Question

所以我做了一堆Doctrine2迁移（https://github.com/doctrine/migrations），但我对我正在尝试的新迁移有疑问。

我一直在深入挖掘库，我发现$this->addSql()用于构建要执行的SQL列表，然后稍后执行。

我想做一些事情，我选择一些数据，迭代行，根据它插入新数据，然后删除我选择的数据。这很容易适用于DBAL库，但我想知道，我可以安全地在迁移中使用protected $connection吗？或者是那么糟糕，因为它会在我的任何$this->addSql() SQL被执行之前执行语句？此外，这似乎会破坏我在代码中看到的dry-run设置。有没有人有这种迁移的经验？有没有最好的做法？

以下是我想要进行的迁移，但我不相信Doctrine Migrations支持这一点：

public function up(Schema $schema)
{
    // this up() migration is autogenerated, please modify it to your needs
    $this->abortIf($this->connection->getDatabasePlatform()->getName() != "mysql");

    $this->addSql("ALTER TABLE article_enclosures ADD is_scrape TINYINT(1) NOT NULL");
    $this->addSql("ALTER TABLE images DROP FOREIGN KEY FK_E01FBE6AA536AAC7");

    // now lets take all images with a scrape and convert the scrape to an enclosure
    // 
    // Select all images where not scrape_id is null (join on article_image_scrape)
    // for each image:
    //     insert into article_enclosures
    //     update image set enclosure_id = new ID
    //     delete from article_image_scrape where id...
    //
    // insert into article_enclosures select article_image_scrapes...

    $sql = "SELECT i.id img_id, e.* FROM images i JOIN article_image_scrapes e ON i.scrape_id = e.id";
    $stmt = $this->connection->prepare($sql);
    $stmt->execute();
    $scrapesToDelete = array();
    while ($row = $stmt->fetch()) {
        $scrapeArticle = $row['article_id'];
        $scrapeOldId = $row['id'];
        $scrapeUrl = $row['url'];
        $scrapeExtension = $row['extension'];
        $scrapeUrlHash = $row['url_hash'];
        $imageId = $row['image_id'];

        $this->connection->insert('article_enclosures', array(
            'url' => $scrapeUrl,
            'extension' => $scrapeExtension,
            'url_hash' => $scrapeUrlHash
        ));

        $scrapeNewId = $this->connection->lastInsertId();

        $this->connection->update('images', array(
            'enclosure_id' => $scrapeNewId,
            'scrape_id' => null
        ), array(
            'id' => $imageId
        ));

        $scrapesToDelete[] = $scrapeOldId;
    }

    foreach ($scrapesToDelete as $id) {
        $this->connection->delete('article_image_scrapes', array('id' => $id));
    }

    $this->addSql("INSERT INTO article_scrapes (article_id, url, extension, url_hash) "
            ."SELECT s.id, s.url, s.extension, s.url_hash"
            ."FROM article_image_scrapes s");

    $this->addSql("DROP INDEX IDX_E01FBE6AA536AAC7 ON images");
    $this->addSql("ALTER TABLE images DROP scrape_id, CHANGE enclosure_id enclosure_id INT NOT NULL");
}

Answer 1

您可以像这样使用$connection

$result = $this->connection->fetchAssoc('SELECT id, name FROM table1 WHERE id = 1');
$this->abortIf(!$result, 'row with id not found');
$this->abortIf($result['name'] != 'jo', 'id 1 is not jo');
// etc..

您应该只读取数据库而不使用连接进行更新/删除，这样就不会破坏干运行选项。

在您的示例中，您应该进行两次迁移。第一个会做两个转换表。第二个将执行“刮刮图像并将刮削转换为外壳”程序。如果出现问题，使用多次迁移更容易还原它们。

Answer 2

仅供参考，最新文档显示此示例使用“postUp”方法更好[/ p>]

http://symfony.com/doc/current/bundles/DoctrineMigrationsBundle/index.html

// ...
use Symfony\Component\DependencyInjection\ContainerAwareInterface;
use Symfony\Component\DependencyInjection\ContainerInterface;

class Version20130326212938 extends AbstractMigration implements ContainerAwareInterface
{

    private $container;

    public function setContainer(ContainerInterface $container = null)
    {
        $this->container = $container;
    }

    public function up(Schema $schema)
    {
        // ... migration content
    }

    public function postUp(Schema $schema)
    {
        $em = $this->container->get('doctrine.orm.entity_manager');
        // ... update the entities
    }
}

Doctrine2迁移使用DBAL而不是$ this-＆gt; addSql

2 个答案: