什么时候
SELECT * from hospital_basic
LEFT JOIN hospital_clinical on hospital_basic.p_id=hospital_clinical.p_id;
它工作得很好,但在创建视图时会出错
CREATE VIEW hospital_view AS
SELECT * from hospital_basic LEFT JOIN
hospital_clinical on hospital_basic.p_id=hospital_clinical.p_id;
1条答案
按热度按时间ej83mcc01#
您需要为正在选择的列放置一个别名,但您不能这样做
select *
因为表定义的列p\u id将被复制