Erlang Mnesia选择结果与读取不同，foldl

Question

我有3个用例从mnesia表返回数据的情况 1.返回表的所有值，以便使用foldl， 2. return 1 row so我使用读取 3.根据标准返回可变数量的记录，所以我使用select。

我想使用相同的代码来管理结果，但select会返回不同的数据结构。我希望有人能够帮助我重组我的选择以便像其他人一样返回。

下面是问题和结果的示例代码。问题是select不像read和foldl那样返回表的记录名称。

-module(testselect2). 

-export([runtest/0]). 

-record(record_a, {b, c, d}). 
-record(record_b, {record_a, e}). 
-record(record_c, {record_b, f, intval}). 
runtest() -> 
mnesia:create_schema([node()]), 
mnesia:start(), 
mnesia:create_table(record_c, [{attributes, record_info(fields, record_c)}]), 

A1 = #record_a{b = "R1", c = "T1", d = "C1"}, 
B1 = #record_b{record_a = A1, e = "E1"}, 
C1 = #record_c{record_b = B1, f = "F1", intval = 100},  

A2 = #record_a{b = "R2", c = "T2", d = "C2"}, 
B2 = #record_b{record_a = A2, e = "E2"}, 
C2 = #record_c{record_b = B2, f = "F2", intval = 200},  

A3 = #record_a{b = "R3", c = "T3", d = "C3"}, 
B3 = #record_b{record_a = A3, e = "E3"}, 
C3 = #record_c{record_b = B3, f = "F3", intval = 300},  

{atomic, Rw} = mnesia:transaction(
       fun() -> 
       mnesia:write(C1), 
       mnesia:write(C2), 
       mnesia:write(C3)      
       end), 


io:fwrite("Result write = ~w~n", [Rw]), 

{atomic, Rr} = mnesia:transaction(
       fun() -> 
         mnesia:read({record_c, B1}) 

       end), 
io:fwrite("Result read = ~w~n", [Rr]), 

{atomic, Rf} = 
      mnesia:transaction(fun() -> 
             mnesia:foldl(fun (Rec, Acc) -> [Rec | Acc] end, 
                [], 
                record_c) 
           end), 
io:fwrite("Result foldl = ~w~n", [Rf]),           

MatchHead = #record_c{record_b='$1', f='$2', intval='$3'}, 
Guard = {'>', '$3', 100}, 
Result = {{'$1', '$2', '$3'}}, 
{atomic, Rs} = mnesia:transaction(
       fun() -> 
        mnesia:select(record_c, [{MatchHead, [Guard], [Result]}]) 
       end), 
io:fwrite("Result select = ~w~n", [Rs]). 

===== 成绩

44> testselect2：的runTest（）。

结果写入= OK

结果读取= [{record_c，{record_b，{record_a，[82,49]，[84,49]，[67,49]}，[69,49]} ，[70,49]，100}]

结果foldl = [{record_c，{record_b，{record_a，[82,49]，[84,49]，[67,49]}，[69,49 ]}，[70,49]，100}，{record_c，{record_b，{record_a，[82,51]，[84,51]，[67,51]}，[69,51]}，[70， 51]，300}，{record_c，{record_b，{record_a，[82,50]，[84,50]，[67,50]}，[69,50]}，[70,50]，200}]结果select = [{{record_b，{record_a，[82,51]，[84,51]，[67,51]}，[69,51]}，[70,51]，300}，...， {{record_b，{record_a，[82,50]，[84,50]，[67,50]}，[69,50]}，[70,50]，200}]

确定

正如你可以看到上面阅读和与foldl记录开始{record_c，{...其中选择缺少record_c，只是有{{...

我已经无法找到一种方法让选择返回相同的结构，以便我的处理代码可以适用于所有3种用例。任何建议将不胜感激。

Answer 1

我不是mnesia专家，但我知道当您使用ETS匹配表达式时，您可以确定结果的样子。您可以使用Result = {{'$1', '$2', '$3'}}创建结果条件，这使得它们以一元组形式出现为三元组，如我们在输出中看到的那样。每ets:select/1，你要使用的特殊变量'$_'返回整个匹配的对象，所以这应该在地方工作，你的Result = ...线：

Result = '$_',