Check out this article: http://www.extremetech.com/computing/122725-what-the-ipad-3s-retina-display-l...
Some Awesomeness from the Folks at the W3C
Dynamic outlines: the 'outline' property
The 'outline-color' accepts all colors, as well as the keyword 'invert'.'Invert' is expected to perform a color inversion on the pixels on the screen.This is a common trick to ensure the focus border is visible, regardless of color background.
- Outlines do not take up space.
- Outlines may be non-rectangular.
Outlines have been around since CSS2.1, but hadn't heard about them.
Enjoy!
Play Star Wars Episode IV
Using a simple telnet address you Terminal will connect in and play Star Wars Episode IV in its ASCII glory:
telnet towel.blinkenlights.nl
jQuery
So I have heard so much about the time saving goodness of JavaScript frameworks that I decided to see about it myself. Wow. I've been wasting time and keystrokes.
It took me about ten seconds to permanently switch to jQuery.
Compare the difference between the following two JavaScript fragments:
Non jQuery: document.getElementById("iframe").src = url;
jQuery: $('#iframe').load(url);
See what I mean?!
Resetting Favicons on Apple Safari Web Browser
Favicons not changing on Safari (after all your hard work)?
Here is the solution (hidden in plain sight).
Reset/Renew Safari favicons:
After you click Reset, reload your page and be happy! Validate CSS3
Are you using CSS3 and wonder why the CSS even has the standard if they won't validate perfectly legal CSS3 code? All you have to do is hack the badge code.
Here is the code provided by W3C for CSS validation badge:
Just add
Here is the code provided by W3C for CSS validation badge:
http://jigsaw.w3.org/css-validator/check/referer
?profile=css3so that it looks like this:
http://jigsaw.w3.org/css-validator/check/referer?profile=css3
Pi in Hexidecimal?
Ya, there's (Python) code for that: def pi():
N = 0
n, d = 0, 1
while True:
xn = (120*N**2 + 151*N + 47)
xd = (512*N**4) + 1024*N**3 +
712*N**2 + 194*N + 15)
n = ((16 * n * xd) + (xn * d)) % (d * xd)
d *= xd
yield 16 * n // d
N += 1
N = 0
n, d = 0, 1
while True:
xn = (120*N**2 + 151*N + 47)
xd = (512*N**4) + 1024*N**3 +
712*N**2 + 194*N + 15)
n = ((16 * n * xd) + (xn * d)) % (d * xd)
d *= xd
yield 16 * n // d
N += 1
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