How to order by an arbitrary condition in SQL

Question

I have the following table:

CREATE TABLE Bable
    (
     id int identity primary key, 
     name varchar(20), 
     about varchar(30)
    );
INSERT INTO Bable (name,about) VALUES
('ООО Name Firm 1','texttexttexttext'),
('ООО Name Firm 2','texttexttexttext'),
('ООО Name Firm 3','texttexttexttext'),
('ООО Name Firm 4','texttexttexttext'),
('ООО Name Firm 5','texttexttexttext'),
('ООО Name Firm $1','texttexttexttext'),
('ООО Name Firm $2','texttexttexttext'),
('ООО Name Firm $3','texttexttexttext'),
('ООО Name Firm 6','texttexttexttext'),
('ООО Name Firm 7','texttexttexttext')

And I can write a query like the following:

SELECT * FROM Bable WHERE about = 'texttexttexttext'

How can I alter this query to return results ordered such that those with names containing "$" appear first, followed by those that do not, with each group then ordered by name ascending?

Structure of the table is here

Solution 1

SELECT *
FROM   Bable
ORDER  BY CASE WHEN name LIKE '%$..' THEN 0 ELSE 1 END,
          Name 

• SQLFiddle Demo

Solution 2

select * from Bable 
order by charindex('$',name,0) desc, name asc

SQL Fiddle Demo

Solution 3

You can do the same with charatindex

SELECT * FROM Bable WHERE about = 'texttexttexttext'
Order by Case When CHARINDEX('$',name)>0 Then 0 Else 1 End,name