我有一个客户端和一个服务器,我使用客户端发送一个字符串,例如 "Hello"
它在服务器端显示为 <<"Hello">>
.有没有办法简单地转换 <<"Hello">>
返回"Hello"
?我尝试使用,
String = io_lib:format("~p",[StringIn]),
lists:flatten(String),
但这使得格式为 <<"Hello">>
变成[[60,60,"\"Hello\"",62,62]]
...
顺便说下代码:
-module(ss1).
-compile(export_all).
-import(lists, [reverse/1]).
client() ->
{ok, Socket} = gen_tcp:connect("localhost", 2345,[binary, {packet, 4}]),
ok = gen_tcp:send(Socket, "Hello"),
receive
{tcp,Socket,String} ->
io:format("Client received = ~p~n",[String]),
io:format("Client result = ~p~n",[String]),
gen_tcp:close(Socket)
end.
server() ->
{ok, Listen} = gen_tcp:listen(2345, [binary, {packet, 4}, {reuseaddr, true}, {active, true}]),
{ok, Socket} = gen_tcp:accept(Listen),
gen_tcp:close(Listen),
loop(Socket).
loop(Socket) ->
receive
{tcp,Socket,String} ->
io:format("Server received String = ~p~n",[String]),
io:format("Server replying String = ~p~n",[String]),
gen_tcp:send(Socket, String),
loop(Socket);
{tcp_closed, Socket} ->
io:format("Server socket closed~n")
end.
最佳答案
原因是这一行:
{ok, Listen} = gen_tcp:listen(2345, [binary, {packet, 4}, {reuseaddr, true}, {active, true}]),
具体来说,在选项中传递的原子 binary
。如果您改为传递 list
,您将获得一个常规字符串。查看documentation对于 gen_tcp:listen/2
。
以下代码不起作用的原因:
String = io_lib:format("~p",[StringIn]),
lists:flatten(String), ...
是 lists:flatten/1
实际上没有改变 String
;它返回一个新列表,该列表是展平其参数的结果。这会奏效:
lists:flatten(io_lib:format("~p", [StringIn])).
关于string - Erlang:将 TCP 发送的字符串转换为正确的形式,例如<<"SomeString">> 到 "SomeString"?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15512689/