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Replacing all recipe ingredient with another.
Posted: Mon Aug 01, 2016 12:13 pm
by kasandraen
Hi, I'va made this script:
Code: Select all
function ReplaceAllIngredientFluidWithFluid(Fluid1 , Fluid2)
for i, recipe in pairs(data.raw.recipe) do
for v, ingredient in pairs(data.raw.recipe[recipe.name].ingredients) do
if ingredient.type == "fluid" then
if ingredient.name == Fluid1 then
ingredient.type = "fluid"
ingredient.name = Fluid2
end
end
end
end
end
to replace all fluid with another one (say change all recipe that uses water to use a new fluid named clean-water) and it works, but I tried changing it to do the same to items, but it doesnt work
Code: Select all
function ReplaceAllIngredientItemWithItem(Item1 , Item2)
for i, recipe in pairs(data.raw.recipe) do
for v, ingredient in pairs(data.raw.recipe[recipe.name].ingredients) do
if ingredient.type == "item" then
if ingredient.name == Item1 then
ingredient.name = Item2
ingredient.amount = ingredient.amount
end
end
end
end
Are you able to see what I did wrong?
or does it even work the same way??
Re: Replacing all recipe ingredient with another.
Posted: Mon Aug 01, 2016 1:24 pm
by daniel34
Recipe ingredients (items) can be confusing, most of them don't have a set type (type is nil) and ingredient.name is also nil most of the time.
You should be able to check if it's an item type by testing if ingredient.type == nil or ingredient.type == "item".
If you have determined that it is an item then check if ingredient.name is nil, in that case use ingredient[1] instead.
Re: Replacing all recipe ingredient with another.
Posted: Wed Aug 03, 2016 2:46 pm
by kasandraen
daniel34 wrote:Recipe ingredients (items) can be confusing, most of them don't have a set type (type is nil) and ingredient.name is also nil most of the time.
You should be able to check if it's an item type by testing if ingredient.type == nil or ingredient.type == "item".
If you have determined that it is an item then check if ingredient.name is nil, in that case use ingredient[1] instead.
Got it working, Thanks!
For future reference if anyone else needs it:
Code: Select all
function ReplaceAllIngredientItemWithItem(Item1 , Item2)
for i, recipe in pairs(data.raw["recipe"]) do
for v, ingredient in pairs(data.raw["recipe"][recipe.name].ingredients) do
if ingredient.name == Item1 then
RemoveFromRecipe(recipe.name,Item1)
AddToRecipe(recipe.name, Item2, ingredient.amount)
elseif ingredient[1] == Item1 then
RemoveFromRecipe(recipe.name,Item1)
AddToRecipe(recipe.name, Item2, ingredient[2])
end
end
end
end
Re: Replacing all recipe ingredient with another.
Posted: Thu Jul 26, 2018 2:52 pm
by Ober3550
Thanks! This was useful for fixing the two different concrete recipes in angels and vanilla!
Re: Replacing all recipe ingredient with another.
Posted: Thu Jul 26, 2018 3:43 pm
by eradicator
For future readers reference i'm gonna leave some code here. Recipes are generally difficult to work with. They can specify inputs and outputs in different formats and places. For example the OPs code does not consider normal= and expensive= difficulties (probably because they didn't exist back then :p).
Here's my attempt at converting any recipe to a "standard" form (as of 0.16.51):
(This is not well tested, please use it as a reference only.)
Code: Select all
--normalizes a recipe to use normal/expensive and fully specified products notation
local function get_normalized_recipe(recipe)
local r = util.table.deepcopy(recipe)
if (not r.normal) and (r.energy_required or r.ingredients or
r.result or r.results)
then
--generates a "normal" node if the recipe uses direct specification
r.normal = {
enabled = r.enabled,
hidden = r.hidden, --TODO: test if hidden belongs here
energy_required = r.energy_required,
ingredients = r.ingredients,
result = r.result,
result_count = r.result_count,
results = r.results
}
r.enabled,r.hidden,r.energy_required,r.ingredients,r.result,r.result_count,r.results
= nil,nil,nil,nil,nil,nil,nil
end
for _,difficulty in pairs{'normal','expensive'} do
if r[difficulty] then
local r = r[difficulty]
--generate a "results" node from singular results and force type specification
if r.result then
r.results = {{type='item',name=r.result,amount=r.result_count or 1}}
elseif r.results then
for _,result in pairs(r.results) do
result.type = result.type or 'item'
end
end
--normalize ingredients to full format
if r.ingredients then
for i,ingredient in pairs(r.ingredients) do
if not ingredient.name then
r.ingredients[i] = {type='item',name=ingredient[1],amount=ingredient[2]}
end
end
end
end
end
return r
end