MySQL查询返回错误的值

Question

我有这个MySQL查询，我希望在前端点击相关字母后获取正确的艺术家。现在也有艺术家以数字开头，所以我想排除它们。即使现在，无论如何都会以数字开头的艺术家进行展示，所以我认为我的MySQL查询是错误的。

$sql = "SELECT artist_id, formated_name FROM artists WHERE formated_name LIKE '".$letter."%' OR formated_name LIKE '0%' OR formated_name LIKE '1%' OR formated_name LIKE '2%' 
            OR formated_name LIKE '3%' OR formated_name LIKE '4%' OR formated_name LIKE '5%' OR formated_name LIKE '6%' OR formated_name LIKE '7%'
            OR formated_name LIKE '8%' OR formated_name LIKE '9%' OR formated_name LIKE '#%'";

有人可以帮助我吗？提前谢谢......

Answer 1

你可以这样做：

$sql = "SELECT artist_id, formated_name FROM artists 
WHERE formated_name LIKE '".$letter."%' 
AND LEFT(formated_name, 1) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9')";

Answer 2

$sql = "SELECT artist_id, formated_name FROM artists 
WHERE formated_name LIKE '".$letter."%' 

这只会返回以您的信件开头的结果。