you switch items from two listing (stack).

undo means reversing the last thing you did. so stack is the most appropriate for this

perhaps try simpler subset of your problem first. maybe change mechanics a bit.

select * from
( select_stmt1 union select_Stmt2) as temp_tbl

not tried but you may try

select SID, count() from student s, sborrow b
where s.SID = b.SID
group by s.SID
order by count(
) desc
limit 1

paste the error would be very helpful

better to host on a vps as you could control everything and not rely on your host

If you want to take out the time component, use date function.

E.g. where date(tbl.datefield) >= date(:dateparam)

the first time you run it, the database if preparing execution plan. afterwards, when it sees the same query again, it will retrieve execution plan from a cache.

mysql also does query caching too, so it is possible the result you got are cached.

could be out of topic, but i strongly suggest PDO

cant you just convert it to a decimal number (e.g. int), add them up, and then convert back to binary?

If Javascript is ok with you, try CopperLicht

i thought he wants to convert the program to a flowchat

we can only give advice on tech problems. for creative, i think you have to resolve yourself

i suggest working on gameplay programming before doing any api stuff. I mean, try working on some logic programming. E.g. implement a number guessing game. Then make up more and more complex game to write.

learning bash is a good investment. very useful if you are into servers and such

  1. is C/C# worth learning if you have basic knowledge of c++ and other programing languages such as python,ruby,lua.......

focus on the platform you want to write code into. for desktop games of course c/c++. for android, then c/c++ is not required

  1. i know math is needed in programing but how advanced does it really get
    enough math for you to solve problems.

3.is learning much languages as possible better

better to focus on 1

  1. isit better to go to college and learn coding or isit possible to learn without a degree

it is best that you love programming. then you dont need to answer this quesiton

If I were you, I would start working on improving your logic skills before diving into more complex game development and working with engines.

start with 2D games and do some simple game like breakout. you may do using ascii graphics first so that you are not overwhelmed learning too many things at once

I recommend 3ds max because it is easy to learn and geared towards gaming. I think they are popular among game developers because it is easy to extend via custom made plugins.

you can export your objects using .obj format and load it on irrlicht. try irredit (editor) to help you easily start going

welcome. btw, i recommend irrlicht. i have great success with it working with 3D.

If just rectable collision detection, just test the for corders if inside the rect.

boolean collide(int x1, int y1, int w1, int h1, 
int x2, int y2, int w2, int h2) {
return 
collide(x1, y1, w1, h1, x2, y2) || 
collide(x1, y1, w1, h1, x2+w2, y2) || 
collide(x1, y1, w1, h1, x2, y2+h2) || 
collide(x1, y1, w1, h1, x2+w2, y2+h2);

}

boolean collide(int x1, int y1, int w, int h, 
int x2, int y2) {
return x2>=x1 && x2 <= x1+w
&& y2>=y1 && y2 <= y1+h;
}

CXF in my opinion is the best framework for this. You may also try JEE, as it only needs annotation to do this. But of course, it is contract last. I prefer contract first Frameworks

idx1 = -1;
idx2 = -1;
for (int i=0; i<arr.length; i++) {
if (num1 == arr[i] && idx1 == -1) { idx1 = i;}
if (num2 == arr[i] && idx2 == -1) { idx2 = i;}
}

if (idx1 != -1 and idx2 != -1) { answer = Math.abs(idx2-idx1); }

I thought bytes are -128 to 127

JamesCherrill commented: Yes. My mistake. +15

I agree have 2 stacks. E.g. do and redo.

doOperation(String str) {
do.add(str);
}

undo() {
String s= do.pop();
redo.add(s);
}

redo() {
String s= redo.pop();
do.add(s);
}

You can access using request.getParameter

You need AJAX request to sent to your backend

I think that is expected. You selected 3 so only 3 will be sent. For others hidden, all will be sent

thank you JorgeM

thank you