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In Mongoose, let's say I have a findOne with two conditions. If I don't get a doc result in my callback, how do I know which condition was the one that failed? Is there a way to write the code so that I can tell?
@sgoldber61 if you need to know, why not break up the search?
@cmccormack How do I break up the search?
@sgoldber61 just nest two searches
Then you can add logic to determine what to do if it's missing one criteria
@cmccormack OK I see. It's a little inelegant to have to search again, but I guess I have to. I'm doing the nightlife coordination app, and I'm interested in storing in a database the bars that have a nonzero amount of people going. If a user clicks on a bar, I need to check if the bar is in the database. If the bar is not in the database, then I need to put it in. But if the bar is in the database, I need to check to see if the user had already clicked "going" by seeing if the user is $in the userid array for the bar.
And I guess I have to search again within the callback...
I don't see why that would be inelegant. You're doing different things, it's not just one query
You could always query just using one criteria and use the response documents to search for whatever else you need and then filter in JS
I need to check if the bar is in the database. If the bar is not in the database, then I need to put it in
maybe using a method like findOneAndUpdate() with upsert: true would handle that, along with the flag to return the modified (or new, if upsert) document, which you could then check on the front end to see if your user is present?
that way you only make a single query
i'm not sure if there's a mongoose equivalent to that method, i've only ever used plain mongo
Yeah I can do that in mongoose, but is it actually legit to check on the front end if the user is present? What if I (say, using something like Postman) ask the back end to have a user "like" a bar but the bar has already been liked?
I could linearly search through the array that has the list of users going/liking the bar using a simple JS for loop, but I take it I'm not supposed to do that.
You can do it in the callback on the backend, doesn't have to be on the frontend
looks like i did 2 separate calls to my db
how i did it if it helps any
Is it better to linearly search through that array of liking users or is it better to make a 2nd call to the db? I've heard that if you linearly search through the array yourself it takes O(n) time, is it the case that making a 2nd call to the db and using $in for teh user array only takes O(log(n)) time because of how mongodb structures itself internally, or is that not true?
But if the bar is in the database, I need to check to see if the user had already clicked "going" by seeing if the user is $in the userid array for the bar.
If you just grab all the bars in the database using @thekholm80 suggestion, you can use the results to determine if they have already clicked 'going'
@sgoldber61 it's only O(n) if unsorted
i've never had more than 1 user, so i've never worried about performance lol. I suppose it's something I should be aware of
So then I put the list of users in a priority queue aka heap? I.e. this: https://en.wikipedia.org/wiki/Heap_(data_structure) Does mongodb, when I add a user... do this efficiently?
Sorry for the endless questions, I'll stop, but I know so little about performance or actually how to properly do common things in an idiomatic way in mongodb/mongoose.
i wish i could help you more. i've learned just enough mongodb to do what i've needed so far, which is barely scratching the surface of what it's capable of
and if they don't release an update for debian 9 i may be ditching mongo altogether
Sometimes I feel like I don't know anything. It's been since last summer since I started this, and I've only gotten to the nightlife coordination app...
How are you doing Kyle? I see you here all the time lol. Are you applying soon?
i've got 3(?) projects left, then I don't know what I'm going to do
probably just hang out here and tell the newbies how good they have it and how hard it was back in my day
and it sounds like you're kicking butt, i think it's more important to understand what you're doing than to speed through the curriculum
@sgoldber61 Probably better to get it done now and worry about performance later
Um... while the Javascript ecosystem is really large, there's so much version changes that it gets worse and worse in our day LOL! Yeah, I applied to a city of Los Angeles thing where they hook up young people with companies.
that sounds awesome, i hope it works out
@cmccormack OK then I'll just manually look through the array. It looks like there won't be zillions of users going anyway. Thanks for all your help!
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@thekholm80 I hope so too thanks.
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@sgoldber61 Yeah there will be a point where that does matter, and at that point you'll have a lot more drive to optimize :)
hopefully at that point you'll have an amazing senior dev working with you to answer all your questions, too
Yeah, and I hope in time that all three of us will be that guy!!!
Mongodb has some great tools to see how long a query takes, if you want you can generate a million user list and test against that
Got it, I will look into timing tools and see how long different ways of doing things takes.
how to set the element for p
@lisazzzs Not exactly sure what your question is -
p element usually looks like this - <p>This is my paragraph</p>
<img class="small-image" src="/images/relaxing-cat.jpg">
is this code right?
