Split string with no literal

       Couple of days ago I had a problem with split function in String class. I was having a string with me and I wanted to get the individual characters from it. Split function in string class requires a separator criteria. In my case there was none. so I used a regular expression to do the work. Here is the code for the same.

string sample = "abcdefg";

Regex regex = new Regex("");          
string[] substrings = regex.Split(sample, sample.Length, 1);
foreach (string s in substrings)
{
   if (!string.IsNullOrEmpty(s))
   {
       Console.WriteLine("The splitted character is: " + s);                    
   }
}

The output of this program is splitted string based on every character.
a, b, c, d, e, f, g The for is to just display the output. You can manipulate it as you wanted to.

Pure Join Table and Entity Framework

A table that contains only foreign keys (sometimes called a pure join table) and represents a many-to-many relationship between two tables in the database will not have a corresponding entity in the conceptual model. When the Entity Data Model tools encounter such a table, the table is represented in the conceptual model as a many-to-many association instead of an entity. 

Performing Inner Joins in LINQ

To start with I have created a database schema with 2 tables as;

1. LeftTable

    2. RightTable

Added the sample data as;

If inner join is applied on these 2 tables the result set will return rows when there is at least one match in both the tables.

Using SQL syntax for inner join we get following results;

SELECT lefttable.lefttablefield1, lefttable.lefttablefield2, lefttable.lefttablefield3

FROM lefttable

INNER JOIN righttable

ON lefttable.lefttablefield1 = righttable.lefttablereferrence

I applied the join syntax which we are using in Dynamic Budgets in most of the cases and got the same result set;

var qry = from leftTable in instance.LeftTable

join rightTable in instance.RightTable on

new { a = leftTable.LeftTableField1 } equals

new { a = rightTable.LeftTablereferrence }

select leftTable;

Found result as;