Question

Using SQL, how could I give values to dates, grouped by each company?

Current Table:

Date            Company          Employees

2012-04-28      Apple, Inc.      7543
2012-04-27      Apple, Inc.      7510
2012-04-26      Apple, Inc.      7484

2012-04-28      Google, Inc.     11303
2012-04-27      Google, Inc.     11300
2012-04-26      Google, Inc.     11280
2012-04-25      Google, Inc.     11278
2012-04-24      Google, Inc.     11254

Desired Table:

Date            company_day      Company          Employees

2012-04-28      3                Apple, Inc.      7543
2012-04-27      2                Apple, Inc.      7510
2012-04-26      1                Apple, Inc.      7484

2012-04-28      5                Google, Inc.     11303
2012-04-27      4                Google, Inc.     11300
2012-04-26      3                Google, Inc.     11280
2012-04-25      2                Google, Inc.     11278
2012-04-24      1                Google, Inc.     11254

The value for company_day begins at the earliest date for each company. company_day is company specific.

Answer 1

I would first get the MIN date for each company.

Then you can JOIN the result of that query and simply display the difference between it and the Date column.

SELECT 
    MyTable.[Date]
    ,DATEDIFF(MyTable.[Date], MinTable.[MinDate]) + 1 AS Company_Day
    ,MyTable.[Company]
    ,MyTable.[Employees]
FROM
    MyTable
JOIN
    (
    SELECT 
        MIN(b.[Date]) AS MinDate
        , b.[Company] 
    FROM 
       MyTable b 
    GROUP BY 
        b.[Company]
    ) AS MinTable
        ON MinTable.[Company] = MyTable.[Company]

Something like that. This code is untested, but should be close.
http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_datediff
http://www.w3schools.com/sql/func_datediff_mysql.asp