SQL Server Syntax - Search Date

I've noticed that if I search 11/08/2011 it will find my date. But my date picker enters the date as 11/8/2011.
Is there a way to convert the date format somehow in the query so that it finds my date the way it is stored in the database?
2011-11-08 00:00:00.000

Use CONVERT(DATETIME, @yourdatetimevalue).

e.g.
You can put into the SQL Server script the date time from datepicker by replacing the string below with your date time value:

SELECT CONVERT(DATETIME, '11/8/2011')

> Where DayRequested like '%" & Request.Form("MyDay") & "%'

Sounds like a very good opportunity for SQL-injection. Please adapt your code.

Anyway, for SQL-queries I prefer the format "yyyy/mm/dd", that almost never fails. The steps I would take are:
1. Convert the Request-value to a (sub)type Date. If that fails (for example when someone tries SQL-injection), report back to the user that he entered an invalid date.
2. Convert the Date to a string in the format "yyyy/mm/dd" by something like:
    Year(yourDate) & "/" & Month(yourDate) & "/" Day(yourDate)
    and use that string in the query.

I tried the following but shomething must be wrong with the syntax. It returns dates like 11/13/2011 but it wont' return days with single digits like 11/8/2011

Where CONVERT(varchar(10),DayRequested,101) like '%' + CONVERT(varchar(10),'%" & Request.Form("MyDay") & "%',101) + '%'

it is because you've converted it into a string rather than to a DATE at first. So in your code above you can change it like this:

Where CONVERT(varchar(10), CONVERT(DATE, DayRequested)) like '%' + CONVERT(varchar(10), CONVERT(DATE, '" & Request.Form("MyDay") & "'), 101) + '%'

You should be fine!!!

Unfortunately this won't return anything; not even the double digit days.

Thank you very much, your solution is perfect. I've learned a lot.