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    remove A (#31157) (compare)

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Stephen James
@sjames1958gm
@mitchell369 You need 64 < char && char < 72 to have two numeric comparisons
mitchell369
@mitchell369
@sjames1958gm so can I use a colon after the first evalution then see if it is <|> ?
ok, I don't havemuch experience with && but I will give that a go
Stephen James
@sjames1958gm
@mitchell369 Just read && as AND
Unwana Essien
@afixoftrix
I don't understand why it doesn’t understand spread operator.
mitchell369
@mitchell369
@sjames1958gm That completely destroy the output lol
Stephen James
@sjames1958gm
@mitchell369 char.charCodeAt() % 26 + 65 what is this trying to do?
Markus Kiili
@Masd925
@sjames1958gm Spread operator is ES6 (ES2015).
mitchell369
@mitchell369
@sjames1958gm Its evaluating the remainder when I pass it through 26 character alphabet, I found that is the most effective method to produce the A => N conversion and Z => M or w/e it is. Adding the 65 get's you back to the Ascii recognized coding for A-Z
Unwana Essien
@afixoftrix
@mitchell369 Ok I just added babel ‘stage-3’ presets and it worked.
Stephen James
@sjames1958gm
@Masd925 Odd, it appears that for some reason babel doesn't work for objects like above with just es2015 preset
Markus Kiili
@Masd925
@sjames1958gm Compatibility of spread also differs depending on where/how it is used.
Stephen James
@sjames1958gm
@mitchell369 Why is the second number 72? and not 91 (one more than Z
mitchell369
@mitchell369
@sjames1958gm right on, good start... should be 92
Stephen James
@sjames1958gm
@mitchell369 BTW, your 64 < char && char < 72 is comparing the string value char not the char code of char
mitchell369
@mitchell369
@sjames1958gm That's a good lead, basically what I see it doing is changing all values that fall below that range into 65 and providing me with the output
Stephen James
@sjames1958gm
@mitchell369 You need to make sure that the types in your variables match the types you are comparing
mitchell369
@mitchell369
@sjames1958gm specifying the the limits within the charCodeAt(), seems to return hex values...
Stephen James
@sjames1958gm
@mitchell369 Your algorithm is close. just that the comparison should be with charCode not char
var c = char.charCodeAt();
return String.fromCharCode(64 < c && c < 91 ? c % 26 + 65 : c);
Ahmad Abdolsaheb
@ahmadabdolsaheb

hey everyone, I am trying to use filter, however, i want it to some times not filter anything. this is currently what i have

cards.filter(card=> card.type == "awards")

i have a variety of cards and sometimes i want filter to include all of the cards. how can I do that?

Ken Haduch
@khaduch

@ahmadabdolsaheb - without knowing all that you're doing, how about not calling the .filter when you don't want to actually filter, or if you make a "control" variable like this:

let doAll = true;
cards = cards.filter(card=> card.type == "awards" || doAll );

So that you can set doAll to the value you want (based on some condition) and then it would always pass the test? That's at least one take at it... You have to catch the return from the .filter, so I just threw that in there...

Ahmad Abdolsaheb
@ahmadabdolsaheb
@khaduch so cards.filter(card=> card.type == true ); does not seem to return all items.
i tried it in js fiddle
array = [1,2,3,5]; console.log(array.filter(num => num == true)); only returns one
Markus Kiili
@Masd925
@ahmadabdolsaheb What are you trying to do? Why use abstract equality == (which nobody remember how it works really).
Ahmad Abdolsaheb
@ahmadabdolsaheb
@Masd925 i have cards with different types. i wanna show a specific card type when a relevant button is pressed. but, sometime, i wanna show all cards.
i am trying to use filter not to filter :)
cards.filter(card=> card.type == "awards")
what should I use instead of "awards and includes everything
or
cards.filter(card=> card.type === "awards")
Markus Kiili
@Masd925
@ahmadabdolsaheb There is no value that would be equal to all strings. cards.filter(card => true) would keep all elements.
cards.filter(card => card.type) would also keep all cards provided that types are non-empty strings.
@ahmadabdolsaheb I think that changing the logic as @khaduch said is best.
Ahmad Abdolsaheb
@ahmadabdolsaheb
@Masd925 @khaduch i think i will change logic since making a control variable didn't work. thanks
CamperBot
@camperbot
ahmadabdolsaheb sends brownie points to @masd925 and @khaduch :sparkles: :thumbsup: :sparkles:
:star2: 3677 | @khaduch |http://www.freecodecamp.org/khaduch
:star2: 4653 | @masd925 |http://www.freecodecamp.org/masd925
Ken Haduch
@khaduch
:point_up: January 15, 2018 9:20 AM @ahmadabdolsaheb - you need to do something like what I posted in my comment above yours - cards = cards.filter(card=> card.type == "awards" || doAll ); - the variable doAll would be set to true or false (or something that evaluated to true or false) and if it did not match the first conditional test, then if doAll evaluated to true, it would add the card to the output array. It is the || "or" condition that makes it work. It is a common way to affect conditional tests. So it will work, but it might not always be the best way to do things, for example, if you have a 1000 element array that you don't need to filter, then maybe it is not too efficient to do this (just thinking of performance).
Jason Luboff
@JLuboff
:wave:
Gersho
@Gersho
:wave:
Jason Luboff
@JLuboff
What up
Markus Kiili
@Masd925
Yo
Jason Luboff
@JLuboff
@Masd925 Yo'
Kaz Baig
@kbaig
:wave:
Jason Luboff
@JLuboff
Whats good?
Kaz Baig
@kbaig
TFW you were sick last week so you didn't get much done at work, but now you're even more sick and need to start working harder to catch up :(
Jason Luboff
@JLuboff
Flu?
Markus Kiili
@Masd925
@JLuboff I finally did a portfolio page.
Jason Luboff
@JLuboff
@Masd925 Impressive!
Kaz Baig
@kbaig
@JLuboff sinus infection into bronchitis about 4 weeks ago that almost ended but now I'm coughing hard again and throat is swollen :(