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How to check if character is a letter in Javascript?
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You can still use a regex and just add more detail as you need it: str.match(/[A-Z|a-z|ü|é]/i); //etc
– Eli
Jun 22 '15 at 21:54
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Find the min/max element of an Array in JavaScript
...ogrammer I require strongly typed questions.
– ChaosPandion
Nov 3 '09 at 18:49
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Just sharing a j...
Getting distance between two points based on latitude/longitude
I tried implementing this formula: http://andrew.hedges.name/experiments/haversine/
The aplet does good for the two points I am testing:
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jQuery removeClass wildcard
...! How can I check for two or more classes to be removed? i.e. sport-, nav- and color- ?
– lowtechsun
Feb 20 '16 at 15:09
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How to draw a custom UIView that is just a circle - iPhone app
...ally just a ball (a 2D circle)? Would I just override the drawRect method? And can someone show me the code for drawing a blue circle?
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What does enumerable mean?
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An enumerable property is one that can be included in and visited during for..in loops (or a similar iteration of properties, like Object.keys()).
If a property isn't identified as enumerable, the loop will ignore that it's within the object.
var obj = { key: 'val' };
console...
ListView inside ScrollView is not scrolling on Android
... ScrollView . I have an Activity which has some EditTexts in the top part and then a tab host with two tabs which have one ListView each. When the EditText views are focused, the soft keyboard comes up and as I have a ScrollView, the content is scrollable. But the problem comes when there are more...
Android Camera Preview Stretched
I've been working on making my custom camera activity on Android, but when rotating the camera, the aspect ratio of the surface view gets messed up.
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Detect IE version (prior to v9) in JavaScript
...a version of Internet Explorer prior to v9. It's just not worth our time and money to support IE pre-v9 . Users of all other non-IE browsers are fine and shouldn't be bounced. Here's the proposed code:
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Why doesn't Dijkstra's algorithm work for negative weight edges?
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Recall that in Dijkstra's algorithm, once a vertex is marked as "closed" (and out of the open set) - the algorithm found the shortest path to it, and will never have to develop this node again - it assumes the path developed to this path is the shortest.
But with negative weights - it might not be...