@nuejinjian102 it's correct as far as the syntax, but not sure what problem you're doing. Are you getting an error?
@cmccormack shouldn't the source be src="images/relaxing-cat.jpg" w/o the extra forward slash?
@nashkell really depends on what you're doing
It's not a syntax problem though, may be a problem if used with an exercise if the image is relative to the current directory and not root
^ that was my reasoning for questioning the forward slash :smile:
@nuejinjian102 if you're doing the "Size your Images" lesson, the name of the class should be
smaller-image
Create a class called smaller-image and use it to resize the image so that it's only 100 pixels wide.
general fyi, FCC.org is back up, and running
把smaller-image class 添加到<img src="/images/relaxing-cat.jpg">里面 怎么写是正确的
@mlixett999 <img class="smaller-image thick-green-border" src="/images/relaxing-cat.jpg">
Hi guys, i need some help here please, i am trying to write a function, that takes in an array of elements of integers,string and float. Removes only strings and rounds up floats, returning only an array elements of integers
This what i wrote, but it doesn't seem to be doing anything
function mySort(nums) {
// empty array to hold only integer elements
const num = [];
// get rid of strings and round nums
for (let i = 0; i < nums.length; i++)
{
// if a number, push to n
if (!isNaN(nums[i]))
{
Math.trunc(num.push[i]);
}
return num;
}
}
const array = [ 90, 45, 66, 'bye', 100.5 ];
console.log(mySort(array));
@Frediflexta First you'd want to fix
num.push[i] it should be num.push(nums[i]). Second you'd want to return num after the loop is over cause rn it returns num after first iteration.
So that means the return should be outside the for loop
Correct
Alright, please can you just hang on a bit
let me get back to you with the correction
Also, thats incorrect usage of
Math.truncsince it returns the truncated value so basically you'd want to do this num.push(Math.trunc(nums[i])); instead
Awesome, cause i just realised it was working, was trying something else
Alright, That works now... one more question please
So, i want to sort the array in ascending order, but the twist is that i want to sort odd numbers first, before sorting even numbers
As in sorting an array like this
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] should return [1, 3, 5, 7, 9, 2, 4, 6, 8]
i know i can use
.sort(function (a,b) {})
but how do i test for the odds in such a way that it will first finish sorting the odd before sorting evens
@ezioda004 You there?
@Frediflexta Yes, this is an ugly solution but it should work
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let arr1 = [];
let arr2 = [];
arr.forEach(x=>x%2 !==0?arr1.push(x):arr2.push(x))
arr1.concat(arr2);
Hahahaha...
If the array isnt sorted first then you need to sort before
Wow.. trying to read the code, so you used a switch operator, and concatenated to arrays?
can you explain the forEach loop please?
@ezioda004 a more efficient way of identifying even/odd is to test for odd using
binary and operation
a & 1 === 1 ? 'odd' : 'even'
where
a is a number
Ternary operator, its short for
So its equivalent of
if else, so it checks elements if its odd it pushes to first array, and if its even it pushes to 2nd array and then concat concatenates them.So its equivalent of
arr.forEach(function(x){
if (x%2 !==0){
arr1.push(x)
}
else {
arr2.push(x)
}
})
@SweetCodingInc Ah, I have yet to learn bitwise operators, funny I was thinking about it yesterday
and a more functional way would be
function mySort(nums) {
return nums
.filter(n => !isNaN(n))
.map(n => Math.round(n))
.sort((a,b) => a-b)
}
const array = [ 90, 45, 66, 'bye', 100.5 ];
const output = mySort(array);
const odds = output.filter(n => n&1);
const evens = output.filter(n => n&1 ? false: true);
console.log(output); // [ 45, 66, 90, 101 ]
console.log(odds); // [ 45, 101 ]
console.log(evens); // [ 66, 90 ]
Wow.. okay, i won't your code went over my head
this is functional programming right?
yes
Okay i just started learning JS january this year, please don't blow my brains
i am still just grasping OOP
Be mercyful to us noobs
There is a lil bug with my solution
@Frediflexta This is more non-ugly solution
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
arr.sort((a, b)=> b % 2 - a % 2 || a - b) //[1, 3, 5, 7, 9, 2, 4, 6, 8, 10]function mySort(nums) {
const num = [];
// get rid of strings and round nums
for (let i = 0; i < nums.length; i++)
{
// if a number, push to n
if (!isNaN(nums[i]))
{
num.push(Math.trunc(nums[i]));
}
}
let odd = [];
let even = [];
// loop through num and push accordinly to the appropriate array
num.forEach( function (x) {
x % 2 !== 0 ? odd.push(x) : even.push(x);
});
// return both arrays added
return odd.concat(even);
}
i understand it better this way
@Frediflexta If you just started, ignore everything I just posted.. You're coding the right way... The approach you're using will take you a long way... fancy, short hands will not...
