Use of SqlParameter in SQL LIKE clause not working

What you want is:

tblCustomerInfo.Info LIKE '%' + @SEARCH + '%'

(or edit the parameter value to include the % in the first place).

Otherwise, you are either (first sample) searching for the literal "@SEARCH" (not the arg-value), or you are embedding some extra quotes into the query (second sample).

In some ways, it might be easier to have the TSQL just use LIKE @SEARCH, and handle it at the caller:

command.Parameters.AddWithValue("@SEARCH","%" + searchString + "%");

Either approach should work.

Just a little careful with a slight difference between Add and AddWithValue methods. I had the problem below, when I used the Add method and put the wrong SqlType parameter.

• nchar and nvarchar can store Unicode characters.
• char and varchar cannot store Unicode characters.

For example:

string query = " ... WHERE stLogin LIKE @LOGIN ";

SqlParameter p = new SqlParameter("@LOGIN", SqlDbType.Char, 255) 
{ 
    Value = "%" + login + "%" 
};

command.Parameters.AddWithValue(p.ParameterName, p.Value); //works fine!!!

command.Parameters.Add(p); // won't work

When I changed the SqlType to NVarChar, the two methods worked fine to me.

SqlParameter p = new SqlParameter("@LOGIN", SqlDbType.NVarChar, 255) 
{ 
    Value = "%" + login + "%" 
};

command.Parameters.AddWithValue(p.ParameterName, p.Value); //worked fine!!!

command.Parameters.Add(p); //worked fine!!!

Instead of using:

const string Sql = 
@"select distinct [name] 
  from tblCustomers 
  left outer join tblCustomerInfo on tblCustomers.Id = tblCustomerInfo.CustomerId  
  where (tblCustomer.Name LIKE '%@SEARCH%' OR tblCustomerInfo.Info LIKE '%@SEARCH%');";

Use this code:

const string Sql = 
@"select distinct [name] 
  from tblCustomers 
  left outer join tblCustomerInfo on tblCustomers.Id = tblCustomerInfo.CustomerId  
  where (tblCustomer.Name LIKE '%' + @SEARCH + '%' OR tblCustomerInfo.Info LIKE '%' + @SEARCH + '%');";