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What is a “batch”, and why is GO used?
I have read and read over MSDN, etc. Ok, so it signals the end of a batch.
6 Answers
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Sum a list of numbers in Python
I have a list of numbers such as [1,2,3,4,5...] , and I want to calculate (1+2)/2 and for the second, (2+3)/2 and the third,
(3+4)/2 , and so on. How can I do that?
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Abstract class in Java
...e. This can be useful if you want people to only implement your interface and no others. However, as a general beginner rule of thumb, if your abstract class only has abstract methods, you should probably make it an interface.
The following is illegal:
interface InterfaceB
{
void interfaceMe...
Difference between abstraction and encapsulation?
What is the precise difference between encapsulation and abstraction?
39 Answers
39
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Android: ScrollView force to bottom
... That doesn't seem to do anything, any suggestions? I tried it on onCreate and later in the onClick of a button.
– RichardJohnn
Aug 2 '10 at 18:50
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NodeJS: Saving a base64-encoded image to disk
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I think you are converting the data a bit more than you need to. Once you create the buffer with the proper encoding, you just need to write the buffer to the file.
var base64Data = req.rawBody.replace(/^data:image\/png;base64,/, "");
requ...
`levels
...The code is simply setting up a sort of pipeline:
Start with dat$product
Convert it to a factor
Change the levels
Store that in res
Personally, I think that line of code is beautiful ;)
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How can I negate the return-value of a process?
...at negates the value a process returns. It should map 0 to some value != 0 and any value != 0 to 0, i.e. the following command should return "yes, nonexistingpath doesn't exist":
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What is the easiest way to get the current day of the week in Android?
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Use the Java Calendar class.
Calendar calendar = Calendar.getInstance();
int day = calendar.get(Calendar.DAY_OF_WEEK);
switch (day) {
case Calendar.SUNDAY:
// Current day is Sunday
break;
case Calendar.MONDAY:
// Current day is Monday
break;
case Calen...
Delete duplicate records in SQL Server?
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You can do this with window functions. It will order the dupes by empId, and delete all but the first one.
delete x from (
select *, rn=row_number() over (partition by EmployeeName order by empId)
from Employee
) x
where rn > 1;
Run it as a select to see what would be deleted:
select *...