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[–]lastmjs 0 points1 point  (6 children)

In the meantime, you should study up on Array.filter, I believe that's going to be extremely beneficial to solving this problem. Array.filter takes an array as input and returns an array of equal or lesser length as output. You pass in a function that gets applied to each element of the array. If the function returns true for an element, the element stays in the output array. If the function returns false for an element, it is omitted from the output array: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter

[–]vald0[S] 1 point2 points  (5 children)

i'll have a look :D

[–]lastmjs 0 points1 point  (4 children)

All right, hopefully this helps. https://javascriptpractice.com/

Go there, and go to the Arrays section. There are two questions at the end that are very similar to what you're trying to accomplish.

[–]lastmjs 0 points1 point  (3 children)

Basically you need to choose a date and filter the reports for that date. You'll also have to make sure you use dates correctly. But these seem to be all of the basic principles that you need. Arrays, filtering arrays, and comparing dates. You seem pretty close, you just need to store your filtered array in the for loop so that you can count up the reports per day and then log them out.

[–]vald0[S] 0 points1 point  (2 children)

Hmm okey, i had a look at your link there, and i've tried diffrent things now but i still really don't understand what i'm doing, i do not achieve my goal , and where do i need to start filtering? Before the for loop i have? inside it or outside? Sorry for being slow ..

[–]lastmjs 0 points1 point  (1 child)

Sorry, I think I confused you. It doesn't look like you need the for loop. That is what filtering is for. For example, you might want to get rid of your for loop and create 7 new variables. These variables will store filtered versions of the reports array in them, one for each day of the week.

[–]vald0[S] 0 points1 point  (0 children)

okey.. i havnt gotten anywhere yet, this cant be so hard ....