SELECT *
FROM PATIENT
WHERE PATIENTNUM = (SELECT PATIENTNUM
FROM CONSULTATION
WHERE CONSULTATIONDATE =
(SELECT DATE_ADD(curdate(), INTERVAL 1 DAY) AS DAYTOM);
这不起作用。
但是当我只插入这个
时SELECT DATE_ADD(curdate(), INTERVAL 1 DAY) AS DAYTOM
它的工作......
为什么?
答案 0 :(得分:1)
我认为正确的方法是
SELECT *
FROM PATIENT
WHERE PATIENTNUM IN (SELECT PATIENTNUM
FROM CONSULTATION
WHERE CONSULTATIONDATE IN (SELECT DATE_ADD(curdate(), INTERVAL 1 DAY) AS DAYTOM));
答案 1 :(得分:0)
你错过了一个支架?
SELECT *
FROM PATIENT
WHERE PATIENTNUM = (SELECT PATIENTNUM
FROM CONSULTATION
WHERE CONSULTATIONDATE = (SELECT DATE_ADD(curdate(), INTERVAL 1 DAY) AS DAYTOM));