^
it is in line with my logic, was just having problems with the check n push
i didn't think of having two arrays n concat'in them
@Frediflexta Also, if you're staying away from functional, avoid using this as well
num.forEach
and even
.push and .sort
and
.concat too
Huh, they are functions provided to make things a bit easier
i think
lol
Not like i am staying away, i just have'nt gotten there yet
Well I think hes talking about FP approach, but there needs to be transition at one point I suppose
of course i will, and when that time comes, i will
@SweetCodingInc Why
.push()?
There is room for improvement
@ezioda004 to precisely understand pre and post increment
I just finished CS50x, so i think i have an okay understanding of lower implementation of data structures
@ezioda004 Have you studied math/calculus ?
@SweetCodingInc Yes, in highschool
Right.. so you know derivatives and intergration.. at least conceptually?
Hey guys, so one small bug, when i run my code with test
[ 90, 45, 66, 'bye', 100.5 ];
i get
[ 45, 90, 66, 100 ]
Yes, hated integrals tho
instead of
[ 45, 66, 90, 100 ]
intergration - higher order function
derivative - lower level, pure function...
combined together it's functional math - thus the name, functional programming...
in order to understand function, you must be aware of both derivation and intergration
@Frediflexta You need to sort array before doing forEach
.push higher order
.concat higher order
.sort higher order
That I understand but how else would you add elements to an array
any thoughts about Wes Bos react and es6 courses?
@ezioda004
This part
num.forEach( function (x) {
x % 2 !== 0 ? odd.push(x) : even.push(x);
});can be implemented in pure fashion as
var oddCount = 0, evenCount = 0;
for(var i=0; i<num.length; num++){
if(x%2 !== 0){
odd[oddCount++] = num[i]
} else {
even[evenCount++] = num[i];
}
}Ah yes but this so much work >.<
@Frediflexta This should work
function mySort(nums) {
const num = [];
// get rid of strings and round nums
for (let i = 0; i < nums.length; i++)
{
// if a number, push to n
if (!isNaN(nums[i]))
{
num.push(Math.trunc(nums[i]));
}
}
num.sort((a, b)=>a-b)
let odd = [];
let even = [];
// loop through num and push accordinly to the appropriate array
num.forEach( function (x) {
x % 2 !== 0 ? odd.push(x) : even.push(x);
});
// return both arrays added
return odd.concat(even);
}
@SweetCodingInc What would you say is the advantage of the latter implementation ?
@ezioda004
Pros:
1) Faster
2) You understand how it works. (purely for the sake of learning)
3) Lesser abstraction
Cons:
1) Longer code
Also, can you sort an array without using native
.sort method?
Alright, Its all good now, it worked! Thanks guys( @ezioda004 , @SweetCodingInc ,), Will be right back
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I know the argument that why not use it if it's available. You don't learn anything that way
if you go to some company for interview that pays really well, all they're gonna be interested in is how well you can navigate through loops and implement search and sort operations
@SweetCodingInc you mean that you should be able to implement codes with and without native methods such as
sort reduce map etc?
@SweetCodingInc I see, so would you say that these algorithms/native ways are used in production? Cause I have no idea.
Also, can you sort an array without using native
.sortmethod?
I think I can I've wrote a sorting algorithm for a challenge a long time ago, it was very big lol
@IsaakNazar Yes... You must be able to produce the same result as those methods using simple loops.
I'd say thats one of the disadvantages of learning JS before C. I havent studied DS and Algorithms but I understand the point youre trying to make and I agree. Its on my todo list, but Im already full atm with learning React and Googles scholarship and C
Hey everyone, can anyone help me with the Recipe Box. How can I edit the recipe name and ingredients in place?
@SweetCodingInc ok :+1:
@ezioda004
I see, so would you say that these algorithms/native ways are used in production? Cause I have no idea.
Dev and prod are completely different things. We're talking in terms of learning and becoming a badass programmer.
