[php]
- SELECT * FROM `_table` WHERE name LIKE '% $s %';
但在discuz的模糊查询中,会有一定的区别
【1】 [php]
- DB::fetch_first("SELECT * FROM %t WHERE name LIKE '% %s %'", array($this->_table,$str));
Discuz! Database Error
(0) SQL string format error! This SQL need "4" vars to replace into.(SQL语句格式错误,该语句需要4个变量)
出现的原因是discuz把模糊查询中的 ‘%’ 也会当成变量,因此出现错误
【2】修改查询语句,修改变量的表达方式,把%放在array的变量里面而不是查询语句中
[php]
- DB::fetch_first("SELECT * FROM %t WHERE name LIKE '%s'", array($this->_table,'%'.$str.'%'));
(1064) You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%xE4xBDxA0xE5xA5xBD%''' at line 1 SELECT * FROM _table WHERE name LIKE ''%你好%''
查询的语句翻译为如下:SELECT * FROM _table WHERE name LIKE ''%你好%''
从显示的SQL语句中会发现,like后面的条件是在双引号内,而语句里一般为单引号,这是因为生成的语句会自动把条件加单引号,所以条件最终变为双引号,解决的方法是:
把select语句里的 '%s' 的单引号去除
[php]
- DB::fetch_first("SELECT * FROM %t WHERE name LIKE %s", array($this->_table,'%'.$str.'%'));