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why my navigation bar menu is not changing upon scrolling
Baillie O'Grady
@baillieo
Mrs-coder what are you trying to achieve?
mrs-coder
@mrs-coder
trying to change the navigation bar menu change upon scrolling . Like if we scroll till contact me, then Contact menu on the navigation bar becomes active
armynational
@armynational
hey so im trying to put a background image on my personal portfolio and everything looks right but its not working any help\
detour26
@detour26
Hey everyone, I am trying to do the Personal portfolio challenge and I cant figure out how to (1) get the background image to fit the screen and (2) get the website to scroll up and down.....Any suggestions? thanks
armynational
@armynational
theres a tutuorial on youtube that has helped ,me alot the channel is called code tutorials360
@detour26
detour26
@detour26
@armynational Thanks, I'll go ahead and take a look
CamperBot
@camperbot
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Mois├ęs Man
@moigithub
This message was deleted
Mario Galindo
@mariogalindov
thanks @TylerMoeller I saw the posibble solutions but I think the best I can do for now is to change the image of the thumbnail, thanks to you too @asparism I tried the max height but I do want all of them to be the same size.
CamperBot
@camperbot
mariogalindov sends brownie points to @tylermoeller and @asparism :sparkles: :thumbsup: :sparkles:
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Tyler Moeller
@TylerMoeller
@mrs-coder What you're looking for is called Scroll Spy. Here are the instructions: http://getbootstrap.com/javascript/#scrollspy
John Nunns
@johnnunns
hey guys just need a little help here:
function getCount(str) {
  var vowelsCount = 0;

  str.split(' ')
  switch (str){
    case "a": vowelsCount + 1;
    break;
    case "e": vowelsCount + 1;
    break;
    case "i": vowelsCount + 1;
    break;
    case "o": vowelsCount + 1;
    break;
    case "u": vowelsCount + 1;
    break;
    };

  return vowelsCount;
}
just need to return the vowel count in a given string
probably not using the split property correctly though
c0d0er
@c0d0er
could anybody help for the following 2 questions?
var arr = [              
    [0,0,0]
];
var x=[...arr, [1,1,1,]]
console.log(x)//[ [ 0, 0, 0 ], [ 1, 1, 1 ] ]
x[0][0]=1;
console.log(x)//[ [ 1, 0, 0 ], [ 1, 1, 1 ] ]
console.log(arr)//[ [ 1, 0, 0 ] ]---> why use spread operator, the arr element is still changed?
var arr = [              
    [0,0,0]
];
var x=arr.concat([[1,1,1]]);
console.log(x)//[ [ 0, 0, 0 ], [ 1, 1, 1 ] ]
x[0][0]=1;
console.log(x)//[ [ 1, 0, 0 ], [ 1, 1, 1 ] ]
console.log(arr)//[ [ 1, 0, 0 ] ]---> why use concat, the arr element is still changed?
Tyler Moeller
@TylerMoeller
@johnnunns Correct, .split(' ') will separate your string by spaces rather than characters. Use str.split('') instead. You could then to a str.split('').map to loop through the string and evaluate each character.
@c0d0er Not sure how to answer that question - I'd have to know the context.
c0d0er
@c0d0er
@TylerMoeller there is no other contexts, you can just run, it will comes the result
Tyler Moeller
@TylerMoeller
Are you asking which is better to use?
.concat() vs. spread operator for a case like that, I mean
@c0d0er Or are you asking why the original arr changes when you're only updating x?
c0d0er
@c0d0er
@TylerMoeller yes
@TylerMoeller i am asking why the original arr changes when you're only updating x?
Tyler Moeller
@TylerMoeller
In JavaScript arrays are objects and variables are references to objects. When you update the variable, you are also updating the object. To prevent it, you need to clone the array first.
@c0d0er
Ben Mayer Davis
@benmayerdavis
has anybody else run into problems with adding icons to your document? I'm trying to use Font Awesome but the icons just won't show up in my document
c0d0er
@c0d0er
@TylerMoeller thanks! how to clone? could you give me a simple pseudo code?
CamperBot
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Tyler Moeller
@TylerMoeller
For a shallow copy, you can just do: var copyOfMyArray = myArray.slice(0);
John Nunns
@johnnunns
without going into high level functions yet @TylerMoeller could you tell me where I'm going wrong here?
function getCount(str) {
var sum=0
var i = str.split(");
switch (i){
case 'a': sum + 1;
break;
case 'e': sum +1;
break;
case 'i': sum + 1;
break;
case 'o': sum + 1;
break;
case 'u': sum + 1;
break;
};

return sum

}
c0d0er
@c0d0er
@TylerMoeller this is a really good one! thanks!
CamperBot
@camperbot
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John Nunns
@johnnunns
It's something wrong in my switch statement it looks to me
Tyler Moeller
@TylerMoeller
@johnnunns str.split('') needs to be two apostrophes, not a double quote, for starters :)
John Nunns
@johnnunns
I changed it back to str.split('') forgot to change before i copy pasted
Tyler Moeller
@TylerMoeller
But yes, your switch statement runs once against the array
John Nunns
@johnnunns
yeah ^ :)
Tyler Moeller
@TylerMoeller
For example, say str='hello world';
You are doing:
  switch (['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']){
  case 'a': sum + 1;
  break;
  case 'e': sum + 1;
  break;
  case 'i': sum + 1;
  break;
  case 'o': sum + 1;
  break;
  case 'u': sum + 1;
  break;
};