在c#中,我可以将字符串值转换为字符串文字,我将在代码中看到它的方式吗?我想用转义序列替换制表符,换行符等。

如果这段代码:

Console.WriteLine(someString);

生产:

Hello
World!

我想要这样的代码:

Console.WriteLine(ToLiteral(someString));

生产:

\tHello\r\n\tWorld!\r\n

当前回答

有趣的问题。

如果你找不到更好的方法,你可以随时替换。 如果你选择它,你可以使用这个c#转义序列列表:

\' - single quote, needed for character literals \" - double quote, needed for string literals \ - backslash \0 - Unicode character 0 \a - Alert (character 7) \b - Backspace (character 8) \f - Form feed (character 12) \n - New line (character 10) \r - Carriage return (character 13) \t - Horizontal tab (character 9) \v - Vertical quote (character 11) \uxxxx - Unicode escape sequence for character with hex value xxxx \xn[n][n][n] - Unicode escape sequence for character with hex value nnnn (variable length version of \uxxxx) \Uxxxxxxxx - Unicode escape sequence for character with hex value xxxxxxxx (for generating surrogates)

这个列表可以在c#常见问题中找到 有哪些字符转义序列可用?

其他回答

如果JSON约定对于你想要转义的未转义字符串足够,并且你已经使用了JSON。NET (Newtonsoft.Json)在你的项目中(它有一个相当大的开销),你可以像下面这样使用这个包:

using System;
using Newtonsoft.Json;

public class Program
{
    public static void Main()
    {
        Console.WriteLine(ToLiteral(@"abc\n123"));
    }

    private static string ToLiteral(string input)
    {
        return JsonConvert.DeserializeObject<string>("\"" + input + "\"");
    }
}
public static class StringEscape
{
  static char[] toEscape = "\0\x1\x2\x3\x4\x5\x6\a\b\t\n\v\f\r\xe\xf\x10\x11\x12\x13\x14\x15\x16\x17\x18\x19\x1a\x1b\x1c\x1d\x1e\x1f\"\\".ToCharArray();
  static string[] literals = @"\0,\x0001,\x0002,\x0003,\x0004,\x0005,\x0006,\a,\b,\t,\n,\v,\f,\r,\x000e,\x000f,\x0010,\x0011,\x0012,\x0013,\x0014,\x0015,\x0016,\x0017,\x0018,\x0019,\x001a,\x001b,\x001c,\x001d,\x001e,\x001f".Split(new char[] { ',' });

  public static string Escape(this string input)
  {
    int i = input.IndexOfAny(toEscape);
    if (i < 0) return input;

    var sb = new System.Text.StringBuilder(input.Length + 5);
    int j = 0;
    do
    {
      sb.Append(input, j, i - j);
      var c = input[i];
      if (c < 0x20) sb.Append(literals[c]); else sb.Append(@"\").Append(c);
    } while ((i = input.IndexOfAny(toEscape, j = ++i)) > 0);

    return sb.Append(input, j, input.Length - j).ToString();
  }
}

有趣的问题。

如果你找不到更好的方法,你可以随时替换。 如果你选择它,你可以使用这个c#转义序列列表:

\' - single quote, needed for character literals \" - double quote, needed for string literals \ - backslash \0 - Unicode character 0 \a - Alert (character 7) \b - Backspace (character 8) \f - Form feed (character 12) \n - New line (character 10) \r - Carriage return (character 13) \t - Horizontal tab (character 9) \v - Vertical quote (character 11) \uxxxx - Unicode escape sequence for character with hex value xxxx \xn[n][n][n] - Unicode escape sequence for character with hex value nnnn (variable length version of \uxxxx) \Uxxxxxxxx - Unicode escape sequence for character with hex value xxxxxxxx (for generating surrogates)

这个列表可以在c#常见问题中找到 有哪些字符转义序列可用?

Hallgrim的回答很好,但是“+”,换行和缩进的添加破坏了我的功能。一个简单的解决方法是:

private static string ToLiteral(string input)
{
    using (var writer = new StringWriter())
    {
        using (var provider = CodeDomProvider.CreateProvider("CSharp"))
        {
            provider.GenerateCodeFromExpression(new CodePrimitiveExpression(input), writer, new CodeGeneratorOptions {IndentString = "\t"});
            var literal = writer.ToString();
            literal = literal.Replace(string.Format("\" +{0}\t\"", Environment.NewLine), "");
            return literal;
        }
    }
}

一个更结构化的方法,包括字符串和字符的所有转义序列,是:

它不会用对应的文字替换Unicode字符。它也不能煮鸡蛋。

public class ReplaceString
{
    static readonly IDictionary<string, string> m_replaceDict
        = new Dictionary<string, string>();

    const string ms_regexEscapes = @"[\a\b\f\n\r\t\v\\""]";

    public static string StringLiteral(string i_string)
    {
        return Regex.Replace(i_string, ms_regexEscapes, match);
    }

    public static string CharLiteral(char c)
    {
        return c == '\'' ? @"'\''" : string.Format("'{0}'", c);
    }

    private static string match(Match m)
    {
        string match = m.ToString();
        if (m_replaceDict.ContainsKey(match))
        {
            return m_replaceDict[match];
        }

        throw new NotSupportedException();
    }

    static ReplaceString()
    {
        m_replaceDict.Add("\a", @"\a");
        m_replaceDict.Add("\b", @"\b");
        m_replaceDict.Add("\f", @"\f");
        m_replaceDict.Add("\n", @"\n");
        m_replaceDict.Add("\r", @"\r");
        m_replaceDict.Add("\t", @"\t");
        m_replaceDict.Add("\v", @"\v");

        m_replaceDict.Add("\\", @"\\");
        m_replaceDict.Add("\0", @"\0");

        //The SO parser gets fooled by the verbatim version
        //of the string to replace - @"\"""
        //so use the 'regular' version
        m_replaceDict.Add("\"", "\\\"");
    }

    static void Main(string[] args){

        string s = "here's a \"\n\tstring\" to test";
        Console.WriteLine(ReplaceString.StringLiteral(s));
        Console.WriteLine(ReplaceString.CharLiteral('c'));
        Console.WriteLine(ReplaceString.CharLiteral('\''));

    }
}