值基本上,该表的设置是这样的:MySQL的像逗号
+-----------+----------+-------------+
| land | city | perimeter |
+-----------+----------+-------------+
| America | Kansas | 1 |
| Britain | Berlin | 4 |
| Japan | Tokyo | 5 |
+-----------+----------+-------------+
我当前的查询:
$query = "SELECT land, city, perimeter FROM agents WHERE land LIKE ? OR city LIKE ? OR perimeter LIKE ?";
$params = array("%China%","%Kansas%","%6%");
此查询的工作,这个它会返回America,Kansas,1
。但是,如果我的params
将等于:
$params = array("%China%","%Beijing,London,Kansas,Bali%","%6%");
这不会返回任何内容。如何在由逗号分隔的值中使用LIKE
以便至少匹配那些以逗号分隔的值存在的项目。
请参阅:http://stackoverflow.com/questions/327274/mysql-prepared-statements-with-a-variable-size-variable-list – ceving 2012-07-26 12:23:35