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parse_mod_info: Filter empty strings to marshmallow doesn't get angry #33

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parse_mod_info: Filter empty strings to marshmallow doesn't get angry #33

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84cf606
Ignore "internal" globals when parsing lua
duk3luk3 Aug 31, 2017
6687cab
Fix mod validation failing for empty string values
duk3luk3 Aug 31, 2017
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2 changes: 2 additions & 0 deletions faf/tools/fa/mods.py
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Original file line number Diff line number Diff line change
Expand Up @@ -47,6 +47,8 @@ def parse_mod_info(zip_file_or_folder):
else:
raise ValueError("Not a directory nor a file: " + str(zip_file_or_folder))

lua_data = {k: v for k, v in lua_data.items() if v is not None and v is not ""}

data, errors = ModInfoSchema().load(dict(lua_data))
if errors:
raise ValueError(errors)
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16 changes: 10 additions & 6 deletions faf/tools/lua/parse.py
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Original file line number Diff line number Diff line change
Expand Up @@ -3,7 +3,6 @@
try:
import lupa


def from_lua(lua_code):
"""
Use Lupa as a parser by actually running the code
Expand All @@ -14,18 +13,23 @@ def from_lua(lua_code):
fa_functions_path = pkg_resources.resource_filename('static', 'lua/fa_functions.lua')

lua = lupa.LuaRuntime()
# we don't care about all of the "internal" globals
# especially not the ones that contain binary stuff
# resulting in UnicodeDecodeError when we try to unfold them
default_globals = dict(lua.globals()).keys()
with open(fa_functions_path, 'r') as fp:
lua.execute(fp.read())

lua.execute(lua_code)

def unfold_table(t, seen=None):
def unfold_table(t):
result = {}
for k, v in t.items():
if not lupa.lua_type(v): # Already a python type
result[k] = v
elif lupa.lua_type(v) == 'table':
result[k] = dict(v)
if k not in default_globals:
if not lupa.lua_type(v): # Already a python type
result[k] = v
elif lupa.lua_type(v) == 'table':
result[k] = dict(v)
return result

return unfold_table(lua.globals())
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