Right way to integrate extra columns into grid?

[print]

Hi,

i search for the right way to bring in some extra columns into my grid.

My first example is a simple image column, like "src/folder/{{ row.id }}.jpg".
I am not sure what's the right way for this, i hope anyone can get me a short tipp?

My second problem are columns with a ManyToOne situation:

 * @GRID\Source(columns="id, name, category")
 */
class MyClass
{
....
     * @GRID\Column(id="category.name", title="category", field="category")
     */
    protected $category;

But my column category does not work/display in my grid with this configuration.

I hope anyone can help?

Thanks!

[ghost]

1. create own class like

use Sorien\DataGridBundle\Grid\Column\Column;

class ImageColumn extends Column
{

public function getFilters()
{
    return array();
}

public function renderCell($value, $row, $router)
{
    return '<div class="image" style="background-image: url(\''.Image::getPath().'/'.$value.'\');"/></div>';
}

public function getType()
{
    return 'image'; //important
}
}

add to DI

    <service id="grid.column.image" class="PathToClass\ImageColumn" public="false">
        <tag name="grid.column.extension" />
    </service>

and use it like this

 * @ORM\OneToOne(targetEntity="Image", cascade={"persist", "remove"})
 * @ORM\JoinColumn(name="main_image_id", referencedColumnName="id")
 * @GRID\Column(type="image", field="mainImage.file", size="64") //set type image
 */
private $mainImage;

dont use id for property mapping

@GRID\Column(title="category", field="category.name")

[Abhoryo]

Without create a class, maybe you can add a simple column and fill it in the template. What do you think @S0RIEN ?

[ghost]

yes, its easier way, when you don't need to store same special data, or create own filterings