How to self-join table in a way that every record is joined with the “previous” record?

Question

I have a MS SQL table that contains stock data with the following columns: Id, Symbol, Date, Open, High, Low, Close.

I would like to self-join the table

Answer 1

You could do something like this:

DECLARE @Today DATETIME
SELECT @Today = DATEADD(DAY, 0, DATEDIFF(DAY, 0, CURRENT_TIMESTAMP))

;WITH today AS
(
    SELECT  Id ,
            Symbol ,
            Date ,
            [OPEN] ,
            High ,
            LOW ,
            [CLOSE],
            DATEADD(DAY, -1, Date) AS yesterday 
    FROM quotes
    WHERE date = @today
)
SELECT *
FROM today
LEFT JOIN quotes yesterday ON today.Symbol = yesterday.Symbol
    AND today.yesterday = yesterday.Date

That way you limit your "today" results, if that's an option.

EDIT: The CTEs listed as other questions may work well, but I tend to be hesitant to use ROW_NUMBER when dealing with 100K rows or more. If the previous day may not always be yesterday, I tend to prefer to pull out the check for the previous day in its own query then use it for reference:

DECLARE @Today DATETIME, @PreviousDay DATETIME
SELECT @Today = DATEADD(DAY, 0, DATEDIFF(DAY, 0, CURRENT_TIMESTAMP));
SELECT @PreviousDay = MAX(Date) FROM quotes  WHERE Date < @Today;
WITH today AS
(
    SELECT  Id ,
            Symbol ,
            Date ,
            [OPEN] ,
            High ,
            LOW ,
            [CLOSE]
    FROM quotes 
    WHERE date = @today
)
SELECT *
FROM today
LEFT JOIN quotes AS previousday
    ON today.Symbol = previousday.Symbol
    AND previousday.Date = @PreviousDay