Sometimes the seemingly easiest pieces of code can end up giving you trouble. This morning I was working on a project where I wanted to display the text (not the value) of a selected radio button.
In the end it was very simple. I added a label to my radio buttons and used the following line of code.
$('input[name=myname]:checked + label').text();
Recently I had the necessity to retrieve a concatenated string of values as a column for a report. This can be achieved quite simply by creating a UDF that returns the concatenated string and calling within the SELECT statement.
However, I do not like using UDF’s within SELECT statements due to performance issues. So I started searching the net for a possible alternative and came across the following:
http://www.kodyaz.com/articles/concatenate-using-xml-path.aspx
Now I have never used the SQL STUFF function or XML Path clause but the combination of the two enabled me to achieve my goal very quickly.
SELECT TOP 1
STUFF(
(
SELECT ', ' + Username
FROM Users
FOR XML PATH('')
), 1, 1, ''
) As ListOfUsers
From Users
A fantastic article from Ryan Singer at 37 Signals.
http://37signals.com/papers/introtopatterns/index
The use of design patterns in web design can help developers tackle the tricky balance between user experience and business objectives.
The OVER clause allows partitioning and ordering of the rowset.
The syntax for this function:
Ranking Window Functions
< OVER_CLAUSE > :: =
OVER ( [ PARTITION BY value_expression, ... [ n ] ]
<ORDER BY_Clause> )Aggregate Window Functions
< OVER_CLAUSE > :: =
OVER ( [ PARTITION BY value_expression, ... [ n ] ] )
As specified above the OVER clause can be used in conjunction with aggregate functions, which can make your code a little shorter.
For example, consider the following bookings table:
SELECT * FROM #Bookings ORDER BY CourseID
| CourseID | DelegateID | BookingDate | BookingCost |
| 1 | 1 | 2011-02-04 | 10.00 |
| 1 | 2 | 2011-02-04 | 10.00 |
| 2 | 1 | 2011-01-04 | 10.00 |
| 3 | 1 | 2011-03-04 | 50.00 |
| 3 | 2 | 2011-03-04 | 50.00 |
Now it is quite simple to return the total cost per delegate by using the SUM aggregate function with the OVER clause.
SELECT CourseID, DelegateID, BookingDate, BookingCost, SUM(BookingCost) OVER (PARTITION BY DelegateID) AS TotalCost FROM #Bookings
| CourseID | DelegateID | BookingDate | BookingCost | TotalCost |
| 1 | 1 | 2011-02-04 | 10.00 | 70.00 |
| 2 | 1 | 2011-01-04 | 10.00 | 70.00 |
| 3 | 1 | 2011-03-04 | 50.00 | 70.00 |
| 1 | 2 | 2011-02-04 | 10.00 | 60.00 |
| 3 | 2 | 2011-03-04 | 50.00 | 60.00 |
An alternative SQL sctipt without using the OVER clause would be:
SELECT CourseID, DelegateID, BookingDate, BookingCost,
(SELECT SUM(BookingCost) FROM #Bookings BK WHERE BK.DelegateID = #Bookings.DelegateID) AS TotalCost
FROM #Bookings
