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How to set a selected option of a dropdown list control using angular JS
...d be selected automatically.
Check out sample on http://docs.angularjs.org/api/ng.directive:select which selects red color by default by simply assigning $scope.color='red'.
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Shortcut for creating single item list in C#
...which is "alpha" and it's based on a very stable and heavily used internal API.)
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Maven does not find JUnit tests to run
... It is also worth noting that you must use org.junit.jupiter.api.Test instead of org.junit.Test when using this plugin or the tests will not be found.
– austin_ce
Feb 13 '19 at 20:10
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Convert all first letter to upper case, rest lower for each word
... Haha, really? That's where it is? I love .NET but some of the API designers are real jerks.
– George Mauer
Dec 21 '09 at 23:37
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Secure random token in Node.js
...the same thing, but using the async version of randomBytes. See nodejs.org/api/…
– Alec Thilenius
Sep 28 '16 at 19:50
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How to find the Number of CPU Cores via .NET/C#?
...essors"]);
}
Processors excluded from Windows:
You can also use Windows API calls in setupapi.dll to discover processors that have been excluded from Windows (e.g. through boot settings) and aren't detectable using the above means. The code below gives the total number of logical processors (I ha...
Android ADB device offline, can't issue commands
...ui
Now my development tools are:
SDK rev 16.0.2
SDK tools rev 21.1
SDK API 17, rev 2
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What's the Best Way to Shuffle an NSMutableArray?
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If you import GameplayKit, there is a shuffled API:
https://developer.apple.com/reference/foundation/nsarray/1640855-shuffled
let shuffledArray = array.shuffled()
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Jackson with JSON: Unrecognized field, not marked as ignorable
... this is a very stupid default setting imho, if you add a property to your api, the whole serialization fails
– headsvk
Oct 5 '16 at 17:00
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Merging objects (associative arrays)
...: 110} // Pseudo JS
(assoc. arrays are objects in js)
look here: http://api.jquery.com/jQuery.extend/
edit: Like rymo suggested, it's better to do it this way:
obj3 = $.extend({}, obj1, obj2);
obj3 == {a: 4, b: 2, c: 110}
As here obj1 (and obj2) remain unchanged.
edit2: In 2018 the way ...