cross-posted from: https://lemmy.nz/post/4294116

I have a file with content like this:

item({
     ["attr"] = {
        ["size"] = "62091";
        ["filename"] = "qBuUP9-OTfuzibt6PQX4-g.jpg";
        ["stamp"] = "2023-12-05T19:31:37Z";
        ["xmlns"] = "urn:xmpp:http:upload:0";
        ["content-type"] = "image/jpeg";
     };
     ["key"] = "Wa4AJWFldqRZjBozponbSLRZ";
     ["with"] = "email@address";
     ["when"] = 1701804697;
     ["name"] = "request";
});

I need to know what format this is, and if there exists a tool in linux already to parse this or if I need to write one myself?

Thanks!

  • 56!@lemmy.ml
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    11 months ago

    I think it’s just normal Lua code.

    Here’s a quick json converter (based on https://stackoverflow.com/a/55575074), assuming you have lua installed:

    local function to_json(obj)
        local result = {}
        for key, value in pairs(obj) do
            if type(value) == "string" then
                value = string.format("\"%s\"", value)
            elseif type(value) == "table" then
                value = to_json(value)
            end
            table.insert(result, string.format("\"%s\":%s", key, value))
        end
        return "{" .. table.concat(result, ",") .. "}"
    end
    
    function item(obj)
        print(to_json(obj))
    end
    
    dofile(arg[1])
    

    It just defines the item function to print json, and executes the data file.

    arg[1], the first command line argument, is the path to the data file:

    $  lua to_json.lua path/to/datafile.list
    

    and pipe the output to something.json or whatever else you want to do.