Question

我对MySQL查询有点困惑，需要一些帮助。

我们假设我们有一张桌子＆＃34;动物＆＃34;使用这些列：

animal_id | animal_species
1         | dog
2         | cat
3         | dog
4         | pig

此外，我们有一张表＆＃34;考试＆＃34;其中包括对这些动物的各种检查：

examination_id | examination_animal_id
1              | 1
2              | 1
3              | 3
4              | 2
5              | 4

最后我们有另一张桌子＆＃34; diagnoses_statistics＆＃34;对于某些检查包含一个或多个诊断：

diagnoses_statistics_id | diagnoses_statistics_examination_id | diagnoses_statistics_diagnosis
1                       | 1                                   | luxation
2                       | 3                                   | thrombopathy
3                       | 3                                   | trauma
4                       | 3                                   | luxation
5                       | 5                                   | trauma

所有表格都包含许多其他列，我只是缩短了它们，因为上面的示例应足以说明我的问题。

我们想知道每个物种有多少相同的诊断。为此，我们可以运行此查询：

SELECT animals.animal_species,
       diagnoses_statistics.diagnoses_statistics_diagnosis,
       COUNT(diagnoses_statistics.diagnoses_statistics_diagnosis) AS diagnoses_count
FROM examinations
JOIN animals ON animals.animal_id = examinations.examination_animal_id
JOIN diagnoses_statistics ON diagnoses_statistics.diagnoses_statistics_examination_id = examinations.examination_id
GROUP BY animals.animal_species,
         diagnoses_statistics.diagnoses_statistics_diagnosis
ORDER BY animals.animal_species,
         diagnoses_statistics.diagnoses_statistics_diagnosis

这给了我们这样的东西：

animal_species | diagnoses_statistics_diagnosis | diagnoses_count
dog            | luxation                       | 2
dog            | thrombopathy                   | 1
dog            | trauma                         | 1
pig            | trauma                         | 1

到目前为止，这么好。现在我们想要添加一个列，其中包含任何包含任何诊断的物种的总和：

animal_species | diagnoses_statistics_diagnosis | diagnoses_count | species_count
dog            | luxation                       | 2               | 2
dog            | thrombopathy                   | 1               | 2
dog            | trauma                         | 1               | 2
pig            | trauma                         | 1               | 1

为了达到这个目的，我尝试了这个代码，它似乎在＆＃34; species_count＆＃34;中产生了正确的值，但它打破了列＆＃34; diagnoses_count＆＃34;：

SELECT animals.animal_species,
       diagnoses_statistics.diagnoses_statistics_diagnosis,
       COUNT(diagnoses_statistics.diagnoses_statistics_diagnosis) AS diagnoses_count,
       species_count.species_count
FROM examinations
JOIN animals ON animals.animal_id = examinations.examination_animal_id
JOIN diagnoses_statistics ON diagnoses_statistics.diagnoses_statistics_examination_id = examinations.examination_id
JOIN
    (SELECT animals.animal_species,
            diagnoses_statistics.diagnoses_statistics_diagnosis,
            COUNT(animals.animal_id) AS species_count
     FROM examinations
     JOIN animals ON animals.animal_id = examinations.examination_animal_id
     JOIN diagnoses_statistics ON diagnoses_statistics.diagnoses_statistics_examination_id = examinations.examination_id
     GROUP BY animals.animal_species,
              diagnoses_statistics.diagnoses_statistics_diagnosis
    ) AS species_count ON species_count.animal_species = animals.animal_species
GROUP BY animals.animal_species,
         diagnoses_statistics.diagnoses_statistics_diagnosis
ORDER BY animals.animal_species,
         diagnoses_statistics.diagnoses_statistics_diagnosis

