我试图将ToVerbatim添加到Hallgrim公认的答案中:

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;
        }
    }
}

private static string ToVerbatim(string input)
{
    string literal = ToLiteral(input);
    string verbatim = "@" + literal.Replace(@"\r\n", Environment.NewLine);
    return verbatim;
}

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;
        }
    }
}

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 + "\"");
    }
}

这是一个完全可行的实现，包括Unicode和ASCII不可打印字符的转义。它没有像Hallgrim的答案那样插入“+”符号。

static string ToLiteral(string input) {
    StringBuilder literal = new StringBuilder(input.Length + 2);
    literal.Append("\"");
    foreach (var c in input) {
        switch (c) {
            case '\"': literal.Append("\\\""); break;
            case '\\': literal.Append(@"\\"); break;
            case '\0': literal.Append(@"\0"); break;
            case '\a': literal.Append(@"\a"); break;
            case '\b': literal.Append(@"\b"); break;
            case '\f': literal.Append(@"\f"); break;
            case '\n': literal.Append(@"\n"); break;
            case '\r': literal.Append(@"\r"); break;
            case '\t': literal.Append(@"\t"); break;
            case '\v': literal.Append(@"\v"); break;
            default:
                // ASCII printable character
                if (c >= 0x20 && c <= 0x7e) {
                    literal.Append(c);
                // As UTF16 escaped character
                } else {
                    literal.Append(@"\u");
                    literal.Append(((int)c).ToString("x4"));
                }
                break;
        }
    }
    literal.Append("\"");
    return literal.ToString();
}

注意，这也转义了所有Unicode字符。如果你的环境支持它们，你可以改变这一部分，只转义控制字符:

// UTF16 control characters
} else if (Char.GetUnicodeCategory(c) == UnicodeCategory.Control) {
    literal.Append(@"\u");
    literal.Append(((int)c).ToString("x4"));
} else {
    literal.Append(c);
}

Hallgrim的回答非常棒。这里有一个小调整，以防您需要用c#正则表达式解析额外的空白字符和换行符。我需要在序列化的JSON值插入到谷歌表的情况下，遇到了麻烦，因为代码正在插入制表符，+，空格等。

  provider.GenerateCodeFromExpression(new CodePrimitiveExpression(input), writer, null);
  var literal = writer.ToString();
  var r2 = new Regex(@"\"" \+.\n[\s]+\""", RegexOptions.ECMAScript);
  literal = r2.Replace(literal, "");
  return literal;