Question

我已将\用作escape character LIKE运算符。我正在逃避以下四个字符

1 % 2 [ 3 ] 4 _

当我将转义字符作为输入传递时，查询不会返回值。我怎样才能使它发挥作用？

数据插入

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

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

CODE

    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;
    }

参考：

Answer 1

像这样修改CustomFormat方法：

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

用于处理LIKE的C＃代码

     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 = @"\_";

要检查的其他事项Use of REPLACE in SQL Query for newline/ carriage return characters

转义转义字符不起作用 - SQL LIKE运算符

2 个答案: