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Is there an equivalent to background-size: cover and contain for image elements?
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400
Solution #1 - The object-fit property (Lacks IE support)
Just set object-fit: cover; on the im...
Regular Expression: Any character that is NOT a letter or number
...To match anything other than letter or number you could try this:
[^a-zA-Z0-9]
And to replace:
var str = 'dfj,dsf7lfsd .sdklfj';
str = str.replace(/[^A-Za-z0-9]/g, ' ');
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CSS: How to position two elements on top of each other, without specifying a height?
...will come across odd and confusing behavior; you probably want to add top: 0; left: 0 to the CSS for both of your absolutely positioned elements. You'll also want to have position: relative on .container_row if you want the absolutely positioned elements to be positioned with respect to their parent...
Any way to declare a size/partial border to a box?
...o declare a size/partial border to a box in CSS? For example a box with 350px that only shows a border-bottom in its firsts 60px . I think that might be very useful.
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How to restart Activity in Android
...heck the current version and call the code snippet above if you're in API 10 or below. (Please don't forget to upvote Ralf's answer!)
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How to handle Objective-C protocols that contain properties?
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edited Apr 22 '16 at 14:10
Dan Rosenstark
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a...
Are there pronounceable names for common Haskell operators? [closed]
...t desugars to >>=)
<$> (f)map
<$ map-replace by 0 <$ f: "f map-replace by 0"
<*> ap(ply) (as it is the same as Control.Monad.ap)
$ (none, just as " " [whitespace])
. pipe to a . b: "b pipe-to a"
!! index
!...
How can I center an absolutely positioned element in a div?
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<body>
<div style="position: absolute; left: 50%;">
<div style="position: relative; left: -50%; border: dotted red 1px;">
I am some centered shrink-to-fit content! <br />
tum te tum
</div>
</div>
</body>
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round up to 2 decimal places in java? [duplicate]
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Well this one works...
double roundOff = Math.round(a * 100.0) / 100.0;
Output is
123.14
Or as @Rufein said
double roundOff = (double) Math.round(a * 100) / 100;
this will do it for you as well.
s...