Zend_Db的Where子句无法按预期工作

时间:2010-08-09 04:12:34

标签: php mysql zend-framework zend-db

我收到一个错误:
消息: SQLSTATE[42000]: Syntax error or access violation: 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 '%'Chinese'%) ORDER BY text_id DESC LIMIT 10' at line 2

由这行代码引起的

 $select = $this->_db->select('')
            ->from(array('t'=>'as_text'))
           ->where('`s`.`name` LIKE %?%',$search) //this is causing error
            ->limit((int)$limit)
            ->order('text_id DESC')
            ->join(array('s'=>'as_source'),'t.source_id = s.source_id',array('s.name as source'));

我的目标是这个sql:

SELECT `t` . * , `s`.`name` AS `source`
FROM `as_text` AS `t`
INNER JOIN `as_source` AS `s` ON t.source_id = s.source_id
WHERE `s`.`name` LIKE '%Chinese%'
ORDER BY `text_id` DESC
LIMIT 10 

我认为这是 - > where bit,因为当我删除它时我会得到10行。

1 个答案:

答案 0 :(得分:1)

编辑: 这对我有用:

        ->from(array('t'=>'as_text')) 
       ->where("s.name LIKE ?",'%'.$search.'%') //this is causing error 
        ->limit((int)$limit) 
        ->order('text_id DESC') 
        ->join(array('s'=>'as_source'),
        't.source_id = s.source_id',
        array('s.name as source'));

如果有效,请告诉我。