Asp.net-mvc – How to trim spaces of model in ASP.NET MVC Web API

asp.net-mvcmodel-bindingtrim

What is the best way to trim all the properties of the model passed to the MVC web api (post method with complex object). One thing simply can be done is calling Trim function in the getter of all the properties. But, I really do not like that.

I want the simple way something like the one mentioned for the MVC here ASP.NET MVC: Best way to trim strings after data entry. Should I create a custom model binder?

Best Answer

To trim all incoming string values in Web API, you can define a Newtonsoft.Json.JsonConverter:

class TrimmingConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(string);
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        if (reader.TokenType == JsonToken.String)
            if (reader.Value != null)
                return (reader.Value as string).Trim();

        return reader.Value;
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        var text = (string)value;
        if (text == null)
            writer.WriteNull();
        else
            writer.WriteValue(text.Trim());
    }
}

Then register this on Application_Start. The convention to do this in FormatterConfig, but you can also do this in Application_Start of Global.asax.cs. Here it is in FormatterConfig:

public static class FormatterConfig
{
    public static void Register(HttpConfiguration config)
    {
        config.Formatters.JsonFormatter.SerializerSettings.Converters
            .Add(new TrimmingConverter());

    }
}

Related Solutions

C++ – How to trim a std::string

EDIT Since c++17, some parts of the standard library were removed. Fortunately, starting with c++11, we have lambdas which are a superior solution.

#include <algorithm> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](unsigned char ch) {
        return !std::isspace(ch);
    }));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(), [](unsigned char ch) {
        return !std::isspace(ch);
    }).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    ltrim(s);
    rtrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

Thanks to https://stackoverflow.com/a/44973498/524503 for bringing up the modern solution.

Original answer:

I tend to use one of these 3 for my trimming needs:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start
static inline std::string &ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
    return s;
}

// trim from end
static inline std::string &rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
    return s;
}

// trim from both ends
static inline std::string &trim(std::string &s) {
    return ltrim(rtrim(s));
}

They are fairly self-explanatory and work very well.

EDIT: BTW, I have std::ptr_fun in there to help disambiguate std::isspace because there is actually a second definition which supports locales. This could have been a cast just the same, but I tend to like this better.

EDIT: To address some comments about accepting a parameter by reference, modifying and returning it. I Agree. An implementation that I would likely prefer would be two sets of functions, one for in place and one which makes a copy. A better set of examples would be:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    ltrim(s);
    rtrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

I am keeping the original answer above though for context and in the interest of keeping the high voted answer still available.

Bash – How to trim whitespace from a Bash variable

A simple answer is:

echo "   lol  " | xargs

Xargs will do the trimming for you. It's one command/program, no parameters, returns the trimmed string, easy as that!

Note: this doesn't remove all internal spaces so "foo bar" stays the same; it does NOT become "foobar". However, multiple spaces will be condensed to single spaces, so "foo bar" will become "foo bar". In addition it doesn't remove end of lines characters.

Best Answer

Related Solutions

C++ – How to trim a std::string

Original answer:

Bash – How to trim whitespace from a Bash variable

Related Topic