为什么来自Datagrip的同一个SQL查询和嵌入在Python代码中的SQL不同有输出？

Question

无论查询是在纯SQL的Datagrip上进行还是在Jupyter上使用嵌套在Python代码中的SQL查询进行的，我都得到了两个不同的输出。在

查询是来自多个用户组的特定状态的计数。
以下是本案例的用户组：

                              ids
grupos                           
0       [160, 161, 365, 386, 471]
1                      [296, 306] 

我的数据库中的表如下所示：

代码：

来自Jupyter:

^{pr2}$

来自Datagrip:

我必须把查询分开，否则结果就会混在一起。在

SELECT COUNT(swipe.eclipse_id), subscriber_hashtag.hashtag_id  FROM subscriber_hashtag
  INNER JOIN eclipse_hashtag ON eclipse_hashtag.hashtag_id = subscriber_hashtag.hashtag_id
  LEFT OUTER JOIN swipe ON subscriber_hashtag.subscriber_id = swipe.subscriber_id
  WHERE (swipe.state= 3 OR swipe.state = 6 or swipe.state=9) AND subscriber_hashtag.subscriber_id = 160 OR subscriber_hashtag.subscriber_id = 161 OR subscriber_hashtag.subscriber_id = 365 OR subscriber_hashtag.subscriber_id = 386 OR subscriber_hashtag.subscriber_id = 471 OR subscriber_hashtag.subscriber_id = 499
    GROUP BY subscriber_hashtag.hashtag_id
      ORDER BY COUNT(swipe.eclipse_id) DESC;

那么

SELECT COUNT(swipe.eclipse_id), subscriber_hashtag.hashtag_id  FROM subscriber_hashtag
  INNER JOIN eclipse_hashtag ON eclipse_hashtag.hashtag_id = subscriber_hashtag.hashtag_id
  LEFT OUTER JOIN swipe ON subscriber_hashtag.subscriber_id = swipe.subscriber_id
  WHERE (swipe.state= 3 OR swipe.state = 6 or swipe.state=9) AND subscriber_hashtag.subscriber_id = 296 OR subscriber_hashtag.subscriber_id = 306
    GROUP BY subscriber_hashtag.hashtag_id
      ORDER BY COUNT(swipe.eclipse_id) DESC;

输出：

来自Jupyter:

[(1500L, 996), (1185L, 592), (480L, 1214), (432L, 329), (375L, 398), (306L, 357), (300L, 473), (288L, 325), (225L, 322), (207L, 321), (207L, 1385), (195L, 1445), (180L, 1049), (108L, 334), (105L, 1183), (90L, 387), (81L, 324), (75L, 617), (72L, 379), (63L, 1331), (54L, 2546), (54L, 2545), (48L, 961), (48L, 962), (45L, 1382), (30L, 1432), (30L, 1429), (27L, 1334), (24L, 1128), (18L, 1376), (18L, 386), (18L, 1345), (18L, 1335), (9L, 1354), (9L, 1356), (9L, 1355), (9L, 1357), (9L, 1361), (9L, 1364), (9L, 1374), (9L, 1375), (9L, 1373)]

来自Datagrip:

您可以看到这两种输出之间没有任何共同点：

对[160, 161, 365, 386, 471]的第一个查询：

3000    397
2967    321
2352    329
2233    960
2000    392
1975    685
1896    337
1536    529
637 328
553 704
240 2545
240 2546
237 652
196 758
196 573
147 483
98  584
98  450
98  448
79  2549
79  2554
79  2552
79  2553
79  2551
79  2550
58  1376
56  428
49  451
49  759
49  449
49  760
34  2580
32  325
29  2547
29  425
25  322
13  594
12  334
9   427
6   323
3   347
3   595
3   345
1   521
1   333

对[296, 306]的第二个查询：

6600    996
5214    592
2880    329
2112    1214
1920    325
1650    398
1500    322
1380    321
1380    1385
858 1445
792 1049
720 334
600 387
540 324
480 379
462 1183
420 1331
360 2546
360 2545
330 617
306 357
300 473
300 1382
180 1334
132 1432
132 1429
120 386
120 1335
120 1376
120 1345
60  1364
60  1374
60  1356
60  1357
60  1355
60  1361
60  1354
60  1375
60  1373
48  962
48  961
24  1128

Answer 1

本质上，它是AND和WHERE子句中OR混合使用的SQL逻辑问题。在DataGrip中，您需要将所有OR子句包装在括号中，以复制包含一个IN()子句的Python版本。以下两种说法应得出相同的结果：

DataGrip调整后的SQL

SELECT COUNT(swipe.eclipse_id), subscriber_hashtag.hashtag_id  
FROM subscriber_hashtag
INNER JOIN eclipse_hashtag 
  ON eclipse_hashtag.hashtag_id = subscriber_hashtag.hashtag_id
LEFT OUTER JOIN swipe 
  ON subscriber_hashtag.subscriber_id = swipe.subscriber_id
WHERE (swipe.state= 3 OR swipe.state = 6 OR swipe.state=9) AND 
      (subscriber_hashtag.subscriber_id = 160 OR 
       subscriber_hashtag.subscriber_id = 161 OR  
       subscriber_hashtag.subscriber_id = 365 OR 
       subscriber_hashtag.subscriber_id = 386 OR 
       subscriber_hashtag.subscriber_id = 471 OR 
       subscriber_hashtag.subscriber_id = 499)
GROUP BY subscriber_hashtag.hashtag_id
ORDER BY COUNT(swipe.eclipse_id) DESC;

Python呈现的SQL

^{pr2}$