In real world projects, of course you would use existing libs and algorithms.
But in order for you to select the right tech, hack around it and make it useful for your usecase, you must have intimate knowledge of how things work
@MukeshAngrish I'll send you a small poc that shows in place edit concept
Ooh, ok but why wont you use the native ways if they are faster? Isnt that a major factor? I guess that varies but still
@ezioda004 In real world projects or while learning?
@MukeshAngrish Check this out - https://codesandbox.io/s/1v526v8y4q
@SweetCodingInc Real world projects, not talking about libraries but say for example sorting, you can make an algorithm for that, so wont you use that instead of
.sort()since its faster? Or is the speed difference is not that much so why bother kinda thing?
@ezioda004 In most projects that's not an issue
not on front end at least
the native js sort is pretty decent
@SweetCodingInc Ah okay, thanks for the info. You always comes around and give me reality check :laughing: , Ive yet to learn so much
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@ezioda004 no worries.. keep at it
badass programmer means to be the best?
@IsaakNazar As best as you can be
best is a relative term
and as best as is bad gramma
Yes badass programmers who wear hoodies and their terminal font color is green like in movies duhhh
:laughing:
ok, if I will solve 3 problems everyday in Codewars, will I become a badass programmer and good at problem solving, your opinions?
and a ton of VMWare themes
@IsaakNazar You're going about it the wrong way
don't put time limit on it
you need to develop your way of thinking
initially it's gonna take you a while
I remember the first time I implemented search function...
took me good 3 months
but then again, I am a very slow learner
@SweetCodingInc I do a lot of codewars too, but I lack DS and Algorithms, complexity and stuff (I know that because Ive seen people talk about it in the main room lol). Which book/course would you suggest for that?
@ezioda004 Take it step by step.. Don't mean to sound insulting, but if you're asking questions like
how do I add an element to array without using .push, may be it's time to master what you do know so far
the DS and algo are extension to your existing capabilities. They won't make much of a sense if you don't master the basics
and there aren't many basics
if/else
loops
search/sort
if/else
loops
search/sort
don't even go to advance ds until you can do the things mentioned above intuitively
after than, the learning curve would be a lot smoother
@SweetCodingInc Thanks, that's quite helpful.
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@SweetCodingInc I didnt know you were talking about native way of implementing it, so I thought there was another way method or something like that, but I'm pretty sure I could have atleast done that. I'm pretty confident with basics and feel like I'm lacking ds and stuff thats why I asked
@ezioda004 I see
If that't the case, this site would be great for you
@SweetCodingInc Thanks, I'll do it after completing CS50. I've done pset0 and pset1 and I can see what you're talking about. Learning and understanding how these method works is important and helpful in becoming "badass" programmer :laughing: I think week 4 has Algorithms, excited for that but anyway thanks again :)
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api offline
:wave:
@kbaig yo
@cmccormack How's it?
I just realized the Casual room is down
@kbaig Good - learned today that you can do Math.floor by simply prepending with
~~, so I have that going for me :)
@cmccormack The bitwise operator is a bit diff but in almost all cases does the same thing as floor
I've used
~ in the past for making nicer indexOf lookups but never used it to coerce to an integer
close enough for what I'm using it for :)
I decided to avoid them :)
psshh elitist
can't play in the dirt with the rest of us worms
:p
Nah it's more about the fact that I don't plan on using JS professionally, so I'm likely to forget a lot of shorthand if I get busy with other projects
Would rather have used the basics well
@kbaig so you're more of a
String(float).split('.')[0] kind of guy?
That's my jam dude @cmccormack
@kbaig
The bitwise operator is a bit diff but in almost all cases does the same thing as floor
and a hell of a lot faster
HAHA
is it faster in JS? I thought JS was slow with bitwise operators
Pfff who needs performant code :shipit:
Not true
could still be faster no clue
@SweetCodingInc The code you gave me, it is editing on character at a time for my code. Can you help me with that
@MukeshAngrish let me check
but did you understand the general idea behind it?
Yeah I understood it, but I think your code works because you have the text called from the state whereas I am calling it from props
@MukeshAngrish Yeah.. so you call that
props.onUpdateRecipe function, which updates your component and resets the state
@MukeshAngrish I'd suggest when you're in edit mode, you replace that edit button with let's say, save button
and call
this.props.onUpdateRecipe on click of save button
@SweetCodingInc The check icon is identical to a save button.
oh right.. then call
this.props.onUpdateRecipe on click of that
instead of inside
handleNameUpdate
@SweetCodingInc How will the edit in
state change to false then?
