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FFmpeg on Android
I have got FFmpeg compiled (libffmpeg.so) on Android. Now I have to build either an application like RockPlayer or use existing Android multimedia framework to invoke FFmpeg.
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Redirect to an external URL from controller action in Spring MVC
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Align image in center and middle within div
...:absolute; width:100%; height:100%">
<img src="http://www.garcard.com/images/garcard_symbol.png">
</div>
JSFiddle
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Extending the User model with custom fields in Django
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The least painful and indeed Django-recommended way of doing this is through a OneToOneField(User) property.
Extending the existing User model
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If you wish to store information related to User, you can use a one-to-one relationship to a model containing the f...
How do I check CPU and Memory Usage in Java?
... edited Aug 21 '19 at 11:01
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answered Sep 16 '08 at 17:22
JeremyJeremy
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Are there any naming convention guidelines for REST APIs? [closed]
...cto standards for naming conventions within the API (eg: URL endpoint path components, querystring parameters)? Are camel caps the norm, or underscores? others?
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How to replace plain URLs with links?
...ur own regexp to parse URLs is a terrible idea. You must imagine this is a common enough problem that someone has written, debugged and tested a library for it, according to the RFCs. URIs are complex - check out the code for URL parsing in Node.js and the Wikipedia page on URI schemes.
There are a...
How can I display just a portion of an image in HTML/CSS?
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@series0ne You could probably combine with background-size
– Stijn de Witt
Jun 19 '15 at 15:13
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Set cURL to use local virtual hosts
...s an option explicitly for this: --resolve
Instead of curl -H 'Host: yada.com' http://127.0.0.1/something
use curl --resolve 'yada.com:80:127.0.0.1' http://yada.com/something
What's the difference, you ask?
Among others, this works with HTTPS. Assuming your local server has a certificate for ya...
How to get an element by its href in jquery?
...attribute selector for that.
var linksToGoogle = $('a[href="http://google.com"]');
Alternatively, if your interest is rather links starting with a certain URL, use the attribute-starts-with selector:
var allLinksToGoogle = $('a[href^="http://google.com"]');
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