Question

我有三个表患者，疾病和提供者。所有三个表都可以通过patient_id和disease_name关联。我知道你需要：CONSTRAINT，FOREIGN KEY和REFERENCES来完成所有三个表的关系。我在第7行收到错误有人能告诉我我做错了什么吗？我无法发布问题的图片，但这里是错误：错误 SQL查询：

MySQL说：文档

＃1064 - 您的SQL语法出错;检查与您的MySQL服务器版本相对应的手册，以便在＆＃39; patient_id＆＃39;）附近使用正确的语法

CREATE TABLE patient(
    patient_id INT (3) NOT NULL AUTO_INCREMENT,
    place VARCHAR (20) NOT NULL,
    profession VARCHAR (30)NOT NULL,
    age INT (11),
    PRIMARY KEY (patient_id),
    CONSTRAINT patient_info FOREIGN KEY (patient_id) REFERENCES patient ('patient_id')
);
CREATE TABLE disease (
    patient_id INT (3) NOT NULL AUTO_INCREMENT,
    disease_name VARCHAR (20) NOT NULL,
    duration_of_stay INT(11),
    CONSTRAINT patient FOREIGN KEY(patient_id) REFERENCES patient ('patient_id'),
    CONSTRAINT disease_info FOREIGN KEY(disease_name) REFERENCES disease ('disease_name') 
);

CREATE TABLE provider(
    disease_id INT (3) NOT NULL AUTO_INCREMENT,
    disease_name VARCHAR (20) NOT NULL,
    doctor_name VARCHAR (20) NOT NULL,
    CONSTRAINT disease_info FOREIGN KEY (disease_name) REFERENCES disease ('disease_name')
);

创建相关表SQL

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