我正在按照http://framework.zend.com/manual/en/learning.quickstart.create-model.html的指南进行操作,这与it技术有关。然而,我遇到了一系列问题。目前我遇到的问题是以下错误:
An error occurred
Application error
Exception information:
Message: Primary key column(s) (id) are not columns in this table ()
Stack trace:
#0 C:\wamp\www\zendtest\quickstart\library\Zend\Db\Table\Abstract.php(982): Zend_Db_Table_Abstract->_setupPrimaryKey()
#1 C:\wamp\www\zendtest\quickstart\library\Zend\Db\Table\Select.php(100): Zend_Db_Table_Abstract->info()
#2 C:\wamp\www\zendtest\quickstart\library\Zend\Db\Table\Select.php(78): Zend_Db_Table_Select->setTable(Object(Application_Model_DbTable_Guestbook))
#3 C:\wamp\www\zendtest\quickstart\library\Zend\Db\Table\Abstract.php(1018): Zend_Db_Table_Select->__construct(Object(Application_Model_DbTable_Guestbook))
#4 C:\wamp\www\zendtest\quickstart\library\Zend\Db\Table\Abstract.php(1326): Zend_Db_Table_Abstract->select()
#5 C:\wamp\www\zendtest\quickstart\application\models\GuestbookMapper.php(58): Zend_Db_Table_Abstract->fetchAll()
#6 C:\wamp\www\zendtest\quickstart\application\controllers\GuestbookController.php(14): Application_Model_GuestbookMapper->fetchAll()
#7 C:\wamp\www\zendtest\quickstart\library\Zend\Controller\Action.php(516): GuestbookController->indexAction()
#8 C:\wamp\www\zendtest\quickstart\library\Zend\Controller\Dispatcher\Standard.php(295): Zend_Controller_Action->dispatch('indexAction')
#9 C:\wamp\www\zendtest\quickstart\library\Zend\Controller\Front.php(954): Zend_Controller_Dispatcher_Standard->dispatch(Object(Zend_Controller_Request_Http), Object(Zend_Controller_Response_Http))
#10 C:\wamp\www\zendtest\quickstart\library\Zend\Application\Bootstrap\Bootstrap.php(97): Zend_Controller_Front->dispatch()
#11 C:\wamp\www\zendtest\quickstart\library\Zend\Application.php(366): Zend_Application_Bootstrap_Bootstrap->run()
#12 C:\wamp\www\zendtest\quickstart\public\index.php(26): Zend_Application->run()
#13 {main}
Request Parameters:
array (
'controller' => 'guestbook',
'action' => 'index',
'module' => 'default',
)
但我确定我的表格有一个名为id的主键:
CREATE TABLE guestbook (
id INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT,
email VARCHAR(32) NOT NULL DEFAULT 'noemail@test.com',
comment TEXT NULL,
created DATETIME NOT NULL
);
CREATE INDEX "id" ON "guestbook" ("id");
不理解为什么会出现错误。
我该如何查看是否选取了正确的数据库。
SHOW CREATE TABLE guestbook
的输出,以便我们可以仔细检查实际结构。 - Tim Fountain