如何在Erlang中删除字符串中的所有空白字符？

Question

我知道erlang中有string:strip.但它的行为对我来说很奇怪.

A = "  \t\n"  % two whitespaces, one tab and one newline
string:strip(A)   % => "\t\n"
string:strip(A,both,$\n)  %  string:strip/3 can only strip one kind of character

我需要一个函数来删除所有前导/尾随空白字符,包括空格,\ t,\n,\ r \n等.

some_module:better_strip(A)    % => []

erlang有一个功能可以做到这一点吗？或者如果我必须自己做,最好的方法是什么？

Answer 1

试试这个:

re:replace(A, "(^\\s+)|(\\s+$)", "", [global,{return,list}]).

Answer 2

尝试这种结构:

re:replace(A, "\\s+", "", [global,{return,list}]).

示例会话:

Erlang R15B01 (erts-5.9.1) [source] [async-threads:0] [hipe] [kernel-poll:false]

Eshell V5.9.1  (abort with ^G)
1> A = "  21\t\n  ".
"  21\t\n  "
2> re:replace(A, "\\s+", "", [global,{return,list}]).
"21"

UPDATE

上面的解决方案也会删除字符串内的空格符号(不仅仅是引导和拖尾).

如果你只需要剥离前导和拖尾,你可以使用这样的东西:

re:replace(re:replace(A, "\\s+$", "", [global,{return,list}]), "^\\s+", "", [global,{return,list}]).

示例会话:

Erlang R15B01 (erts-5.9.1) [source] [async-threads:0] [hipe] [kernel-poll:false]

Eshell V5.9.1  (abort with ^G)
1> A=" \t \n 2 4 \n \t \n  ".
" \t \n 2 4 \n \t \n  "
2> re:replace(re:replace(A, "\\s+$", "", [global,{return,list}]), "^\\s+", "", [global,{return,list}]).
"2 4"

Answer 3

时下使用 string:trim

1> string:trim("  \t\n").
[]