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!
    5·
    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.