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DataObjectManager Module /

Discuss the DataObjectManager module, and the related ImageGallery module.

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ImageDataObjectManager in SiteConfig


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arnhoe

Community Member, 6 Posts

25 May 2012 at 11:56pm

Edited: 25/05/2012 11:56pm

Hello,

I have managed to get myself a many_many relationship in the siteconfig, my goal is that, the user is able to upload a image and select one or more items from the menu.

Now, I would like to filter in the CMS on a menu item, and then it will show the images this menu item has. But I am not sure if I created my pages in the way it should be. Is there anyone that would like to check my pages?

<?php
class BannerImages extends DataObject
{
	static $db = array('Title' => 'Text');

	static $has_one = array(
		'SiteConfig' => 'SiteConfig',
		'myImage' => 'BetterImage'
	);

	static $many_many = array('Images' => 'SiteTree');

	public function getCMSFields_forPopup()
	{
		$images = DataObject::get('SiteTree', 'ShowInMenus = 1');
		return new FieldSet(new TextField('Title'), new FileIFrameField('myImage'), new CheckboxSetField('Images', 'Images', $images));
	}

}

<?php

class CustomSiteConfig extends DataObjectDecorator
{

	function extraStatics()
	{
		return array('has_many' => array('Images' => 'BannerImages'));
	}

	public function updateCMSFields(FieldSet &$fields)
	{

		$manager = new ImageDataObjectManager($this -> owner, 'Images', 'BannerImages', 'myImage', array('Title' => 'Title'), 'getCMSFields_forPopup');

		$manager -> setParentClass("SiteConfig");
		$manager -> setPluralTitle('Images');
		$manager -> setRelationAutoSetting(false);
		$manager -> setFilter('ID', // Name of field to filter
		'Filter by Page', // Label for filter
		Dataobject::get("SiteTree") -> toDropdownMap("ID", "Title") // Map for filter (could be $dataObject->toDropdownMap(), e.g.)
		);

		$fields -> addFieldToTab("Root.Images", $manager);

	}

}

And I am calling my code in the following way, I am really certain that this isnt the right way, but I tried DataObject::get, but I cannot call the CurrentPageID to equal with the SiteTreeID.

public function BannerImages()
	{
		// build a new sql query
		$sqlQuery = new SQLQuery();
		$sqlQuery -> select = array('a.SiteTreeID AS sitetreeid, a.BannerImagesID as bannerimagesid, b.myImageID as Image, c.Filename as ImageURL');
		$sqlQuery -> from = array("`bannerimages_images` a LEFT JOIN `bannerimages` b on a.BannerImagesID = b.ID LEFT JOIN `file` c on b.myImageID = c.ID");
		$sqlQuery -> where = array("SiteTreeID = " . $this -> ID . "");
		$sqlQuery -> distinct = true;
		$sqlQuery -> orderby = "RAND()";
		$sqlQuery -> limit = "1";

		$rawSQL = $sqlQuery -> sql();
		$result = $sqlQuery -> execute();

		// create a new data object set
		$myDataObjectSet = new DataObjectSet();

		// push our result into the data object set
		foreach ($result as $row) {
			$myDataObjectSet -> push(new ArrayData($row));
		}

		return $myDataObjectSet;

	}

I hope you guys can help me, so I will be able to help any other people in the future with this matter.