Escaping the escape character does not work – SQL LIKE Operator

Question

Я использовал \ как escape character за LIKE оператор. I am escaping following four characters

1 % 2 [ 3 ] 4 _

When I pass the escape character as input, the query does not return a value. Как я могу заставить это работать?

Data Insert

DECLARE @Text VARCHAR(MAX)
SET @Text = 'Error \\\ \\  C:\toolbox\line 180'

INSERT INTO Account (AccountNumber,AccountType,Duration,ModifiedTime) 
VALUES (198,@Text,1,GETDATE())

КОД

    static void Main(string[] args)
    {

        string searchValue1 = @"Error \\\ \\  C:\toolbox\line 180";
        string searchValue2 = @"55555";

        string result1 = DisplayTest(searchValue1);
        string result2 =  DisplayTest(searchValue2);

        Console.WriteLine("result1:: " + result1);
        Console.WriteLine("result2:: " + result2);
        Console.ReadLine();

    }}


     private static string DisplayTest(string searchValue)
    {
        searchValue = CustomFormat(searchValue);


        string test = String.Empty;
        string connectionString = "Data Source=.;Initial Catalog=LibraryReservationSystem;Integrated Security=True;Connect Timeout=30";

        using (SqlConnection connection = new SqlConnection(connectionString))
        {
            connection.Open();
            string commandText = @"SELECT AccountType,* 
                              FROM Account 
                              WHERE AccountType LIKE @input ESCAPE '\'";
            using (SqlCommand command = new SqlCommand(commandText, connection))
            {
                command.CommandType = System.Data.CommandType.Text;
                command.Parameters.AddWithValue("@input", "%" + searchValue + "%");

                using (SqlDataReader reader = command.ExecuteReader())
                {
                    if (reader.HasRows)
                    {
                        while (reader.Read())
                        {

                            test = reader.GetString(0);
                        }
                    }
                }
            }
        }

        return test;
    }


    private static string CustomFormat(string input)
    {
        input = input.Replace(@"%", @"\%");
        input = input.Replace(@"[", @"\[");
        input = input.Replace(@"]", @"\]");
        input = input.Replace(@"_", @"\_");
        //input = input.Replace(@"\", @"\\");
        return input;
    }

ССЫЛКА:

1. Как я могу избежать квадратных скобок в предложении LIKE?
2. How to escape a string for use with the LIKE operator in SQL Server?

Answer 1

Изменить ваш CustomFormat метод как это:

private static string CustomFormat(string input)
{
    input = input.Replace(@"\", @"\\"); 
    input = input.Replace(@"%", @"\%");
    input = input.Replace(@"[", @"\[");
    input = input.Replace(@"]", @"\]");
    input = input.Replace(@"_", @"\_");
    return input;
}

Answer 2

C# код для обработки как

     public static string WildcardFormatSpecialCharacter(string source)
    {
        string formattedResult = string.Empty;
        if (!String.IsNullOrEmpty(source))
        {
            //Escape the escape character
            formattedResult = source.Replace(DataLayerConstants.EscapeCharacter, DataLayerConstants.EscapeCharacterWithEscape);
            //The %
            formattedResult = formattedResult.Replace(DataLayerConstants.Percentage, DataLayerConstants.PercentageWithEscape);
            //The [
            formattedResult = formattedResult.Replace(DataLayerConstants.OpenSqaureBracket, DataLayerConstants.OpenSqaureBracketWithEscape);
            //The ]
            formattedResult = formattedResult.Replace(DataLayerConstants.CloseSqaureBracket, DataLayerConstants.CloseSqaureBracketWithEscape);
            //The _
            formattedResult = formattedResult.Replace(DataLayerConstants.Underscore, DataLayerConstants.UnderscoreWithEscape);
        }
        return formattedResult;
    }


    public const string EscapeCharacter = @"\";
    public const string EscapeCharacterWithEscape = @"\\";
    public const string Percentage = "%";
    public const string PercentageWithEscape = @"\%";
    public const string OpenSqaureBracket = "[";
    public const string OpenSqaureBracketWithEscape = @"\[";
    public const string CloseSqaureBracket = "]";
    public const string CloseSqaureBracketWithEscape = @"\]";
    public const string Underscore = "_";
    public const string UnderscoreWithEscape = @"\_";

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