Question

mysql里 select * from table where title like ‘%a%’的参数化查询怎么写？

$sql = "select * from table where title like ?";
$rs = $db->query($sql, array("'%t%'"));

这么些好像无效。求正解！

Answer 1

在zendframework的model层你可以这么写
$sql = $this->select()->where("title like ? ", "%t%");
$rs = $this->fetchAll($sql);
要不你就直接写
$sql = "select * from table where title like '%t%' ";
$rs = mysql_fetch_object($sql);
你也可以尝试替换字符串
$sql = "select * from table where title like '?' "
$sql = str_replace('?', '%t%', $sql)

Answer 2

select * from table where title like ‘%@a%’

然后你这个

MySqlParameter[] paras = new MySqlParameter[]
{
new MySqlParameter("@a", ticketId)
};

http://feiger.cn作答

Answer 3

楼主 俺今天试成功了 mysql中模糊查询的参数化： …… like CONCAT('%',?para1,'%') 酱紫就喔了

Answer 4

$sql = "select * from table where title like '%?%'";
$rs = $db->query($sql, array(t));

追问

试过不行~

追答

你用的是什么语言 这种语法我没见过！~

就是你把前面SQL语句里的问号替换成你的那个变量就可以了！~