它应该很简单,但是不起作用
SELECT * FROM profile WHERE name LIKE = 'H%';
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= 'H%'' at line 1
表中的列
SHOW COLUMNS FROM profile;
+-------+-------------------------------------------------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------------------------------------------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | NULL | |
| birth | date | YES | | NULL | |
| color | enum('blue','red','green','brown','black','white') | YES | | NULL | |
| foods | set('lutefisk','burrito','curry','eggroll','fadge','pizza') | YES | | NULL | |
| cats | int(11) | YES | | NULL | |
我应该怎么做?
最佳答案
在您的选择语句中删除=
所以;
SELECT * FROM profile WHERE name LIKE 'H%';
关于mysql - 如何在MySQL中使用H搜索名称?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/45344914/