Question

我的数据库中有三个表，一个用于患者，一个用于温度，一个表包含患者的id和温度的id作为主键。

病人表;

+----+------+--------+------------+
| id | name | gender | birth      |
+----+------+--------+------------+
|  1 | Hind | F      | 2015-01-01 |
|  2 | Saad | M      | 2016-01-01 |
+----+------+--------+------------+

patient_info表：

+------+------+
| id_t | id_p |
+------+------+
|    1 |    1 |
|    1 |    2 |
|    2 |    3 |
|    2 |    4 |
+------+------+

温度表：

+----+-------------+
| id | temperature |
+----+-------------+
|  1 |        37.9 |
|  2 |          37 |
|  3 |          38 |
|  4 |        38.4 |
+----+-------------+

所以我想要的是用温度检索所有患者，这就是我所做的：

mysql> SELECT p.*, t.* FROM patient p INNER JOIN patient_info pi ON pi.id_p = p.id INNER JOIN temperature t ON t.id = pi.id_t;
+----+------+--------+------------+----+-------------+
| id | name | gender | birth      | id | temperature |
+----+------+--------+------------+----+-------------+
|  1 | Hind | F      | 2015-01-01 |  1 |        37.9 |
|  2 | Saad | M      | 2016-01-01 |  1 |        37.9 |
+----+------+--------+------------+----+-------------+

但这不是我想要的，我需要知道所有Hind和Saad的温度谢谢大家

Answer 1

我希望你真的能在这样的事情之后 - 尽管将温度存储在一个单独的表格中并没有意义......

DROP TABLE IF EXISTS patient;

CREATE TABLE patient
(patient_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,name VARCHAR(12) NOT NULL
,gender CHAR(1) NOT NULL
,dob DATE NOT NULL
);

INSERT INTO patient VALUES
(101 ,'Hind','F','2015-01-01'),
(102 ,'Saad','M','2016-01-01');


DROP TABLE IF EXISTS patient_temperature;

CREATE TABLE patient_temperature
(patient_id INT NOT NULL
,temperature_id INT NOT NULL
,PRIMARY KEY(patient_id,temperature_id)
);

INSERT INTO patient_temperature VALUES
(101,1),
(101,2),
(102,3),
(102,4);

DROP TABLE IF EXISTS temperature;

CREATE TABLE temperature
(temperature_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,temperature DECIMAL(5,2) NOT NULL
);

INSERT INTO temperature VALUES
(1,37.9),
(2,37),
(3,38),
(4,38.4);

SELECT p.*
     , t.*
  FROM patient p 
  JOIN patient_temperature pt 
    ON pt.patient_id = p.patient_id 
  JOIN temperature t 
    ON t.temperature_id = pt.temperature_id;
+------------+------+--------+------------+----------------+-------------+
| patient_id | name | gender | dob        | temperature_id | temperature |
+------------+------+--------+------------+----------------+-------------+
|        101 | Hind | F      | 2015-01-01 |              1 |       37.90 |
|        101 | Hind | F      | 2015-01-01 |              2 |       37.00 |
|        102 | Saad | M      | 2016-01-01 |              3 |       38.00 |
|        102 | Saad | M      | 2016-01-01 |              4 |       38.40 |
+------------+------+--------+------------+----------------+-------------+

如何通过mysql检测所有患者的温度信息？

1 个答案: