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Avoid synchronized(this) in Java?
Whenever a question pops up on SO about Java synchronization, some people are very eager to point out that synchronized(this) should be avoided. Instead, they claim, a lock on a private reference is to be preferred.
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Clicking URLs opens default browser
...ge loaded, it has to work like a normal browser and open the link in the same WebView . But it's opening the default browser and loading the page there?
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How to delete all Annotations on a MKMapView
...l the annotations besides the user location?
– kevin Mendoza
Jun 13 '10 at 22:17
1
You do not nee...
Entity Framework Timeouts
I am getting timeouts using the Entity Framework (EF) when using a function import that takes over 30 seconds to complete. I tried the following and have not been able to resolve this issue:
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Remove CSS “top” and “left” attributes with jQuery
Im building a draggable map that when the map is dragged the element is given a 'left' and 'top' attribute with values for each as so...
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Why does the PHP json_encode function convert UTF-8 strings to hexadecimal entities?
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Aha. Thanks! I should have read the documentation more carefully. Thanks.
– David Jones
May 11 '13 at 14:47
3
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How do I add a margin between bootstrap columns without wrapping [duplicate]
...div class="col-md-4" id="server_1">
<div class="server-action-menu">
Server 1
</div>
</div>
</div>
CSS
.server-action-menu {
background-color: transparent;
background-image: linear-gradient(to bottom, rgba(30, 87, 153, 0.2) 0%, rgba(1...
What is meant by Resource Acquisition is Initialization (RAII)?
What is meant by Resource Acquisition is Initialization (RAII)?
7 Answers
7
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angularJS: How to call child scope function in parent scope
How can call a method defined in child scope from its parent scope?
4 Answers
4
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JSON Array iteration in Android/Java
...value = settings.getJSONObject(i).getString("value");
String name = settings.getJSONObject(i).getString("name");
applicationSettings.put(name, value);
}
2.) make a JSONArray of names
JSONArray names = json.names();
JSONArray values = json.toJSONArray(names)...
