Trimming Blanks from String
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| Revision as of 22:13, 11 September 2006 (edit) 213.171.204.166 (Talk) (use the much-nicer form of a literal space character) ← Previous diff |
Revision as of 16:11, 22 November 2006 (edit) (undo) Kaiserpanda (Talk | contribs) m Next diff → |
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| [[Category:CookBook]][[Category:StringRecipes]][[Category:Regular_Expressions]] | [[Category:CookBook]][[Category:StringRecipes]][[Category:Regular_Expressions]] | ||
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Revision as of 16:11, 22 November 2006
Problem
You need to ignore leading or trailing space in a string.
Solution
The string module to the rescue again, in the form of the strip command:
1> string:strip(" Some text ").
"Some text"
2> string:strip(" Some text ", left).
"Some text "
3> string:strip(" Some text ", right).
" Some text"
4> string:strip(" Some text ", both).
"Some text"
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Note that the default behavior of 'strip' is to remove spaces from both ends.
However, this solution only handles one type of character; only the '$\s' (space character) will typically be removed. What if you wish to remove all forms of whitespace?
One approach is to use the regular expression library to find (and replace) the unwanted spaces. This should be possible using a single expression and a back reference, but since the standard regular expression implementation seems to only support replacement, the following will work:
1> {_,S,_} = regexp:sub(" foo bar ", "^[ \t]*", ""),
2> regexp:sub(S, "[ \t]*$", "").
{ok,"foo bar",1}
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Or as a function:
-module(cookbook).
-export([trim_whitespace/1]).
trim_whitespace(Input) ->
{_,LS,_} = regexp:sub(Input, "^[ \t]*", ""),
{_,RS,_} = regexp:sub(LS, "[ \t]*$", ""),
RS.
1> c('cookbook').
{ok,cookbook}
2> cookbook:trim_whitespace(" foo bar ").
"foo bar"
19>
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