查询可以在phpMyAdmin和Chrome中工作,但不能在Safari中工作
这个查询在Chrome和phpMyAdmin中运行,但在Safari中失败:我不知道如何进行故障排除查询可以在phpMyAdmin和Chrome中工作,但不能在Safari中工作,php,mysql,Php,Mysql,这个查询在Chrome和phpMyAdmin中运行,但在Safari中失败:我不知道如何进行故障排除 SELECT r.like_dislike FROM ratings r LEFT JOIN lessons l ON l.lesson_id = r.lesson_id LEFT JOIN subjects s ON s.subject_id = l.subject_id W
SELECT r.like_dislike FROM ratings r
LEFT JOIN lessons l ON
l.lesson_id = r.lesson_id
LEFT JOIN subjects s ON
s.subject_id = l.subject_id
WHERE l.teacher_id = 142
有时r.like\u unlike为空,所以我运行了这个,但仍然存在相同的问题:
SELECT IFNULL(r.like_dislike,0) FROM ratings r
LEFT JOIN lessons l ON
l.lesson_id = r.lesson_id
LEFT JOIN subjects s ON
s.subject_id = l.subject_id
WHERE l.teacher_id = 142
质问与问题无关browers@Dagon你是对的,这与用户是否登录有关