也许我这样做完全错了。

你能帮我解决一下这个问题吗？

非常感谢您的帮助。

Answer 1

你非常接近。我想这就是你的意思。请注意子查询中的COUNT(DISTINCT ...)：

SELECT
    a.animal_species,
    d.diagnoses_statistics_diagnosis,
    COUNT(1) as diagnoses_count,
    spc.species_count
FROM
    animals a
    JOIN examinations e ON
        e.examination_animal_id = a.animal_id
    JOIN diagnoses_statistics d ON
        d.diagnoses_statistics_examination_id = e.examination_id
    JOIN (SELECT 
              a2.animal_species,
              COUNT(DISTINCT a2.animal_id) as species_count
          FROM
              animals a2
              JOIN examinations e2 ON
                  e2.examination_animal_id = a2.animal_id
              JOIN diagnoses_statistics d2 ON
                  d2.diagnoses_statistics_examination_id = e2.examination_id
          GROUP BY
              a2.animal_species         
         ) spc ON
         spc.animal_species = a.animal_species
GROUP BY
    a.animal_species,
    d.diagnoses_statistics_diagnosis,  
    spc.species_count

您可以在rextester.com/XXQ94494处尝试一下。 @p.Salmon的帽子提示，用于构建数据集的代码。

Answer 2

我不知道如何获得狗的物种数量2，因为这个物种有3个诊断。如果3是正确的，您可以使用子查询。

DROP TABLE IF EXISTS ANIMALS;
CREATE TABLE ANIMALS (animal_id INT, animal_species VARCHAR(3));
INSERT INTO ANIMALS VALUES
(1         , 'dog'),
(2         , 'cat'),
(3         , 'dog'),
(4         , 'pig');

DROP TABLE IF EXISTS EXAMINATIONS;
CREATE TABLE EXAMINATIONS (examination_id INT, examination_animal_id INT);
INSERT INTO EXAMINATIONS VALUES
(1              , 1),
(2              , 1),
(3              , 3),
(4              , 2),
(5              , 4);

DROP TABLE IF EXISTS diagnoses_statistics;
CREATE TABLE diagnoses_statistics (diagnoses_statistics_id INT, diagnoses_statistics_examination_id INT, diagnoses_statistics_diagnosis VARCHAR(20));
INSERT INTO diagnoses_statistics VALUES
(1                       , 1                                   , 'luxation'),
(2                       , 3                                   , 'thrombopathy'),
(3                       , 3                                   , 'trauma'),
(4                       , 3                                   , 'luxation'),
(5                       , 5                                   , 'trauma');

SELECT A.ANIMAL_SPECIES,  D.diagnoses_statistics_diagnosis, COUNT(*) DIAGNOSES_COUNT
    ,(SELECT COUNT(DISTINCT diagnoses_statistics_diagnosis)
    FROM ANIMALS A1
    JOIN EXAMINATIONS E ON E.EXAMINATION_ANIMAL_ID = A1.ANIMAL_ID
    JOIN diagnoses_statistics D ON D.diagnoses_statistics_EXAMINATION_ID = E.EXAMINATION_ID
    WHERE A1.ANIMAL_SPECIES = A.ANIMAL_SPECIES) SPECIES_COUNT
FROM ANIMALS A
JOIN EXAMINATIONS E ON E.EXAMINATION_ANIMAL_ID = A.ANIMAL_ID
JOIN diagnoses_statistics D ON D.diagnoses_statistics_EXAMINATION_ID = E.EXAMINATION_ID
GROUP BY A.ANIMAL_SPECIES,  D.diagnoses_statistics_diagnosis
ORDER BY a.animal_species,
         d.diagnoses_statistics_diagnosis

结果

+----------------+--------------------------------+-----------------+---------------+
| ANIMAL_SPECIES | diagnoses_statistics_diagnosis | DIAGNOSES_COUNT | SPECIES_COUNT |
+----------------+--------------------------------+-----------------+---------------+
| dog            | luxation                       |               2 |             3 |
| dog            | thrombopathy                   |               1 |             3 |
| dog            | trauma                         |               1 |             3 |
| pig            | trauma                         |               1 |             1 |
+----------------+--------------------------------+-----------------+---------------+
4 rows in set (0.00 sec)

具有多个计数的MySQL查询

2 个答案: