Langstrings Options

[decompile]

Hey,

I found Langstrings today and must say its so easy to use but i have one question to it.

strings.ini

[EliminatedTarget]
en = Eliminated $target with $weapon

Syntax: Select all

strings = LangStrings('strings')
message = (strings["EliminatedTarget"][language]

But how can i replace the $target and $weapon ?

This example was taken from ES

[L'In20Cible]

Syntax: Select all

strings['EliminatedTarget'].get_string(language=None, target='foo', weapon='bar')

[decompile]

Thank you perfect.

[decompile]

Syntax: Select all

strings['EliminatedTarget'].get_string(language=None, target='foo', weapon='bar')

Thank you perfect.

Is it also possible to use the keywords from list?

for example

if the options are:

Syntax: Select all

options = ["target" : "joker", "weapon" : "AK47"]
strings['EliminatedTarget'].get_string(language=None, options)

And can i still use $target and $weapon for my keywords in the ini or is there another syntax for that?

[L'In20Cible]

Well, no. You cannot assign them an order cause that can change from a language to another and they need to be translated wherever they are in the sentence so they have to be named keywords.

Actually, there is 2 styles currently implemented to convert keywords and this is going to change. The Template.substitute way (which is the one using the $ prefix) is going to be removed. You will have to use the str.format pattern (minus the indexing part like I mentionned above) so your file will become:

Syntax: Select all

[EliminatedTarget]
en = Eliminated {target} with {weapon}

Anyways, you could also use ** combined with a dict storing your keywords:

Syntax: Select all

keywords = {'target': 'foo', 'weapon': 'bar'}
string = strings['EliminatedTarget'].get_string(language, **keywords)

Since you mentionned list, you could always use pairs with the dict constructor:

Syntax: Select all

keywords = dict([('target', 'foo'), ['weapon', 'bar')])

But heh, prefer the dictionary.