如何将元组对列表转换为Erlang中的记录？

Question

假设我有这个:

-record(my_record, {foo, bar, baz}).

Keyvalpairs = [{foo, val1},
               {bar, val2},
               {baz, val3}].

Foorecord = #my_record{foo=val1, bar=val2, baz=val3}.

如何将Keyvalpairs转换为Foorecord？

Answer 1

最简单的事情是:

Foorecord = #my_record{foo=proplists:get_value(foo, Keyvalpairs), 
      bar=proplists:get_value(bar, Keyvalpairs),
      baz=proplists:get_value(baz, Keyvalpairs)}.

如果这太重复,你可以这样做:

Foorecord = list_to_tuple([my_record|[proplists:get_value(X, Keyvalpairs)
      || X <- record_info(fields, my_record)]]).