MS Drops Licensing Restrictions from Web Server 2008 226
Channel Guy writes "According to a report from CRN, Microsoft plans to allow users of the Web Server SKU in Windows Server 2008 to 'run any type of database software with no limit on the number of users, provided they deploy it as an Internet-facing front-end server.' The previous limit was 50 users. Microsoft's partners expect the changes to go a long way toward making Windows Web Server 2008 more competitive with the LAMP stack, against which Microsoft has been making headway in recent months."
Apache responds (Score:5, Funny)
The downside... (Score:4, Funny)
Re:The downside... (Score:5, Funny)
Re:Still have to pay for the OS (Score:5, Funny)
Re:Still have to pay for the OS (Score:3, Funny)
Re:For most of those hosting, the cost is negligab (Score:2, Funny)
Last time I checked, Slashdot [slashdot.org] was still using Linux.
So... Is it a hobby? Or a small-time outfit? :-D
Re:The downside... (Score:4, Funny)
I swear it was in the article. Why are you all looking at me? What?!
Re:Still have to pay for the OS (Score:5, Funny)
I hear they've got the sewing market all stitched up, though.
Re:Netcraft? (Score:3, Funny)
Thank you for telling us without using profanity, or threatening to beat anybody up
Re:so what (Score:4, Funny)
That's Mr. Lysdexic to you, buddy.
Re:so what (Score:2, Funny)
Re:Eight different versions of Windows Server (Score:5, Funny)
That means there will be either 8 or 9 editions of Windows 7, depending on weather it is a geometric or arithmetic progression.
If we attempt to count Windows 2000 (1 desktop, 3 server editions, according to Wikipedia), then we get 1, 4, 6 for desktop versions and a resulting polynomial formula of 0.5(x^2)+4.5x-3 (where x is 1 for 2000, 2 for XP and 3 for Vista) meaning Windows 7 will have (if we take x as 4) 23 editions.
If we instead use x=version no. (5 for 2000, 5.1 for XP and 6 for Vista) then we get the formula -27.778(x^2)+310.56x-857.33 then Windows 7 would have -44.532 editions.
For servers, 1, 2, 3 numbering gives a formula of -2.5(x^2)+12.5x-7 with Sever 7 having 3 editions. With version numbering (and assuming that Server 2008 releases with a 6.0 version number), we get -25(x^2)+280x-772 and Server 7 having -37 editions (assuming it has 7.0 version number).
However, it is best to disregard formulas with negative x^2 coefficients, since they will all eventually result in negative values, therefore 23 versions of Windows 7 seems the most reasonable answer here, unless we take negative edition counts as complete Microsoft failure (CMF).
Re:Point seems have been missed here (Score:3, Funny)
So what you're saying is that slashdot was designed to withstand a slashdotting? Now that's forward thinking!
Re:Still have to pay for the OS (Score:2, Funny)