRUE

但我希望只有两个记录XYZABC1230和XYZABC12312。我该怎么办？

编辑：

对于所有相同的序列号，我只需要一个结果。日期并不重要，但我必须将其用于限制。

Answer 1

--This is for MSSQL Server
    with CTE as(
    select *,(row_number() over (partition by MODEM_SERIAL_NUMBER order by MODEM_SERIAL_NUMBER)) 'ROW_N'  from T_RECORDED_TEST   
    where OFFICE_ID='1710011001123' and RECORD_DATE > '2017-11-01' and RECORD_DATE < '2017-12-08' and TEST_RESULT = 'TRUE')
    select MODEM_MODEL,MODEM_SERIAL_NUMBER,OFFICE_ID,RECORD_DATE,RECORD_ID,TEST_RESULT  from CTE where ROW_N=1

Answer 2

使用GROUP BY函数和MAX(ROWID)，假设您希望组中的最新行按值。

SELECT * FROM T_RECORDED_TEST 
WHERE ROWID IN (
    SELECT MAX(ROWID)
    FROM T_RECORDED_TEST
    where office_id='1710011001123' 
    and record_date > '2017-11-01' 
    and record_date < '2017-12-08' 
    and test_result = 'true' 
    GROUP BY MODEM_SERIAL_NUMBER;
)

Answer 3

*是罪魁祸首。如果您只对MODEM_SERIAL_NUMBER感兴趣，可以使用如下的distinct。

SELECT DISTINCT
    ( modem_serial_number )
FROM
    t_recorded_test
WHERE
    office_id = '1710011001123'
    AND   record_date > '2017-11-01'
    AND   record_date < '2017-12-08'
    AND   test_result = 'true';

如何获取sql中两个日期之间所有列的不同行？

3 个答案: