Php 帮助连接Zend中的两个表
我有两张桌子。用户和地址。我在zend像这样加入他们:Php 帮助连接Zend中的两个表,php,zend-framework,join,Php,Zend Framework,Join,我有两张桌子。用户和地址。我在zend像这样加入他们: $table = new Model_User_DbTable(); $select = $table->select(); $select->setIntegrityCheck( false ); $select->join( 'address', 'address.id = user.address_id', array( 'city' => 'address.city' ) ); $row = $table-&
$table = new Model_User_DbTable();
$select = $table->select();
$select->setIntegrityCheck( false );
$select->join( 'address', 'address.id = user.address_id', array( 'city' => 'address.city' ) );
$row = $table->fetchAll( $select );
return $row;
但上面的查询返回的是地址中的所有地址,而不是用户表中的数据。当我删除$select->join'address',address.id=user.address_id',数组'city'=>'address.city';然后它只显示用户表数据
如何获取两个表的数据
谢谢尝试使用
$select = $table->select(Zend_Db_Table::SELECT_WITH_FROM_PART);
尝试使用
$select = $table->select(Zend_Db_Table::SELECT_WITH_FROM_PART);
未经测试:
$table = new Model_User_DbTable();
$resultSet = $table->fetchAll($table->select()
->setIntegrityCheck(false)
->from(array('u'=>'user'))
->join(array('a'=>'address'), 'u.address_id = a.id', array('city'=>'address_city'))
);
未经测试:
$table = new Model_User_DbTable();
$resultSet = $table->fetchAll($table->select()
->setIntegrityCheck(false)
->from(array('u'=>'user'))
->join(array('a'=>'address'), 'u.address_id = a.id', array('city'=>'address_city'))
);
打印SQL查询进行调试打印SQL查询进行调试