我(至少对我来说)有一个复杂的查询,是用这个网站的示例构建的。我添加的最后一件事是 IF 语句。如果没有 IF 语句,它将使用 IF 语句的 TRUE 部分。我希望你们能在这里帮助我。这是查询:
SELECT
t.ID, t.start_time, t.end_time, t.start_date, t.end_date, t.balance,
u1.first_name AS teacher_one_first_name, u1.last_name AS teacher_one_last_name,
u2.first_name AS teacher_two_first_name, u2.last_name AS teacher_two_last_name,
company.name, company.post_city, tag, lvl,
IF(
t.balance=1,
(
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE lead_follow=0 AND company_lesson_block_ID=t.ID) AS lead,
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE lead_follow=1 AND company_lesson_block_ID=t.ID) AS follow
),
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE company_lesson_block_ID=t.ID) AS total_student
)
FROM company_lesson_block AS t
LEFT JOIN company_lvl ON company_lvl.ID = t.lvl_ID
LEFT JOIN tag ON tag.ID = t.style_ID
LEFT JOIN company ON company.ID=t.location_ID
LEFT JOIN user AS u1 ON t.teacher_one_ID=u1.ID
LEFT JOIN user AS u2 ON t.teacher_two_ID=u2.ID
WHERE t.company_ID='1' AND location_ID='1' AND company_season_ID='1'
ORDER BY start_date ASC
我收到的错误消息是:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AS lead, (SELECT count(user_ID) FROM company_less' at line 10
任何帮助/提示均适用
最佳答案
您想要返回 2 列,但 IF
只能返回 1 列,因此您需要 2 个 IF
:
........................
IF(
t.balance=1,
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE lead_follow=0 AND company_lesson_block_ID=t.ID),
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE company_lesson_block_ID=t.ID)
) AS ????,
IF(
t.balance=1,
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE lead_follow=1 AND company_lesson_block_ID=t.ID),
(SELECT count(user_ID)
FROM company_lesson_block_student
WHERE company_lesson_block_ID=t.ID)
) AS ????
........................
您必须在IF
的右括号后设置别名。
也许您必须重新考虑这个逻辑,因为在 FALSE
的情况下,相同的值会返回两次。
关于mysql - mysql使用IF语句和子查询查询错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57944903/