Php 无法找出mysql查询中的错误

我现在正在尝试一个小时，但仍然无法找出此查询中的问题：/

SELECT * FROM question ORDER BY question_id DESC LIMIT 3 WHERE topic_name = (SELECT * FROM topic WHERE subject_name = 'Maths')

这就是错误：

1064-您的SQL语法有错误；检查与您的MySQL服务器版本对应的手册，以了解要使用的正确语法 在“WHERE topic\u name”附近=从主题WHERE subject\u name中选择*= “一号线的数学

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你的陈述都有问题。正确的说法是：

SELECT *
FROM question
WHERE topic_name = (SELECT * FROM topic WHERE subject_name = 'Maths')
ORDER BY question_id DESC LIMIT 3 ;

但是，由于子查询的原因，这将不起作用。也许你的意思是：

SELECT *
FROM question
WHERE topic_name in (SELECT topic_name FROM topic WHERE subject_name = 'Maths')
ORDER BY question_id DESC LIMIT 3 ;

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您的WHERE子句需要在ORDER BY子句之前：

SELECT * FROM question WHERE topic_name IN (SELECT topic_name FROM topic WHERE subject_name = 'Maths') ORDER BY question_id DESC LIMIT 3

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指定哪列并将结果限制为一列，或改用IN运算符

SELECT * FROM question 
WHERE topic_name = (SELECT topic FROM topic WHERE subject_name = 'Maths')
ORDER BY question_id DESC LIMIT 1 

SELECT * FROM question 
WHERE topic_name in(SELECT topic FROM topic WHERE subject_name = 'Maths')
ORDER BY question_id DESC LIMIT 3 

事实上，这不是最好的建议。改为进行内部联接：

SELECT * FROM question q
INNER JOIN topic t 
ON q.topic_name = t.topic
WHERE subject_name = 'Maths'
ORDER BY question_id LIMIT 3 

如果您想保留=我更改了=以解决此问题，我可以在子查询中建议一个最小值、最大值或TOP 1吗