@MukeshAngrish Ideally you should have maintained a state for that component, but for now you can use element ref like this
@SweetCodingInc I think what I can do is on the click on check icon, I can call the
handleNameUpdate and inside it, I can change the editand call the this.props.onUpdateRecipe. But the only problem here is I need to pass either the input element's text or target value
{this.state.edit && (
<input
ref={(input) => { this.recipeTitle = input; }}
className = "recipe-name-input"
onChange = { this.handleNameUpdate }
value = { this.props.name } />
)}
and then
{this.state.edit && (
<i
className = "fas fa-check edit-name"
id = { "edit-name-" + this.props.index }
onClick = { () => {
this.setState({ edit : false });
this.props.onUpdateRecipe(this.props.index, this.recipeTitle.value, this.props.ingredients);
}}>
</i>
)}
man you folks are at it early today
@thekholm80 to continue my 'we're all worms' metaphor, we're just trying to avoid the early birds :)
@MukeshAngrish Yes.. You could do that.. but with the code in your current state it would be a pain in the ass
:D
so easier way would be to use ref
@thekholm80 Ser Kyle :wave:
@SweetCodingInc The code doesn't work.😬
@MukeshAngrish hold on.. will take a look in another 30 mins
I have a meeting to attend now
hi! is it possible to add new objects on a map by doubleclicking it? i am using google maps javascript api
@MukeshAngrish - I have been doing a React course (currently have stepped away from it for a few weeks, but...) I wonder if the way that React handles updates to the screen is not impacting the way your function (editing the recipe title is the current issue, correct?) requires more interaction to do live editing? For example, editing a form in an application that is being used in the course, the input elements have "onChange" handlers set up that set the state for the input element, which, if I recall correctly, would then trigger an event to have React update the text. I might not be remembering all of that correctly, but I'm just looking at the code for the project and sort of remembering how it worked.
@MukeshAngrish - this article deals with that problem (if it is the problem...)
@khaduch I don't know if I completely understand your point but here's what I can say about it. As you said, in React the application is changed with the help of handlers that set the state of the components.
Now the simple solution to what I'm doing is that I can declare the state of each component and keep the copy of the data that is passed through props to that component in its state. That is what @SweetCodingInc suggested initially. But there is one big rule in React. "There should only one source of truth". That is, no two components should have the same data in their states. I know that in a small project like this, the rule can be violated without any consequences. But I don't want to.
The problem I'm currently facing is that React is re-rendering on every key stroke, i.e., on every character change. And I want it to update after I'm finished with editing the text.
Now the simple solution to what I'm doing is that I can declare the state of each component and keep the copy of the data that is passed through props to that component in its state. That is what @SweetCodingInc suggested initially. But there is one big rule in React. "There should only one source of truth". That is, no two components should have the same data in their states. I know that in a small project like this, the rule can be violated without any consequences. But I don't want to.
The problem I'm currently facing is that React is re-rendering on every key stroke, i.e., on every character change. And I want it to update after I'm finished with editing the text.
morning fCC
:wave: @khaduch
@DarrenfJ - howdy Darren - how are things today?
the usual.. workyworking and busy.. and you?
@DarrenfJ - just puttering around at the moment, reading some React info about handling input elements (refreshing my partially-acquired knowledge from my React course.) The perpetual quest to try and learn a bunch of stuff... :)
@DarrenfJ @khaduch :wave:
@thekholm80 :wavy_dash: :wave: :wavy_dash:
@thekholm80 hey brutha! :D
@khaduch ack! yah... have to revisit React.. been on the backburner on back-end stuff too.. node blah blah etc etc.. too much to learn
@DarrenfJ :wave:
@thekholm80 :wave:
@khaduch :wave:
@sjames1958gm hello hello!
@sjames1958gm - :wave: :wave:
@sjames1958gm :wave:
@MukeshAngrish best way to go for that is to
debounce the update. Means any time an input comes in, you trigger a future update of the state - like 500ms after the change. If another input change comes in until then, you cancel the last timeout (delay) and create a new one. That way the state will only be set after you are done with writing or make a pause. You could also put it in a temporary state when a change comes in and only put it into the right state when the focus on the input is lost
hi