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Firebug says “No Javascript on this page”, even though JavaScript does exist on the page
...functionality but Firebug per se is never going to be fixed for Firefox 50 and beyond
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How to merge specific files from Git branches
I have 2 git branches branch1 and branch2 and I want to merge file.py in branch2 into file.py in branch1 and only that file.
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How can I install Apache Ant on Mac OS X?
I tried to install Apache Ant on my Mac and I followed the next steps :
8 Answers
8
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Javascript calculate the day of the year (1 - 366)
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Edit: The code above will fail when now is a date in between march 26th and October 29th andnow's time is before 1AM (eg 00:59:59). This is due to the code not taking daylight savings time into account. You should compensate for this:
var now = new Date();
var start = new Date(now.getFullY...
Difference between len() and .__len__()?
...y calling an object's __len__ method. __something__ attributes are special and usually more than meets the eye, and generally should not be called directly.
It was decided at some point long ago getting the length of something should be a function and not a method code, reasoning that len(a)'s mean...
Placement of the asterisk in pointer declarations
I've recently decided that I just have to finally learn C/C++, and there is one thing I do not really understand about pointers or more precisely, their definition.
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What is a callback?
What's a callback and how is it implemented in C#?
11 Answers
11
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How to remove application from app listings on Android Developer Console
Is there any way to unpublish and then permanently remove an application from the list of applications on Android Developer Console?
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throw new std::exception vs throw std::exception
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The conventional way to throw and catch exceptions is to throw an exception object and to catch it by reference (usually const reference). The C++ language requires the compiler to generate the appropriate code to construct the exception object and to pro...
How do I find duplicates across multiple columns?
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Duplicated id for pairs name and city:
select s.id, t.*
from [stuff] s
join (
select name, city, count(*) as qty
from [stuff]
group by name, city
having count(*) > 1
) t on s.name = t.name and s.city = t.city
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