使用TableGateway在ZF2中左键连接

Question

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 ;*

第二个表:

*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 ;*

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

编辑:

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));
}

}

有了它,它执行以下查询:

SELECT blogs_settings.*, users. usernameAS usernameFROM blogs_settingsINNER JOIN usersON blogs_settings.owner_id= users.user_id

但是resultSet不包含已加入的"用户"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:

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;
}

执行的查询是:

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`

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,它没有.

这是整个对象的转储:

Zend\Db\ResultSet\ResultSet Object
(
[allowedReturnTypes:protected] => Array
    (
        [0] => arrayobject
        [1] => array
    )

[arrayObjectPrototype:protected] => Object\Model\BlogsSettings Object
    (
        [blog_id] => 
        [interest_id] => 
        [owner_id] => 
        [title] => 
        [meta_description] => 
        [meta_keywords] => 
        [theme] => 
        [is_active] => 
        [date_created] => 
    )

[returnType:protected] => arrayobject
[buffer:protected] => 
[count:protected] => 1
[dataSource:protected] => Zend\Db\Adapter\Driver\Pdo\Result Object
    (
        [statementMode:protected] => forward
        [resource:protected] => PDOStatement Object
            (
                [queryString] => 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`
            )

        [options:protected] => 
        [currentComplete:protected] => 
        [currentData:protected] => 
        [position:protected] => -1
        [generatedValue:protected] => 0
        [rowCount:protected] => 1
    )

[fieldCount:protected] => 6
[position:protected] => 
)

起来......任何想法？

Answer 1

如果你正在使用TableGateway,你可以选择这样的连接

$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;

Answer 2

添加@samsonasik的答案并在其评论中解决问题.您将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,将其作为原始SQL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']);
}