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我用这种语法声明了 Line 变量：

Line =“1;nabil;chouaib;法国;27”,

我想声明 5 个变量：Id 、 Firsname 、 lastname 、 address 、 age

Id 应包含 1 Firsname 应包含 nabil 姓应包含 chouaib 地址应包含 france 年龄应包含 27

所以我必须解析变量Line的内容

在这种情况下，分隔符是;

我有这个功能：

test()->

    ForEachLine = fun(Line,Buffer)-> 
                          io:format("Line: ~s~n",Line),


                          [IdStr, FirstnameStr, LastnameStr, AddressTr, AgeStr] = string:tokens(Line, ";"),
   Buffer end,

 InitialBuffer = [],

 csv:parse("/var/lib/mysql/test/person1.csv",ForEachLine,InitialBuffer).

但是当我执行测试功能 T 有这个错误： 

1> model:test().
Line: 1;nabil;chouaib;france;27
** exception error: no match of right hand side value [["1;nabil;chouaib;france;27"]]
     in function  model:'-test/0-fun-0-'/2
     in call from csv:start_parsing/3

问题与此行有关：

[IdStr, Firstname, Lastname, Address, AgeStr] = string:tokens(Line, ";")

我认为Line的类型不是字符串（但在错误之前的控制台中显示此行 Line: 1;nabil;chouaib;france;27）

所以我应该只知道Line的类型

csv.erl 模块是：

%%% --- csv parser in Erlang. ------
%%% To help process large csv files without loading them into
%%% memory. Similar to the xml parsing technique of SAX

-module(csv).
-compile(export_all).

parse(FilePath,ForEachLine,Opaque)->
    case file:open(FilePath,[read]) of
        {_,S} ->
            start_parsing(S,ForEachLine,Opaque);
        Error -> Error
    end.

start_parsing(S,ForEachLine,Opaque)->
    Line = io:get_line(S,''),
    case Line of
        eof -> {ok,Opaque};
        "\n" -> start_parsing(S,ForEachLine,Opaque);
        "\r\n" -> start_parsing(S,ForEachLine,Opaque);
        _ -> 
            NewOpaque = ForEachLine(scanner(clean(clean(Line,10),13)),Opaque),
            start_parsing(S,ForEachLine,NewOpaque)
    end.

scan(InitString,Char,[Head|Buffer]) when Head == Char -> 
    {lists:reverse(InitString),Buffer};
scan(InitString,Char,[Head|Buffer]) when Head =/= Char ->
    scan([Head|InitString],Char,Buffer);
scan(X,_,Buffer) when Buffer == [] -> {done,lists:reverse(X)}.
scanner(Text)-> lists:reverse(traverse_text(Text,[])).

traverse_text(Text,Buff)->
    case scan("",$,,Text) of
        {done,SomeText}-> [SomeText|Buff];
        {Value,Rem}-> traverse_text(Rem,[Value|Buff])
    end.

clean(Text,Char)-> 
    string:strip(string:strip(Text,right,Char),left,Char).
4

1 回答 1

7

您可以使用string:tokens/2函数将字符串拆分为字符串列表，并模式匹配列表以提取变量：

1> [IdStr, Firstname, Lastname, Address, AgeStr] = string:tokens(Line, ";").
["1","nabil","chouaib","france","27"]

然后可以使用list_to_integer/1函数将 IdStr 和 AgeStr 转换为整数：

2> Id = list_to_integer(IdStr).
1
3> Age = list_to_integer(AgeStr).
27
于 2013-01-26T20:04:49.547 回答