Zend Framework：消息：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法错误;

Question

我收到此错误

Message: 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 '' at line 1 
Stack trace: 
#0 H:\Documents\IIS_Server_Root\zendframework\Zend\Db\Statement.php(300): Zend_Db_Statement_Pdo->_execute(Array) 
#1 H:\Documents\IIS_Server_Root\zendframework\Zend\Db\Adapter\Abstract.php(479): Zend_Db_Statement->execute(Array) 
#2 H:\Documents\IIS_Server_Root\zendframework\Zend\Db\Adapter\Pdo\Abstract.php(238): Zend_Db_Adapter_Abstract->query(Object(Zend_Db_Table_Select), Array) 
#3 H:\Documents\IIS_Server_Root\zendframework\Zend\Db\Table\Abstract.php(1526): Zend_Db_Adapter_Pdo_Abstract->query(Object(Zend_Db_Table_Select)) 
#4 H:\Documents\IIS_Server_Root\zendframework\Zend\Db\Table\Abstract.php(1342): Zend_Db_Table_Abstract->_fetch(Object(Zend_Db_Table_Select)) 
#5 H:\Documents\IIS_Server_Root\my.localhost\ahaweb\application\models\Tagjoin.php(28): Zend_Db_Table_Abstract->fetchAll(Object(Zend_Db_Statement_Pdo)) 
#6 H:\Documents\IIS_Server_Root\my.localhost\ahaweb\application\models\Tag.php(44): Model_Tagjoin->getTags('12') 
#7 H:\Documents\IIS_Server_Root\my.localhost\ahaweb\application\models\Bookmark.php(30): Model_Tag->getTags('12') 
#8 H:\Documents\IIS_Server_Root\my.localhost\ahaweb\application\controllers\UserController.php(69): Model_Bookmark->getUserBookmark(1, '12') 
#9 H:\Documents\IIS_Server_Root\zendframework\Zend\Controller\Action.php(513): UserController->editAction() 
#10 H:\Documents\IIS_Server_Root\zendframework\Zend\Controller\Dispatcher\Standard.php(295): Zend_Controller_Action->dispatch('editAction') 
#11 H:\Documents\IIS_Server_Root\zendframework\Zend\Controller\Front.php(954): Zend_Controller_Dispatcher_Standard->dispatch(Object(Zend_Controller_Request_Http), Object(Zend_Controller_Response_Http)) 
#12 H:\Documents\IIS_Server_Root\zendframework\Zend\Application\Bootstrap\Bootstrap.php(97): Zend_Controller_Front->dispatch() 
#13 H:\Documents\IIS_Server_Root\zendframework\Zend\Application.php(366): Zend_Application_Bootstrap_Bootstrap->run() 
#14 H:\Documents\IIS_Server_Root\my.localhost\ahaweb\public\index.php(26): Zend_Application->run() 

在这段代码

$tj = new Model_Tagjoin();     
    $stmt = $tj->select('*')->setIntegrityCheck(false) 
      ->joinInner("tags", "tags.tag_id = ".$this->_name.".tag_id",array("tag_name")) 
      ->where($this->_name.".bmk_id = ?", $bmk_id)->query();     
    $r = $tj->fetchAll($stmt); 

当我print_f的$stmt值我得到

SELECT `tagjoins`.*, `tags`.`tag_name` FROM `tagjoins` 
INNER JOIN `tags` ON tags.tag_id = tagjoins.tag_id WHERE (tagjoins.bmk_id = '12') 

，如果我直接与MySQL使用它工作正常。

请帮忙。我的代码有什么问题？

Answer 1

我不知道如何使用printf，但正如我在手册中读过的那样，您应该使用assemble()方法来查看您的最终sql查询。

这样的事情。

$tj = new Model_Tagjoin();     
$stmt = $tj->select('*')->setIntegrityCheck(false) 
     ->joinInner("tags", "tags.tag_id = ".$this->_name.".tag_id",array("tag_name")) 
     ->where($this->_name.".bmk_id = ?", $bmk_id)->query()->assemble(); 
exit($stmt);     
$r = $tj->fetchAll($stmt); 

然后尝试在phpMyAdmin之类的查询。 请注意，如果此通话失败，您可能要删除该行的​​部分。