file - Erlang 从文件中读取前 5 行

Question

例如，我有一个包含 10 个文本字符串的 txt 文件。如何使用 erlang 读取此文本的前 5 个字符串？

谢谢你。

Answer 1

可能你想要 file:open/2 的组合和 file:read_line/1 启用缓冲。

押韵:

$ cat mary_lamb.txt
Mary had a little lamb,
little lamb, little lamb,
Mary had a little lamb,
whose fleece was white as snow.
And everywhere that Mary went,
Mary went, Mary went,
and everywhere that Mary went,
the lamb was sure to go.

源文件:

$ cat ./read_n_lines.erl
-module(read_n_lines).
-export([read_n_lines/2]).

read_n_lines(Filename,NumLines) ->
     {ok, FileDev} = file:open(Filename, 
          [raw, read, read_ahead]),
     Lines = do_read([],FileDev, NumLines),
     file:close(FileDev),
     Lines.

do_read(Lines, _, 0) ->
     lists:reverse(Lines);
do_read(Lines, FileDev, L) ->
     case file:read_line(FileDev) of
          {ok, Line} ->
               do_read([Line|Lines], FileDev, L - 1);
          eof ->
               do_read(Lines, FileDev, 0)
     end.

raw , 在 Modes ，传递给 file:open/2 , 允许更快地访问文件，因为不需要 Erlang 进程来处理文件。

示例运行:

$ erl
1> c(read_n_lines).
{ok,read_n_lines}
2> Lines = read_n_lines:read_n_lines("./mary_lamb.txt", 5).
["Mary had a little lamb,\n","little lamb, little lamb,\n",
 "Mary had a little lamb,\n",
 "whose fleece was white as snow.\n",
 "And everywhere that Mary went,\n"]
3> length(Lines).
5
4> read_n_lines:read_n_lines("./mary_lamb.txt", 666).
["Mary had a little lamb,\n","little lamb, little lamb,\n",
 "Mary had a little lamb,\n",
 "whose fleece was white as snow.\n",
 "And everywhere that Mary went,\n",
 "Mary went, Mary went,\n",
 "and everywhere that Mary went,\n",
 "the lamb was sure to go."]
5> 

要从字符串中删除换行符，可以使用 string:strip/1,2,3 :

5> lists:map(fun(X) -> string:strip(X, right, $\n) end, Lines).
["Mary had a little lamb,","little lamb, little lamb,",
 "Mary had a little lamb,",
 "whose fleece was white as snow.",
 "And everywhere that Mary went,"]
6>