Peter54 wrote On 06/10/2008 02:05 AM:
> First off I'm not just new to Cake I'm new to programing. I can get my
> head around relatively simple stuff and can generalize once I have a
> concept but even though I know this is either obvious, or been
> answered a thousand times I can't see it anywhere or figure out how it
> works.
>
> I have a two tables Recipes and Ingredients and a JoinTable
> Ingredients_Recipes that contains the foreign keys plus some other
> data.
>
> In my controller I have the following code:
> $recipe = $this->Recipe->find(array('ShowDate'=>$today));
> $this->set('todays_recipe', $recipe);
>
> Which in debug send all the relevant elements of the all tables to the
> view:
> $___dataForView = array(
> "todays_recipe" => array(
> "Recipe" => array(
> "recipe_id" => "20",
> "RecipeName" => "Banana Bread",
> ),
> "Ingredient" => array(
> array(
> "ingredient_id" => "1",
> "IngredientName" => "Eggs",
> "IngredientsRecipe" => array(
> "id" => "1",
> "recipe_id" => "20",
> "ingredient_id" => "1",
> "Amount" => "2",
> )
>
> In my view code
> echo @"The dinner for $today is {$todays_recipe[Recipe]['RecipeName']}
> ';
>
> and I can generalize that to display anyting in the recipe array ie
> serves = $todays_recipe[Recipe]['Serves']";
>
> But when I try this with the Ingredients using - echo "this
> {$today_recipe[Ingredient]['IngredientName']}";
>
> I get undefined constants or undefined variables.
>
> I assume I have to declare the ingredients array in the controller or
> assign it to a variable but how? I don't know and at the risk of
> appearing stupid or irritating I thought I'd ask.
>
Hi,
Isn't it a quote problem ? U do $today_recipe[Ingredient] instead of
$today_recipe['Ingredient'].
I hope it's because of it ;)
--
Simon COURTOIS
{EPITECH.} tek4 | (LINAGORA) Developer | [ADITAM] Project Manager
10, rue Brillat-Savarin 75013 Paris | 01 45 42 72 30 - 06 72 44 67 81
http://www.happynoff.fr
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