使用TableGateway在ZF2中左键连接

时间：2013-02-11 14:07:54

标签： mysql frameworks zend-framework2 tablegateway

我有一张桌子：

*CREATE TABLE IF NOT EXISTS `blogs_settings` (
  `blog_id` int(11) NOT NULL AUTO_INCREMENT,
  `owner_id` int(11) NOT NULL,
  `title` varchar(255) NOT NULL,
  `meta_description` text NOT NULL,
  `meta_keywords` text NOT NULL,
  `theme` varchar(25) NOT NULL DEFAULT 'default',
  `is_active` tinyint(1) NOT NULL DEFAULT '1',
  `date_created` int(11) NOT NULL,

  PRIMARY KEY (`blog_id`),
  KEY `owner_id` (`owner_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;*

第二张表：

*CREATE TABLE IF NOT EXISTS `users` (
  `user_id` int(11) NOT NULL AUTO_INCREMENT,
  `username` varchar(255) NOT NULL,
  `email` varchar(255) NOT NULL,
  `password` varchar(128) NOT NULL,
  `sex` tinyint(1) NOT NULL,
  `birthday` date NOT NULL,
  `avatar_id` int(11) DEFAULT NULL,
  `user_level` tinyint(1) NOT NULL DEFAULT '1',
  `date_registered` int(11) NOT NULL,
  `is_active` tinyint(1) NOT NULL DEFAULT '0',
  `is_banned` tinyint(1) NOT NULL DEFAULT '0',

  PRIMARY KEY (`user_id`),
  KEY `is_active` (`is_active`),
  KEY `user_level` (`user_level`),
  KEY `is_banned` (`is_banned`),
  KEY `username` (`username`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;*

我如何从blogs_settings表中选择所有字段，并在blogs_settings.owner_id = users.user_id上使用ZF2中的TableGateway仅加入users表中的'username'字段。提前致谢。非常感谢您的帮助。

修改 

namespace Object\Model;

use Zend\Db\TableGateway\TableGateway;
use Zend\Db\Sql\Select;

class BlogsSettingsTable {

protected $tableGateway;
protected $select;

public function __construct(TableGateway $tableGateway) {
    $this->tableGateway = $tableGateway;
    $this->select = new Select();
}

public function getBlogs($field = '', $value = '') {
    $resultSet = $this->tableGateway->select(function(Select $select) {
                $select->join('users', 'blogs_settings.owner_id = users.user_id', array('username'));
            });

    return $resultSet;
}

public function getBlog($blogID) {
    $id = (int) $blogID;

    $rowset = $this->tableGateway->select(array('blog_id' => $id));
    $row = $rowset->current();

    if (!$row) {
        throw new Exception('Could not find row with ID = ' . $id);
    }

    return $row;
}

public function addBlog(BlogsSettings $blog) {
    $data = array(
        'owner_id' => $blog->owner_id,
        'title' => $blog->title,
        'meta_description' => $blog->meta_description,
        'meta_keywords' => $blog->meta_keywords,
        'theme' => $blog->theme,
        'is_active' => $blog->is_active,
        'date_created' => $blog->date_created,
    );

    $this->tableGateway->insert($data);
}

public function deleteBlog($blogID) {
    return $this->tableGateway->delete(array('blog_id' => $blogID));
}

}

这样，它执行以下查询：

选择blogs_settings。*，users。username AS username来自blogs_settings内部加入users ON blogs_settings。{ {1}} = owner_id。users

但是resultSet不包含已加入的“users”表中的用户名字段。但是，当我在phpmyadmin中运行查询时，一切正常，我加入了'users'表中的'username'字段。有什么问题？

编辑2 好的，我现在尝试了以下内容：

user_id

执行的查询是：

public function getBlogs() {
    $select = $this->tableGateway->getSql()->select();
    $select->columns(array('blog_id', 'interest_id', 'owner_id', 'title', 'date_created'));
    $select->join('users', 'users.user_id = blogs_settings.owner_id', array('username'), 'left');

    $resultSet = $this->tableGateway->selectWith($select);

    return $resultSet;
}

当我将它运行到phpmyadmin时，它会加入users表中的用户名字段。在zf2中时，它不会。

这是整个对象的转储：

SELECT `blogs_settings`.`blog_id` AS `blog_id`, `blogs_settings`.`interest_id` AS `interest_id`, `blogs_settings`.`owner_id` AS `owner_id`, `blogs_settings`.`title` AS `title`, `blogs_settings`.`date_created` AS `date_created`, `users`.`username` AS `username` FROM `blogs_settings` LEFT JOIN `users` ON `users`.`user_id` = `blogs_settings`.`owner_id`

向上......任何想法？

7 个答案:

答案 0 :(得分：13)

