t1FROM 中间少了一个空格，这里不能修改答案....

SET sql_mode = '';

DELETE t1FROM tdb_goods AS t1
    LEFT JOIN (
        SELECT goods_id, goods_name
	FROM tdb_goods
	GROUP BY goods_name
	HAVING COUNT(goods_name) > 1
	) AS t2
    ON t1.goods_name = t2.goods_name
WHERE t1.goods_id > t2.goods_id;

试试这个？

DELETE  t1 FROM tdb_goods AS t1 LEFT JOIN (SELECT min(goods_id) AS goods_id,goods_name FROM tdb_goods GROUP BY goods_name HAVING count(goods_name)>=2 ) AS t2 ON t1.goods_name=t2.goods_name WHERE t1.goods_id > t2.goods_id;

你的t2表有问题，MySQL 5.7.5后only_full_group_by成为sql_mode的默认选项之一，这可能导致一些sql语句失效，因为执行group by语句实际上把两行纪录或两行以上记录合并成一行，搜索引擎不知道该返回哪一条（因为这几行记录只有goods_name相同，而其他字段是不相同的，所以你要明确指定返回哪一行，我这里就写的min(goods_id),表示返回goods_id最小的那一行），他教学应该用的是MySQL 5.7.5以前的版本，所以没这个问题

>delete t1 from tdb_goods as t1 left join
    -> (select goods_id,goods_name from tdb_goods group by goods_name having count(goods_name) >= 2) as t2
    -> on t1.goods_name = t2.goods_name where t1.goods_id > t2.goods_id;

最后的where条件写错了,是">"大于不是"="等于

下面AS与“）”留个空格呢？     ） AS

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HAVING count(goods_name)>=2 )AS t2 

然后下面条件“=”改成“>”

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WHERE t1.goods_id=t2.goods_id;