如果你正在使用TableGateway，你可以像这样选择连接

$sqlSelect = $this->tableGateway->getSql()->select();
$sqlSelect->columns(array('column_name'));
$sqlSelect->join('othertable', 'othertable.id = yourtable.id', array(), 'left');

$resultSet = $this->tableGateway->selectWith($sqlSelect);
return $resultSet;

答案 1 :(得分：13)

添加@samsonasik的答案并在评论中解决问题。您无法从该语句返回的内容中获取连接值。该语句返回不具有连接行的模型对象。您需要将它作为SQL执行，并将其作为原始SQL进行准备，并将每个结果行作为数组而不是对象返回：

$sqlSelect = $this->tableGateway->getSql()->select();
$sqlSelect->columns(array('column_name_yourtable'));
$sqlSelect->join('othertable', 'othertable.id = yourtable.id', array('column_name_othertable'), 'left');

$statement = $this->tableGateway->getSql()->prepareStatementForSqlObject($sqlSelect);
$resultSet = $statement->execute();
return $resultSet;

//then in your controller or view:

foreach($resultSet as $row){
    print_r($row['column_name_yourtable']);
    print_r($row['column_name_othertable']);
}

答案 2 :(得分：2)

您必须在BlogsSetting Model中包含用户名字段，该字段用作BlogsSettingTable（TableGateway）中的模型

class BlogsSetting {
    public $blog_id;
    public $interest_id;
    public $owner_id;
    public $title;
    public $meta_description;
    public $meta_keywords;
    public $theme;
    public $is_active;
    public $date_created;
    public $username;

    public function exchangeArray($data)
    {
        // Create exchangeArray
    }
}

希望这有帮助

答案 3 :(得分：2)

加入和其中条款与 tableGateway 的确切需要。

public function getEmployeefunctionDetails($empFunctionId) {
    $empFunctionId = ( int ) $empFunctionId;
    //echo '<pre>'; print_r($this->tableGateway->getTable()); exit;
    $where = new Where();
    $where->equalTo('FUNCTION_ID', $empFunctionId);

    $sqlSelect = $this->tableGateway->getSql()->select()->where($where);    

    $sqlSelect->columns(array('FUNCTION_ID'));
    $sqlSelect->join('DEPARTMENTS', 'DEPARTMENTS.DEPARTMENT_ID = EMPLOYEE_FUNCTIONS.DEPARTMENT_ID', array('DEPARTMENT_ID','DEPARTMENT_NAME'), 'inner');
    $sqlSelect->join('ROLES', 'ROLES.ROLE_ID = EMPLOYEE_FUNCTIONS.ROLE_ID', array('ROLE_ID','ROLE_NAME'), 'inner');

    //echo $sqlSelect->getSqlString(); exit;
    $resultSet = $this->tableGateway->selectWith($sqlSelect);

    if (! $resultSet) {
        throw new \Exception ( "Could not find row $empFunctionId" );
    }
    return $resultSet->toArray();
}

答案 4 :(得分：1)

在继承自AbstractTableGateway的类中，你可以像这样使用Select with Closure：

use Zend\Db\Sql\Select;
...
public function getAllBlockSettings()
{
    $resultSet = $this->select(function(Select $select) {
        $select->join('users', 'blogs_settings.owner_id = users.user_id', array('username'));
    });

    return $resultSet;
}

答案 5 :(得分：1)

试一试：

namespace Object\Model;

use Zend\Db\TableGateway\AbstractTableGateway;
use Zend\Db\Sql\Select;

class BlogsSettingsTbl extends AbstractTableGateway {
    public function __construct($adapter) {
        $this->table = 'blogs_settings';
        $this->adapter = $adapter;
        $this->initialize();
    }

    public function fetchAll() {
        $where = array(); // If have any criteria
        $result = $this->select(function (Select $select) use ($where) {
                    $select->join('users', 'blogs_settings.owner_id = users.user_id', array('username'));
                    //echo $select->getSqlString(); // see the sql query
                });
        return $result;
    }
}

在Module.php中添加'getServiceConfig（）'：

'Object\Model\BlogsSettingsTbl' => function($sm) {
    $dbAdapter = $sm->get('Zend\Db\Adapter\Adapter');
    $table = new BlogsSettingsTbl($dbAdapter); // <-- also add this to 'USE' at top
    return $table;
},

答案 6 :(得分：0)

由于OP没有接受任何答案，我会尽力给出解决方案。 我面临与OP状态相同的解决方案，解决它的唯一方法是将此行添加到模型类中（在这种情况下，这可能是'Blogsetttings.php'）。

$this->username= (!empty($data['username'])) ? $data['username'] : null;

您应该将以上行添加到exchangeArray（）方法中。 希望它有所帮助

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